ASBE计算书完整(可供学习)
完整计算书模板
内容摘要本设计主要进行了建筑设计和结构设计两个部分。
在建筑设计中确定了建筑方案，绘制了建筑的平面图、立面图、剖面图、楼梯详图、墙身大样图、部分节点详图。
结构设计分为手算部分和电算部分，手算先确定了计算简图（即取6轴线处的一榀框架进行计算），接着计算了地震作用下的各层重力荷载代表值，再利用顶点位移法求出自震周期，按底部剪力法计算水平地震作用下各层地震力的大小，最后得出水平地震作用下结构内力。
接着计算了竖向荷载作用下的结构内力，然后进行了地震，恒荷，活荷作用所得内力的几种组合。
最后利用内力组合的结果进行了结构构件的配筋。
此外还通过电算完成了楼梯、雨蓬、地基的设计。
关键词: 框架结构结构设计抗震设计AbstractThis design includes two parts-architecture design and structure. Identified In the architectural design the program of the building is Identified, then we draw a building floor plan, elevation, section, details of the staircase and the walls, some details of the node. Structural design is divided into hand calculate part and computer part. At first we should determine the calculation of diagram (that is, take the 6 axis framewor to calculate), then calculated the representative value of the gravity load in the earthquake, then use the vertex displacement method to get the vibration cycle of the building, according to the method of the bottom shear to calculate seismic forces on each floor the size of the earthquake came to the conclusion that the level of seismic structural internal force. Then calculated the vertical load of the structure of the internal force, and then proceed to the earthquake, constant load, live from the Netherlands the role of the several combinations of internal forces. Finally, the results of combination of internal forces of structuralreinforcement elements. In addition, through the computer to complete the staircase, foundation design.Keywords : frames structural design anti-seismic design目录第一部分建筑设计说明书 (1)第一章、工程概况及设计说明 (1)1.1工程概况 (1)1.2设计说明 (1)1.3抗震 (1)第二章、设计资料及设计成果 (2)2.1设计资料 (2)2.2设计成果 (2)第二部分结构设计计算书 (3)第一章、结构布置及梁柱板尺寸的确定 (3)1.1结构布置 (3)1.2确定梁柱截面尺寸 (3)第二章、结构计算简图及梁柱刚度计算 (5)2.1梁柱线刚度的计算 (5)2.2确定计算简图 (6)2.3柱的侧移刚度计算 (6)第三章、重力荷载代表值和水平地震作用计算 (8)3.1等效重力荷载代表值的计算 (8)3.2 水平地震力作用侧移计算 (10)3.3水平地震力作用内力计算 (12)第四章、竖向恒载作用下的内力计算 (19)4.1计算单元的选择确定 (19)4.2荷载计算 (19)4.3梁端固端弯矩计算 (24)4.4弯矩分配 (26)第五章、内力组合 (38)5.1框架梁内力组合 (38)5.2框架柱内力组合 (44)第六章、梁柱截面设计 (46)6.1框架梁截面设计 (46)6.2框架柱截面设计 (47)第七章、楼板设计 (51)7.1板的计算简图及计算信息 (51)7.2配筋计算 (52)7.3跨中挠度计算 (56)7.4裂缝宽度验算 (57)第八章、楼梯设计 (63)8.1梯段板设计 (63)8.2平台板设计 (66)8.3平台梁设计 (74)第九章、基础设计 (76)9.1基础选型 (76)9.2基本资料 (77)9.3计算过程 (78)9.4计算结果 (83)9.5地质勘察报告 (84)致谢 (86)参考文献 (87)附表 (88)第一部分建筑设计说明书第一章、工程概况及设计说明1.1工程概况工程名称：长春市朝阳区住宅小区工程位置：长春市工程总面积：建筑平面为平面矩形，面积2209.68m2，均为住宅层数：六层层高：底层为4300mm，其余标准层为3000mm场地类别：2类结构形式：现浇整体框架耐火等级：二级设计使用年限：50年承重方案：纵横向承重，横向承重为主，荷载由板传递给次梁然后以集中荷载的方式传递给主梁，或者楼板直接传递给主梁。
单向板楼盖设计计算书
单向板楼盖设计计算书解：一、板的设计：１：确定板厚及次梁截面1 板厚ｈｈ≥1/40＝2000／40＝50ｍｍ，并应不小于民用建筑楼面最小厚度70ｍｍ，取ｈ＝80ｍｍ。
②次梁截面b×h次梁截面高h按1/20～1/12初估，选h＝450ｍｍ，截面宽b＝（1/2～1/3）ｈ，选b＝200mm。
图1：板的计算简图４）弯矩及配筋计算取板的截面有效高度h0＝h-20＝60mm，并考虑②～⑤轴线间的弯矩折减，可列表计算如下（表1）。
对四周与梁整浇的单向板跨中和中间支座处，弯矩可减少20％（如上表1中括号数字），也可直接采用钢筋面积折减，即将非折减处的钢筋面积直接乘以0.8这样计算简便些，且误差不大，偏于安全。
表1：板的配筋设计注：①括号内数字用于②～⑤轴间。
2 ρminbh＝75mm2。
在一般情况下，楼面所受动力荷载不大，可采用分离式配筋。
本例采用的配筋图如图2所示。
图2：板配筋平面图（分离式）板配筋选择法：板的配筋用间距ｓ表示。
在求出钢筋面积Ａs后，习惯上是通过查“板配筋表”得到间距的（附表１１）；通过该表的编制可知间距ｓ、钢筋直径ｄ、计算面积Ａｓ的关系是：故在算出钢筋面积后，可将其存储于计算器内，然后选择常用直径ｄ与２８相乘，再平方；然后除以Ａｓ即得间距s，此法快速方便，可称为板配筋的“28d”法。
二、次梁设计（按塑性理论计算）1 计算简图的确定②主梁截面尺寸选择主梁截面高度h为（1/14～1/18）l0＝430～750mm，选h=600mm；由b=(1/2~1/3)h，选b=250mm；则主梁截面尺寸b×h=250mm×600mm③计算简图根据平面布置及主梁截面尺寸，可得出次梁计算简图（图２．２８）。
中间跨的计算跨度l0 l0＝5.75mm；边跨的计算跨度l0＝ln+a/2＝5.88mm＜1.025 ln＝5.9m。
边跨和中间跨计算跨度相差2.2％，可按等跨连续梁计算。
图2：次梁计算简图的确定（a）尺寸及支承情形；（b）计算简图２）内力及配筋计算①正截面承载力计算次梁跨中截面按T形截面计算，翼缘计算宽度按规定（附表１２）有：以此作为判别Ｔ形截面类别的依据。
石材幕墙计算书示例
幕墙示例1北立面80m处石材幕墙设计计算书计算:校核:幕墙公司名称二〇〇四年八月十四日目录一、风荷载计算 (1)1. 风荷载标准值: (1)2. 风荷载设计值: (1)二、石材计算 (1)1. 石材面板荷载计算: (1)2. 石材面板强度计算: (1)3. 石材剪应力计算: (2)4. 石材挂件剪应力计算: (2)三、立柱计算 (2)1. 立柱材料预选: (2)2. 选用立柱型材的截面特性: (3)3. 立柱的强度计算: (4)4. 立柱的稳定性验算: (4)5. 立柱的刚度计算: (5)6. 立柱抗剪计算: (5)四、立梃与主结构连接计算 (6)1. 立柱与主结构连接计算: (6)五、预埋件计算 (7)1. 预埋件受力计算: (7)2. 预埋件面积计算: (7)3. 预埋件焊缝计算: (8)六、横梁计算 (8)1. 选用横梁型材的截面特性: (8)2. 横梁的强度计算: (9)3. 横梁的刚度计算: (10)4. 横梁的抗剪强度计算: (11)七、横梁与立柱连接件计算 (12)1. 横梁与角码连接计算: (12)2. 角码与立柱连接计算: (12)北立面80m处石材幕墙设计计算书一、风荷载计算1.风荷载标准值:μz=0.616×(z/10)0.44=1.53794μf=0.5×35(1.8×(0.22-0.16))×(z/10)-0.22=0.464568βgz=к×(1+2×μf) = 1.63977Wk=βgz×μz×μs×W0 (JGJ102-2003 5.3.2)=1.63977×1.53794×1.5×0.45=1.70226kN/m22.风荷载设计值:W=rw×Wk=1.4×1.70226=2.38316kN/m2二、石材计算1.石材面板荷载计算:B: 该处石板幕墙分格宽: 0.6mH: 该处石板幕墙分格高: 0.6mA: 该处石板板块面积:A=B×H=0.36m2GAK: 石板板块平均自重:t : 石板板块厚度: 25mmGAK=2.8×t/1000=0.07kN/m2αmax: 水平地震影响系数最大值:0.16qEAk=βE×αmax×GAK (JGJ102-2003 5.3.4)=5×0.16×0.07=0.056kN/m2qEA=rE×qEAk=0.0728kN/m2水平荷载设计值：Sz=W+ψE×qEA=2.38316+0.5×0.0728=2.41956kN/m22.石材面板强度计算:校核依据：σ≤4N/mm2a: 短边计算长度: 0.4mb: 长边计算长度: 0.6mt: 石材厚度: 25mmm: 四点支撑板弯矩系数, 按短边与长边的边长比(a/b=0.666667) 查表得: 0.136767Sz: 风荷载标准值: 2.41956kN/m2按四点支撑板计算,应力设计值为：σ=6×m×Sz×b2×103/t2 (JGJ133-2001 5.5.4)=6×0.136767×2.41956×0.62×103/252=1.90607N/mm21.90607N/mm2≤4N/mm2强度满足要求3.石材剪应力计算:校核依据: τ≤ 2N/mm2n: 连接边上的钢销数量: 2β: 应力调整系数: 查表5.5.5 得到 1.25c: 槽口宽度: 7mms: 单个槽底总长度: 100mmτ=Sz×a×b×β/n/(t-c)/s (JGJ133-2001 5.5.7-1)=2.41956×0.6×0.6×1.25/2/(25-7)/100×103=0.302446N/mm20.302446N/mm2≤2N/mm2石材剪应力满足要求4.石材挂件剪应力计算:校核依据: τp ≤ 125N/mm2Ap: 挂件截面面积: 19.6mmτp: 挂件承受的剪应力τp=Sz×a×b×β/2/n/Ap (JGJ133-2001 5.5.5-1)=2.41956×0.6×0.6×1.25/2/2/19.6×103=13.8878N/mm213.8878N/mm2≤125N/mm2石材剪应力满足要求三、立柱计算1.立柱材料预选:(1)风荷载线分布最大荷载集度设计值(矩形分布)Bl: 幕墙左分格宽: 0.6mBr: 幕墙右分格宽: 1.2mqwk=Wk×(Bl+Br)/2=1.70226×(0.6+1.2)/2=1.53203kN/mqw=1.4×qwk=2.14485kN/m(2)分布水平地震作用设计值GAkl: 立柱左边玻璃幕墙构件(包括面板和框)的平均自重: 700N/m2 GAkr: 立柱右边玻璃幕墙构件(包括面板和框)的平均自重: 700N/m2 qEAkl=5×αmax×GAkl=0.56kN/m2qEAkr=5×αmax×GAkr=0.56kN/m2qek=(qEkl×Bl+qEkr×Br)/2=(0.56×0.6+0.56×1.2)/2=0.504kN/mqe=1.3×qek=0.6552kN/m(3)立柱弯矩:Mw: 风荷载作用下立柱弯矩(kN.m)Hvcal: 立柱计算跨度: 4mMw=qw×Hvcal2/8=2.14485×42/8=4.2897kN·mME: 地震作用下立柱弯矩(kN·m):ME=qe×Hvcal2/8=0.6552×42/8=1.3104kN·mM: 幕墙立柱在风荷载和地震作用下产生弯矩(kN·m)采用SW+0.6SE组合M=Mw+0.5×ME=4.9449kN·m(4)W: 立柱抗弯矩预选值(cm3)W=M×103/γ/215.0=4.9449×103/1.05/215=21.9043cm3(5)Ivcal: 立柱惯性矩预选值(cm4)Ivcal=5×105×(qwk+0.5×qek)×Hvcal3/384/206000/0.004 =5×105×(1.53203+0.5×0.504)×43/384/206000/0.004 =180.424cm4选定立柱惯性矩应大于: 180.424cm42.选用立柱型材的截面特性:选用立柱型材名称: C20型材强度设计值: 215N/mm2型材弹性模量: E=206000N/mm2X轴惯性矩: Ix=1913.71cm4Y轴惯性矩: Iy=143.63cm4X轴上部抵抗矩: Wx1=191.371cm3X轴下部抵抗矩: Wx2=191.371cm3y轴左部抵抗矩: Wy1=73.68cm3y轴右部抵抗矩: Wy2=25.8764cm3型材截面积: A=32.8275cm2型材计算校核处壁厚: t=11mm型材截面面积矩: Ss=114.726cm3塑性发展系数: γ=1.05C203.立柱的强度计算:校核依据: N/A+M/γ/w≤ｆa (JGJ102-2003 6.3.7)Hv: 立柱长度GAkl: 幕墙左分格自重: 700N/m2GAKr: 幕墙右分格自重: 700N/m2幕墙自重线荷载:Gk=(GAkl×Bl+GAkr×Br)/2000=(700×0.6+700×1.2)/2000=0.63kN/mNk: 立柱受力:Nk=Gk×Hv=2.52kNN: 立柱受力设计值:N=1.2×Nk=3.024kNσ=N×10/A+M×103/1.05/Wx2=3.024×10/32.8275+4.9449×103/1.05/191.371=25.5301N/mm225.5301N/mm2≤ｆa=215N/mm2立柱强度满足要求4.立柱的稳定性验算:校核依据: N/φ/A+M/(γ×w×(1-0.8×N/Ne))≤ｆa (JGJ102-2003 6.3.8-1)立柱临界轴压力计算: Ne=π2×E×A/1.1/λ2 (JGJ102-2003 6.3.8-2)λ : 立柱长细比iv: 立柱回转半径iv =√(Iz/A)=√(1913.71/32.8275)=7.63518cmλ = Hvcal/iv= 4/7.63518×100= 52.3891φ: 轴心受压稳定系数查表6.3.8求得 0.844249Ne = π2×E×A/1.1/λ2= π2×206000×32.8275/1.1/52.38912/10=2210.7N: 立柱受力设计值:3.024kNσs: 立柱计算强度(N/mm2)σs=N/φ/A+M/(γ×Wx2×(1-0.8×N/Ne))=3.024×10/32.8275/0.844249+4.9449×103/(1.05×191.371×(1-0.8×3.024/2210.7)) =25.727N/mm225.727N/mm2≤ｆa=215N/mm2立柱稳定性满足要求5.立柱的刚度计算:校核依据: Umax≤L/250Dfmax: 立柱最大允许挠度:Dfmax=Hvcal/250×1000=4/250×1000=16mmUmax: 立柱最大挠度Umax=5×(qwk+qek×ψE)×Hvcal4×108/384/E/Ix=5×(1.53203+0.504×0.5)×44×108/384/206000/1913.71=1.50848mm≤16mm立柱最大挠度Umax为: 1.50848挠度满足要求6.立柱抗剪计算:校核依据: τmax≤[τ]=125N/mm2(1)Qw: 风荷载作用下剪力设计值(kN)Qw=γw×qwvk×Hvcal/2=1.4×1.53203×4/2=4.2897kN(2)QE: 地震作用下剪力设计值(kN)QEk=γE×qevk×Hvcal/2=1.3×0.504×4/2=1.3104kN(3)Q: 立柱所受剪力:采用Qw+0.5QE组合Q=Qw+0.5×QE=4.2897+0.5×1.3104=4.9449kN(4)立柱剪应力:τ=Q×Ss×100/Ix/t=4.9449×114.726×100/1913.71/11=2.69495N/mm22.69495N/mm2≤125N/mm2立柱抗剪强度可以满足四、立梃与主结构连接计算1.立柱与主结构连接计算:连接处角码材料 : 钢-Q235Lct: 连接处角码壁厚: 5mmDv: 连接螺栓直径: 12mmDe: 连接螺栓直径: 10.36mm采用SG+SW+0.5SE组合Nw: 连接处风荷载总值(kN):Nw=Qw×2=8.57939kNNE: 连接处地震作用(kN):NE=QE×2=2.6208kNNh: 连接处水平总力(N):Nh=Nw+0.5×NE=8.57939+0.5×2.6208=9.88979kNNg: 连接处自重总值设计值(N):Ng=γG×(GAKVl×Bl+GAKVr×Br)/2000×Hvcal=1.2×(700×0.6+700×1.2)×4/2000=3.024kNN: 连接处总合力(N):N=(Ng2+Nh2)0.5=(3.0242+1.947562)0.5×1000=5796.26NNb=2×3.14×De2×140/4 (GBJ17-88 7.2.1-1) =2×3.14×10.362×140/4=23603NNnum: 立梃与建筑物主结构连接的螺栓个数:Nnum=N/Nb=0.245573个取2个Ncbl: 立梃型材壁抗承压能力(N):Ncbl=Dv×2×325×t×Nnum (GBJ17-88 7.2.1) =12×2×325×11×2=171600N5796.26N ≤ 171600N立梃型材壁抗承压能力满足Ncbg=Dv×2×325×Lct×Nnum (GBJ17-88 7.2.1)=12×2×325×5×2=78000N5796.26N ≤ 78000N角码型材壁抗承压能力满足五、预埋件计算1.预埋件受力计算:V: 剪力设计值: 3024NN: 法向力设计值: 9889.79Ne2: 螺孔中心与锚板边缘距离: 60mmM: 弯矩设计值(N·mm):M=V×e2=3024×60=181440N·m2.预埋件面积计算:Nsnum: 锚筋根数: 4根锚筋层数: 2层Kr: 锚筋层数影响系数: 1混凝土级别：混凝土-C40锚筋强度设计值:fy=210N/mm2d: 钢筋直径: Φ12mmαv: 钢筋受剪承载力系数:αv=(4-0.08×d)×(fc/fy)0.5 (JGJ102-2003 C.0.1-5)=(4-0.08×12)×(1/210)0.5=0.91441αv 取 0.7t: 锚板厚度: 10mmαb: 锚板弯曲变形折减系数:αb=0.6+0.25×t/d (JGJ102-2003 C.0.1-6)=0.6+0.25×10/12=0.808333Z: 外层钢筋中心线距离: 120mmAs: 锚筋实际总截面积:As=Nsnum×3.14×d2/4=4×3.14×122/4=452.389mm2锚筋总截面积计算值:As1=(V/Kv/Kr+N/0.8/Kb+M/1.3/Kr/Kb/Z)/fy (JGJ102-2003 C.0.1-1) =(3024/0.7/1+9889.79/0.8/0.808333+181440/1.3/1/0.808333/120)/210 =100.249mm2As2=(N/0.8/Kb+M/0.4/Kr/Kb/Z)/fy (JGJ102-2003 C.0.1-2)=(9889.79/0.8/0.808333+181440/0.4/1/0.808333/120)/210=95.0942mm2100.249mm2≤452.389mm295.0942mm2≤452.389mm24根Φ12锚筋满足要求A : 锚板面积: 30000 mm2幕墙示例1北立面80m处石材幕墙设计计算书0.5fcA=0.5×19×30000=285000NN=9889.79N≤285000N锚板尺寸满足要求3.预埋件焊缝计算:Hf:焊缝厚度8mmL :焊缝长度100mmHe = Hf×0.7 = 5.6mmLw = L - 10 = 90mmσm=6×M/(2×He×Lw2×1.22) (GBJ17-88 7.1.2)=9.83607N/mm2σn =N/(2×He×Lw×1.22) (GBJ17-88 7.1.2)=8.04205N/mm2τ=V/(2×He×Lw) (GBJ17-88 7.1.2)=3N/mm2σ:总应力σ=((σm+σn)2+τ2)0.5 (GBJ17-88 7.1.2-3)=18.128118.1281N/mm2≤160N/mm2焊缝强度满足!六、横梁计算1.选用横梁型材的截面特性:选用横梁型材名称: L63X5型材强度设计值: 215N/mm2型材弹性模量: E=206000N/mm2X轴惯性矩: Ix=9.57401cm4Y轴惯性矩: Iy=36.7729cm4X轴上部抵抗矩: Wx1=5.07957cm3X轴下部抵抗矩: Wx2=13.334cm3y轴左部抵抗矩: Wy1=13.334cm3y轴右部抵抗矩: Wy2=5.07957cm3型材截面积: A=6.14323cm2型材计算校核处壁厚: t=5mm型材截面面积矩: Ss=5.17618cm3塑性发展系数: γ=1.05L63X52.横梁的强度计算:校核依据: Mx/γWx+My/γWy≤ｆa=215 (JGJ102-2003 6.2.4) (1)横梁在自重作用下的弯矩(kN·m)Hh: 幕墙分格高: 2mBh: 幕墙分格宽: 1.2mGAkh: 横梁自重: 700N/m2Ghk=700×Hh/1000=1.4kN/mGh: 横梁自重荷载线分布均布荷载集度设计值(kN/m)Gh=γG×Ghk=1.68kN/mMhg: 横梁在自重荷载作用下的弯矩(kN·m)Mhg=Gh×Bh2/8=1.68×1.22/8=0.3024kN·m(2)横梁在风荷载作用下的弯矩(kN·m)横梁上部风荷载线分布最大荷载集度标准值(三角形分布)横梁下部风荷载线分布最大荷载集度标准值(梯形分布)分横梁上下部分别计算Hhu: 横梁上部面板高度 2mHhd: 横梁下部面板高度 1mqwku=Wk×Bh/2=1.70226×1.2/2=1.02136kN/mqwkd=Wk×Hhd/2=0.85113kN/m风荷载线分布最大荷载集度设计值qwu=γw×qwku=1.4299kN/mqwd=γw×qwkd=1.19158kN/mMhw: 横梁在风荷载作用下的弯矩(kN·m)Mhwu=qwu×Bh2/12=1.4299×1.22/12=0.171588kN·mMhwd=qwd×Bh2×(3-Hhd2/Bh2)/24=1.19158×1.22×(3-12/1.22)/24=0.164836kN·mMhw=Mhwu+Mhwd=0.336423kN.m(3)地震作用下横梁弯矩qEAk: 横梁平面外地震荷载:GAkhd: 横梁下部面板自重: 700N/m2qEAku=βE×αmax×700/1000 (JGJ102-2003 5.3.4) =5×0.16×700/1000=0.56kN/m2qEAkd=βE×αmax×700/1000 (JGJ102-2003 5.3.4) =0.56kN/m2qek: 水平地震作用线分布最大荷载集度标准值水平地震作用线分布最大荷载集度标准值(三角形分布)qeku=qEAku×Bh/2=0.56×1.2/2=0.336kN/mqekd=qEAkd×Hhd/2=0.28kN/mγE: 地震作用分项系数: 1.3qEu=γE×qeku=0.4368kN/mqEd=γE×qekd=0.364kN/mMhe: 地震作用下横梁弯矩:Mheu=qEu×Bh2/12=0.4368×1.22/12=0.052416kN·mMhed=qEd×Bh2×(3-Hhd2/Bh2)/24=0.364×1.22×(3-12/1.22)/24=0.0503533kN·mMhe=Mheu+Mhed=0.102769kN.m(4)横梁强度:σ: 横梁计算强度(N/mm2):采用SG+SW+0.5SE组合σ=(Mhg/Wx2+Mhw/Wy2+0.5×Mhe/Wy2)×103/γ=(0.3024/13.334+0.336423/5.07957+0.5×0.102769/5.07957)×103/1.05 =94.31N/mm294.31N/mm2≤ｆa=215N/mm2横梁正应力强度满足要求3.横梁的刚度计算:校核依据: Umax≤L/250横梁承受分布线荷载作用时的最大荷载集度：qwk ：风荷载线分布最大荷载集度标准值(kN/m)qek ：水平地震作用线分布最大荷载集度标准值(kN/m)Uhu : 横梁上部水平方向由风荷载和地震作用产生的弯曲:Uhu=(qwku+ψE×qeku)×Bh4/(120×E×Iy)=(1.02136+0.5×0.336)×1.24/(120×206000×36.7729)×108=0.271307mmUhd : 横梁下部水平方向由风荷载和地震作用产生的弯曲:Uhd=(qwkd+ψE×qekd)×Bh4×(25/8-5×(Hhd/2/Bh)2+2×(Hhd/2/Bh)4)/E/Iy/240=(0.85113+0.5×0.28)×1.24×(25/8-5×(1/2/1.2)2+2×(1/2/1.2)4)/206000/36.7729/240×108=0.26195mmUhg : 自重作用产生的弯曲:Uhg=5×Ghk×Bh4×108/384/E/Ix=5×1.4×1.24×108/384/206000/9.57401=1.9166mm综合产生的弯曲为:U=((Uhu+Uhd)2+Uhg2)0.5=1.9894mmDu=U/Bh/1000=0.00165783≤ 1/250挠度满足要求4.横梁的抗剪强度计算:校核依据: τmax≤125N/mm2(1)Qwk: 风荷载作用下横梁剪力标准值(kN)需要分别计算横梁上下部分面板的风荷载所产生的剪力标准值横梁上部风荷载线分布呈三角形分布横梁下部风荷载线分布呈梯形分布Qwku=qwku×Bh/4=1.02136×1.2/4=0.306407kNQwkd=qwkd×Bh/2×(1-Hhd/Bh/2)=0.85113×1.2/2×(1-1/1.2/2)=0.297896kN(2)Qw: 风荷载作用下横梁剪力设计值(kN)Qw=γw×(Qwku + Qwkd)=0.846023kN(3)QEk: 地震作用下横梁剪力标准值(kN)QEku=qeku×Bh/4=0.336×1.2/4=0.1008kNQEkd=qekd×Bh/2×(1-Hhd/Bh/2)=0.28×1.2/2×(1-1/1.2/2)=0.098kN(4)QE: 地震作用下横梁剪力设计值(kN)γE: 地震作用分项系数: 1.3QE=γE×(QEku+QEkd)=0.27832kN(5)Q: 横梁所受剪力:采用Qw+0.5QE组合Q=Qw+0.5×QE=0.985183kN(6)τ: 横梁剪应力τ=Q×Ss×100/Iy/t (JGJ102-2003 6.2.5-2)=0.985183×5.17618×100/36.7729/5=2.7735N/mm22.7735N/mm2≤ 125N/mm2横梁抗剪强度满足要求七、横梁与立柱连接件计算1.横梁与角码连接计算:Q: 连接部位受总剪力:采用Sw+0.5SE组合Q=Qw+0.5×QE=0.846023+0.5×0.27832=0.985183kN普通螺栓连接的抗剪强度计算值: 140N/mm2D : 螺栓公称直径: 6mmDe: 螺栓有效直径: 5.06mmNvbh=3.14×De2×140/4 (GBJ17-88 7.2.1-1) =3.14×5.062×140/4=2815.26NNnum=Q/Nvbh=0.349944横梁与角码连接螺栓取2个Ncb: 连接部位幕墙横梁型材壁抗承压能力计算:Ncb=D×t×325×Nnum=6×5×325×2=19500N985.183N ≤19500N横梁与角码连接强度满足要求2.角码与立柱连接计算:Gh: 自重荷载(N):Gh=1.68kNN: 连接处组合荷载:N=(Gh2+Q2)0.5=(1.682+0.9851832)0.5=1.94756kNNnum2=N/Nvbh=0.691786立柱与角码连接螺栓取2个Lct1: 角码壁厚:4mmNcbj=D×Lct1×133×Nnum2 (GBJ17-88 7.2.1) =6×4×133×2=6384N985.183N ≤ 6384N立柱与角码连接强度满足要求。
埋件计算书
对于封闭式建筑物，按外表面风压的正负情况取 − 0.2 或 0.2 μs_s22 = μs_s21 = −2
面板上的最大负风压:
WLgs2 = max(βgz ⋅ μs_s22 ⋅ μz ⋅ W0) = 2.28 ⋅ kPa
雪荷载的计算
基本雪压：
S0 = 0.3 ⋅ kPa
屋面积雪分布系数： μr = 2
8107.9
75924.
378 -1571.1 -5646.3
10325.
第 9 页，共 27 页
（6）主梁埋件计算：
幕墙结构计算书
1. A点支座的强度分析: 最大支座反力: FX = RA = 603.23 ⋅ kN FY = VA = 244.01 ⋅ kN
反力取自NODE31、34较大值
水平支反力 竖直支反力
100 53165. -0.14876E+06 -58171.
103 -0.10142E+06 -50856. -33360.
137 -0.20012E+06 24878. 0.10747E+06
140 0.11646E+06 46427. 0.11161E+06
146 -0.11987E+06 50446. 0.11475E+06
活荷载，可不与雪荷载和风荷载同时组合。
因此取二者较大值进行计算。
SL = max(Sk , SL1) = 0.6 ⋅ kPa
竖向地震作用的计算：
动力放大系数：
βE = 5
水平地震影响系数： αmax = 0.12 作用在面板上的地震作用标准值：
qEk = βE ⋅ αmax ⋅ Gk = 0.338 ⋅ kPa
完整word版单向板计算书绝对详细
一、设计资料某建筑现浇钢筋混凝土楼盖，建筑轴线及柱网平面见图14-37。
层高4.5m。
楼面可变荷载标5kN/，其分项系数1.3。
楼面面层为30mm厚现制水磨石，下铺70mm厚水泥石灰准值焦渣，梁板下面用20mm厚石灰砂浆抹灰梁、板混凝土均采用C25级；钢筋直径≥12mm时，采用HRB335钢，直径<12mm，采用HPB235钢。
二、结构布置楼盖采用单向板肋形楼盖方案，梁板结构布置及构件尺寸见图14-37。
图14-37 单向板肋形楼盖结构布置三、板的计算板厚80mm。
板按塑性内力重分布方法计算，取每m宽板带为计算单元，有关尺寸及计算简图如图14-38所示。
图14-38 板的计算简图1.荷载计算0.65kN/30mm现制水磨石0.98 kN/14kN/×0.07m＝水泥焦渣70mm25kN/钢筋混凝土板80mm 2 kN/＝0.08m×．17kN/0.34 kN/×0.02m＝20mm石灰砂浆＝3.97 kN/ 恒载标准值5.0 kN/活载标准值＝11.26 kN/ 5.0＝1.2×3.97+1.3×＝荷载设计值p11.26 kN/每米板宽p＝ 2.内力计算计算跨度板厚＝h200mm×450mm ，次梁h＝80mm b×120+＝＝边跨2600－100－2420mm＝2400mm ＝2600中间跨－200 /2400 ＝0.83<10％，故板可按等跨连续板计算。
—跨度差（24202400）板的弯矩计算截面位置弯矩系数M＝边跨跨中=5.99 ×11.26×支座B 4.67－=×11.26×中间跨跨中×11.26×=4.05中间C支座×11.26×=－4.053.配筋计算，＝60mm20＝，，b＝1000mmh＝80mm80－,=1.27＝11.9,=210＝1－＝＝实配钢筋）kN·M（m 截面位置）（．10@140，5130.1515.990.140 边跨跨中5610.109 0.116 3944.67 B支座－8/10@140460①－②8@140，3400.1 4.050.095 轴④－⑤线间359 间中跨跨中6/8@140，轴④②－2690.076 0.0794.05×0.8 线间2813400.095 0.14.05 间①－②－中，8@140C 轴⑤④－支座线间3590.076 0.079 269②－④轴－，6/8@140 4.05×0.8 线间281其中均小于0.35，符合塑性内力重分布的条件。
截面尺寸配筋As
有明显屈服点钢筋配筋时的 b 限值
5.3.4 界限受压区高度与最小配筋率
n 最大配筋率和最大受弯承载力
p 最大配筋率与界限受压区高度的关系
1 fcbx f y As
1
fc fy
x h0
As bh0
s
p 《规范》确定的最大配筋百分率
max
As max bh0
1
fc fy
梁常用的混凝土强度等级是C20、 C25、C30、C35、C40等
h 1 16 1 10l0 c
b 1 2 1 3h
h<300mm 时 d≥8mm;h≥300mm 时 ， d≥10mm
钢筋净距≥25mm，≥钢筋直径
混凝土保护层(到最外侧钢筋边 缘的距离)≥20mm
钢筋净距≥25mm，≥钢筋直径
p 梁的最大受弯承载力
M max
1
fcbxb
(h0
xb 2
)
1 fcbh02b (1 0.5b )
5.3.4 界限受压区高度与最小配筋率
防止设计为超筋梁的条件
x xb 或 b
max
M u max
1 fcbxb (h0
xb 2
)
1
fcbh02b
(1
p 等效矩形应力图形的形心位置应与抛物线加矩形应力图形的总形心位 置相同，即压应力合力的作用点位置不变。
ecu
sc
a1fc
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单筋矩形截面 截面应变分布
【2024版】塔架计算书
可编辑修改精选全文完整版塔架计算书一、主要要求：1、型钢格构式塔架，自立式铁塔。
2、上层标高16.0m，自重120Kg，水平后座力4.12kN。
下层标高13.5m，自重120Kg，水平后座力2.2kN。
3、南京大厂镇江边。
二、设计概况：1、抗震设防烈度7度，设计基本地震加速度0.1g，设计地震分组为第一组.2、基本风压0.4kN/㎡，地面粗糙度为A类(空旷地带)，工程的安全等级为一级(参照《高耸结构设计规范》设计)。
3、按照《高耸规范》第3.4.2条，本塔架结构不必进行构件截面的抗震验算，仅需满足抗震构造要求。
4、荷载的组合，按《高耸规范》第2.0.5条，取用下式：1.2G+1.4W+1.4×0.7L式中，G为自重等永久荷载W为风荷载L为活荷载5、考虑到平时检修使用时人员的上下，采用大型角钢格构式塔架，尺寸如下：三、塔架构件选择说明：1、满足大型格构式柱的构造要求：斜缀条与水平缀条的夹角宜在40°~70°内，水平缀条不小于L63×5，斜缀条不小于L75×6。
2、节点板的厚度由构造决定，选用10mm厚钢板，焊脚尺寸取8mm。
3、除塔架柱脚处的水平缀条连在柱分肢的外侧，其他所有缀条。
包括斜缀条和水平缀条均连在柱分肢的内侧，塔身外表平整，便于运输；根据业主要求，塔架用螺栓连接。
4、塔架可以在工厂分段制作，现场进行拼接。
5、格构式柱（塔架）采用分离式柱脚，柱脚底板由计算确定，且应不小于20mm厚；锚栓直径亦由计算确定，且应不小于20mm，孔径为螺栓直径的1.5倍，垫板孔径比螺栓大2mm。
四、风荷载的计算：按《高耸规范》执行。
W=βZμSμZμrω0式中：ω0=1.1× 0.4=0.44kN/㎡(1.1为工程重要性一级要求，0.4为南京的基本风压)βZ为风振系数：根据荷载规范GB50009-2001附录E，高耸结构的基本自振周期T1=(0.007~0.013)H，本工程为钢结构，取T1=0.013× 16.0=0.208sec;另根据《高耸规范》第3.2.7条，T1<0.25sec时不考虑风振影响，即βZ=1.0μS为风荷载体型系数，取2.6(偏于安全取规范的高值)μZ为风压高度变化系数，按高度16m的取值为1.52μr为风荷载重现期调整系数，为1.2W =1.0×2.6×1.52×1.2×0.44=2.09kN/㎡fA=3757平方毫米W xmin=68744(mm)3I x=6888100(mm)4I x0=10935600(mm)4I y0=2840600(mm)3W x0=110466(mm)3W y0=50467(mm)3I x =42.8mm I x0=54mm I y0=27.5mm Z 0=39.8mm G=29.492Kg/m角钢L100×10：肢宽L=100mm ，肢厚t f =14mmA=1926.1(mm)2W xmin =25060(mm)3I x =1795100(mm)4I x0=2846800(mm)4I y0=743500(mm)3W x0=4260(mm)3W y0=18540(mm)3I x =305mm I x0=384mm I y0=196mm Z 0=28.4mm G=15.12Kg/m六、计算格构式柱的柱身1500mm 高的材料重量及总重：分肢角钢：L140×14， 29.492×1.6×4=188.8 Kg L100×10水平角钢：15.12×1.6×4=96.8 Kg L100×10斜向角钢：15.12×1.8×4=108.9 Kg 节点板：0.3×0.6×0.01×7800×4=56.2 Kg 合计：188.8+96.8+108.9+56.2=450.7 Kg考虑计入爬梯及附属设备等，1600mm 高柱重取1.1×450.7=495.77 Kg 柱全高重：495.77×10(节)=4957.7Kg=49.58 kN七、求塔架内力：控制截面在塔底风荷载沿高度的线载=1.60×2.09=3.344 kN/m塔底轴力设计值： N=49.58×1.2=59.50kN弯矩设计值：M=1/2×3.344(风)×16.02×1.4+(4.12×16.0+2.2×13.5)(后座力)×1.4×0.7=599.2+93.7=692.9 kN ·m 剪力设计值：V=3.344×16.0×1.4+(4.12+2.2)×1.4×0.7=74.91+6.2=81.1 kN查规范〈〈钢结构设计规范〉〉知，格构式柱的轴心受压构件的截面分类为b类。
(完整版)围合计算书
(完整版)围合计算书1. 引言围合计算在许多领域中扮演着重要的角色，例如建筑工程、农业、生物学等。
围合计算能够帮助我们计算出物体或区域的尺寸、面积、体积等参数，为实际工作和研究提供了便利。
本文档将介绍围合计算的基本概念、方法和一些实际应用。
2. 围合计算的基本概念围合计算是指计算出物体或区域的尺寸、面积、体积等参数的过程。
在围合计算中，我们通常使用数学方法进行计算。
以下是一些常用的围合计算的基本概念：2.1 周长周长是指围绕物体或区域的封闭曲线的长度。
在计算周长时，我们可以使用不同的方法，例如直接测量、使用公式计算等。
2.2 面积面积是指物体或区域内部的平面部分的大小。
计算面积时，我们通常使用不同的方法，例如测量、几何公式、数值积分等。
2.3 体积体积是指物体或区域所占据的空间大小。
计算体积时，我们可以使用不同的方法，例如测量、几何公式、数值积分等。
3. 围合计算方法围合计算的方法可以根据不同的需求和情况选择。
以下是一些常用的围合计算方法：3.1 几何方法几何方法是指使用几何图形的性质和公式进行围合计算的方法。
常用的几何方法包括确定点、线、面的位置和关系，使用面积和体积公式等。
3.2 数值方法数值方法是指使用数值计算的方法进行围合计算。
数值方法通常包括数值积分、数值逼近等技术，可以更精确地计算出面积和体积等参数。
3.3 统计方法统计方法是指使用统计学原理和方法进行围合计算。
统计方法通常包括抽样、假设检验、回归分析等技术，可以对大规模的数据进行面积和体积估计。
4. 围合计算的实际应用围合计算在许多领域中有着广泛的应用。
以下是一些围合计算的实际应用：4.1 建筑工程围合计算在建筑工程中可以帮助计算建筑物的周长、面积和体积，来进行设计、施工和材料采购的规划。
4.2 农业围合计算在农业中可以帮助计算农田的面积和周长，来进行土地利用规划、农作物种植和灌溉设计等。
4.3 生物学围合计算在生物学中可以帮助计算生物个体的尺寸、面积和体积，来进行生物分类、生物繁殖和生态研究等。
(完整版)栅栏计算书
(完整版)栅栏计算书1. 需求分析栅栏计算是一种用于对数据进行安全加密和解密的方法。
在进行栅栏计算之前，我们需要明确以下几个需求：- 输入的数据应该是一个字符串，包含字母、数字和特殊字符。
- 用户需要指定栅栏的高度。
- 用户可以选择加密还是解密操作。
2. 功能设计2.1 加密在加密操作中，我们首先需要将输入字符串按照栅栏的高度，分成多行。
然后，将每一行中的字符依次取出，按照从上到下、从左到右的顺序，组成一个新的字符串。
最后，将新的字符串返回给用户。
具体的加密过程如下：伪代码function encrypt(data: string, height: int) -> string:rows = create_rows(data, height)result = ''for column in range(height):for row in rows:result += row[column]return result2.2 解密在解密操作中，我们需要将加密后的字符串按照栅栏的高度，分成多列。
然后，将每一列中的字符依次取出，按照从左到右、从上到下的顺序，组成一个新的字符串。
最后，将新的字符串返回给用户。
具体的解密过程如下：伪代码function decrypt(data: string, height: int) -> string: columns = create_columns(data, height)result = ''for row in range(len(columns[0])):for column in columns:result += column[row]return result3. 技术实现3.1 数据结构我们可以使用二维列表来表示栅栏。
每一行代表一条栅栏，每个元素代表一个字符。
3.2 输入与输出我们需要实现一个用户命令行交互的界面，通过命令行让用户输入字符串和栅栏的高度，并选择加密还是解密操作。
散货船稳性计算书BV
CONTENTS1.PREAMBLE (3)2.GENERAL DESCRIPTION (4)3.LOADING MARK (6)4. INTACT STABILITY CRITERIA (7)5.DAMAGE STABILITY CRETERIA (10)6.ASSESSMENT OF COMPLIANCE WITH STABILITY CRITERIA (10)7.GUIDELINE FOR TRIM & STABILITY CALCULATION (10)8.CAPACITY TABLE OF TANKS (14)9.FREE SURFACE EFFECT CAPSIZING MOMENT (16)10.FLOODING ANGLE CURVE (17)11.MINIMUM INITIAL METACENTRIC HEIGHT (18)12.MISCELLANEOUS CONSUMABLES ON DEPARTURE (19)13.MISCELLANEOUS CONSUMABLES ON ARRIVAL (20)14. TRIM & STABILITY CALCULATION (21)1.PREAMBLEThis Booklet is prepared for the ship's Master in obtaining information and suitable instructions as a guidance to the stability of the ship under varying conditions of service.Relevant requirements in MSC Resolution A749(18) of IMO, and the relevant Class requirements of BV are to be referred to in the usage of this manual.This Booklet comprises following contents. General information and instructions are given for calculation and evaluation of stability of the ship accompanied by a number of loading conditions. Data, such as those of free surface moment of tanks (initial and at large inclination), wind capsizing lever, immersing and flooding angles, limit height of center of gravity, etc., and those of the maximum still water bending moments.General hydrostatic data of the ship, such as displacement, deadweight, center of buoyancy, center of flotation, metacenter, displacement per centimeter of draught and so on, are tabulated against the vessel’s mean draught. Cross stability data, excluding the buoyancy effects of timber deck cargoes or the similar, are provided therein.It is necessary to ensure a satisfactory safety of the ship at any time during each her voyage. Therefore, prior loading operation, the Master shall make a calculation in order to verify that no unacceptable stress in the ship's structure, no insufficient stability, nor inappropriate floating state will occur during the forthcoming voyage.2.GENERAL DESCRIPTIONDimensions:Length overall: ~110.80 MLength between perpendiculars: 104.52 MFreeboard Length 105.02 MBreadth moulded: 18.20 MDepth moulded: 9.00 MDesign draught: 5.50MScantling draught: 6.82MTonnage (ICLL1969):Gross tonnage: 5612Net tonnage: 2867Light ship:Weight: 3050-6.28Longitudinal center of gravityfrom ⊗Vertical center of gravity from7.04baseline:Freeboard:Type of loadline assignment: ICLL 1966Ordinary FreeboardDraught Letter Freeboard Moulded(mm) (m) S 2195 6.820T 2053 6.962W 2337 6.678F 2046 6.969TF 1904 7.111& STABILITY CALCULATION JZ403.101.005JS Page53.LOADING MARKLOADING MARK4.INTACT STABILITY CRITERIAThis vessel is required to comply with the stability requirement of IMO, which are quated hereinfrom IMO Resolution A.749 (18) “Code On Intact Stability For All Types Of Ships Covered ByIMO Instruments“, Chapter 3-DESIGN CRETERIA APPLICABLE TO ALL SHIPS.4.1 General Intact Stability CriteriaThe vessel should comply with the stability criteria as stated below:4.1.1 The area under the righting lever curve (GZ curve) should not be less than 0.055 meter-radians up to θ=30°angle of heel and not less than 0.09 meter-radians up to θ=40°or the angle of flooding θf, if this angle is less than 40°. Additionally, the area under the righting lever curve (GZ curve ) between the angle of heel of 30° and 40° or between 30° and θf if this angle is less than 40°, should not be less than 0.03 meter-radians.4.1.2 The righting lever GZ should be at least 0.20 m at an angle of heel equal to or greater than 30°.4.1.3 The maximum righting arm should occur at an angle of heel preferable exceeding 30° but not less than 25°.4.1.4 The initial metacentric height GMo should not be less than 0.15 m.4.1.5 Provision should be made for a safe margin of stability at all stages of the voyage, regard being given to additions of weight, such as those due to absorption and icing, and to losses of weight, such as those due to consumption of fuel and stores.4.1.6 See also general recommendations of an operational nature given in Section 2.5 of IMO Resolution A.749 (18) Code On Intact Stability For All Types Of Ships Covered By IMO Instruments“,4.2. Severe Wind And Rolling Criterion (Weather Criterion)4.2.1 Beside the stability criteria, this vessel is also to comply with the weather criterion recommended as follows.4.2.2.1 The ability of a ship to withstand the combined effects of beam wind and rolling should be demonstrated for each standard condition of loading, with reference to the figure as follows:•The ship is subjected to a steady wind pressure acting perpendicular to the ship's centreline which results in a steady wind heeling level (l w1).•From the resultant angle of equilibrium (θo), the ship is assumed to roll owing to wave action to an angle of roll (θ1) to windward. Attention should be paid to the effect of steady wind so that excessive resultant angles of heel are avoided;•The ship is then subjected to a gust wind pressure which results in a gust wind heeling lever (l w2);•Under these circumstances, area "B" should be equal to or greater than area "A";•Free surface effects should be accounted for in the standard conditions of loading;θo = a ngle of heel under action of steady windθ1 = angle of roll to windward due to wave actionθ2 = angle of downflooding (θf ) or 50° or θc , whichever is less,where:θf = angle of heel at which openings in the hull, superstructures ordeckhouses which cannot be closed weathertight immerse. In applying thiscriterion, small openings through which progressive flooding cannot takeplace and need not be considered as open.θc = angle of second intercept between wind heeling lever l w2 and GZ curves.4.2.2.2 The wind heeling levers l w1and l w2 referred to are constant values at all angles ofinclination and should be calculated as follows:l P A Z g w 11000=..Δ(m) and l w2 = 1.5 l w1 (m)where:P = 504 N/m 2. The value of P used for ships in restricted service may be reducedsubject to the approval of the Administration;A = projected lateral area of the portion of the ship and deck cargo above thewaterline (m 2);Z = vertical distance from the centre of A to the centre of the underwaterlateral area or approximately to a point at one half the draught (m);Δ = displacement (t) g = 9.81 m/s 24.2.2.3 The angle of roll (θ1) referred to should be calculated as follows:θ11092=⋅⋅⋅ k 1 r s X X (degrees)where:X 1 = factor as shown in table 1X 2 = factor as shown in table 2k = factor as follows:k =1.0 for round-bilged ship having no bilge or bar keelsk =0.7 for a ship having sharp bilgesk =as shown in table 3 for a ship having bilge keels, a bar keel or bothr = 0.73 + 0.6 OG/dwith: OG = distance between the centre of gravity and the waterline (m) (+ if centre ofgravity is above the waterline, - if it is below)d = mean moulded draught of the ship (m)s = factor as shown in table 4.Table 1Table 2Table 3Table 4Values of Factor X1Values offactor X1Values offactor kValues offactor sB/d X1C B X2A k.100 k T sL.B≤2.4 1.00 ≤0.45 0.75 0.0 1.00 ≤6 0.1002.5 0.98 0.50 0.82 1.0 0.98 7 0.0982.6 0.96 0.55 0.89 1.5 0.95 8 0.0932.7 0.95 0.60 0.95 2.0 0.88 12 0.0652.8 0.93 0.65 0.97 2.5 0.79 14 0.0532.9 0.91 ≥0.70 1.003.0 0.74 16 0.0443.0 0.90 3.5 0.72 18 0.0383.1 0.88 ≥4.0 0.70 ≥20 0.0353.20.863.30.843.4 0.82≥3.5 0.80(Intermediate values in tables 1-4 should be obtained by linear interpolation.)Rolling period TCBGM=2(seconds)where:C = 0.373+0.023(B/d)-0.043(L/100)The symbols in the above tables and formula for the rolling period are defined as follows:L = waterline length of the ship (m)B = moulded breadth of the ship (m)d = mean moulded draught of the ship (m)C B = block coefficientAk = total overall area of bilge keels, or area of the lateral projection of the bar keel, or sum of these areas (m2)GM = metacentric height corrected for free surface effect (m).4.3. Effect Of Free Surface Of Liquids In TanksFor all conditions, the initial metacentric height and the stability curves should be corrected for the effect of free surfaces of liquids in tanks in accordance with the following assumptions:4.3.1 Where water ballast tanks are to be discharged and filled during the course of a voyage, the free surface effects should be calculated, to take account of the most onerous transitory stage relating to such operations, consistent with any operating instruction.In calculating the free surface effects in tanks containing consumable liquids, it should be assumed that for each type of liquid, at least one transverse pair or a single centreline tank has a free surface and the tank or combination of tanks taken into account should be those where the effect of free surfaces is the greatest.4.3.2 For the purpose of determining this free surface correction, the tanks assumed slack should be those which develop the greatest free surface moment, Mf.s. at a 30° inclination when in the 50 per cent full condition.4.3.3 The values of Mf.s. for each tank may be derived from the formula:Mf s V b k....=γδwhere:M f.s. is the free surface moment at any inclination, in metre-tonnesV is the tank total capacity in m3b is the tank maximum breadth in metresγ is the specific weight of liquid in the tank in t/m3δ is equal to (V/ blh) (the tank block coefficient)h is the tank maximum height in metresl is the tank maximum length in metresk is the dimensionless coefficient to be determined from the following table according to the ratio b/h. The intermediate values are determined by interpolation.4.3.4 Small tanks, which satisfy the following condition using the value of k corresponding to the angle of inclination of 30°, need not be included in computation:(Mf.s. / Δmin)< 0.01 mwhereΔmin = minimum ship displacement in tonnes (metric tonnes)4.3.5 The usual remainder of liquids in the empty tanks is not taken into account in computation.5.DAMAGE STABILITY CRETERIABeside the intact stability criteria, this vessel must also meet the requirements for damage stability imposed by SOLAS Part1.ChapterII-1.Part B-1 (Subdivision and damage stability of cargo ship) and relevant IACS requirements on following basisAt summer draft 6.820 m, the KG value is taken as 6.732 m, corresponding to 1.00 m GM value; while at partial draft 4.956 m, the KG value is taken as 6.751 m, corresponding to a 1.20 m GM value. These two points gives additional limits to the minimum GM and maximum KG derived from intact stability requirements. Detailed min. GM and max. KG curve can be found in this booklet.6.ASSESSMENT OF COMPLIANCE WITH STABILITY CRITERIAFor the purpose of assessing quickly whether the stability criteria set in proceeding paragraphs are met, a comprehensive limit KG (center of gravity above keel line) is given in this Manual with tabular values.The limit KG value is the maximum permissible KG corrected for free surface effect of liquids in tanks, regarding both the IMO regulations for intact stability and damage stability of cargo ships.Its usage is demonstrated in the following paragraph.7.GUIDELINE FOR TRIM & STABILITY CALCULATION7.1 DISPLACEMENT CALCULATION FROM DRAFT READING1. Read the drafts on the fore, aft and midship draft scales, obtaining T F', T A' and T⊗respectively.If measured port and starboard draughts are different, arithmetical mean of them is to be adopted.2. Correct the observed draught in order to obtain the draughts at FP, AP and midship. This is carried out using the table. in which the correction values can be obtained via apparent trim, as stated below:Apparent trim = TA' - TF'Thus, the corrected drafts are:Draft at midship: T⊗=T’⊗ + T⊗Draft at F.P. TF = TF' + TFDraft at A.P. TA = TA' + TATrim =TA-TFDeflection of the keel h =(TF+TA)/2 - T⊗If h >0, it is in hogging condition, and h<0, in sagging condition.3. Read displacement in table "DISPLACEMENT BY TRIM" with trim=TA-TF, and Draft=(TF+TA)/2-0.75h, obtaining displacement D (t)7.2 TRIM CALCULATIONThis calculation is preceded with "TRIM AND STABILITY CALCULATION SHEET" shown on following page.1. Put the weight of cargo, fuel, fresh water, ballast water, provision, crew and their effective, etc. into the column "Weight".2. Sum up the above mentioned weights to obtain DEADWEIGHT.3. Sum up the DEADWEIGHT and LIGHT SHIP obtain DISPLACEMENT4. Put the vertical and longitudinal center of gravity of each item into column "VCG" or "LCG", respectively.If the LCG is forward of midship, then LCG takes a "+" sign, and if after midship , takes "-" sign.5. Multiply the weights by their LCG, and put the products into "MOMENT abt. MS".6. Sum up the moments abt. MS. of each item, and divide the sum by DISPLACEMENT, we have LCG of the ship.7. Find TM, LCB, LCF and MTC in HYDROSTATIC TABLE according to DISPLACEMENT. Trim is found as:t = (LCG - LCB) * DISP/100/MTC *( p/1.025)Draft at FP:TF= TM + (0.5 - LCF/LPP ) * tDraft at AP:TA= TM - (0.5 + LCF/LPP ) * twhere:TM = mean draft, almost equal to draft at LCFLCF = Longitudinal position of the center of flotation.8. Evaluate basic working environment of the propeller by immersion rate, shown below:Submersion rate = I / Dp * 100 %where: Dp is the diameter of the propeller, h is the height of shaft centerline, above the ship's baseline.I = submersion of the propeller, shown on the figure below:For this vessel, 100% propeller immersion rate is achieved at 5.90 m draft at A.P.7.3 STABILITY CALCULATION"Stability" includes initial stability indicated in the form of initial metacentric height GMo and stability at large inclination expressed by "Righting Lever Curve" (or called GZ curve)This calculation is proceeded with use of "TRIM AND STABILITY CALCULATION SHEET" on following page, the same as for "TRIM CALCULATION". The procedure is described hereafter:1. The columns "WEIGHT" and "DISPLACEMENT" are obtained from trim calculation explained on the previous page.2. Put the vertical center of gravity of each loading weight into the column "VCG".3. Multiply the weight by VCG and put the result into column MOMENT about BL.4. Sum up the above moment abt. BL and put this sum into the bottom of this column.5. Divide the sum by displacement, thus the vertical center of gravity of the ship is obtained.6. Collect IFSM data from table "Initial Free Surface Moment", put them into the last column. Calculate the their sum and write it below the cell marked IFSM.7. Calculate initial metacentric height GMo as follows:Initial metacentric height GM = TKM - KGInitial metacentric height corrected by free surfaceGMo = GM - IFSM/DISPWhere: TKM is the transverse metacentric height above base line, obtained from "Hydrostatic Table"8. Static stability curve can be obtained as:Ls = Lf - KG *sin θ - Mfs/DISPwhere: Lf--form righting lever when assuming the VCG of the ship is equal to zero. Numerical values are tabulated in "CROSS STABILITY TABLE”.Mfs is the heeling moment due to liquid shifting,9. Integrate Ls from 0o to θ obtain dynamical stability lever. Trapezoid method of integral may be adopted.10. Plot the Ls against angle of heel, we have the Righting Lever Curve.11. Calculate capsizing lever due to steady wind Lw1, and capsizing lever due to gust windLw2, equilibrium angle θo, roll angle θ1 according to formula described on page 1- .12. Integrate area A and area B.13. Judge the stability according to intact stability criteria stated on above page.SIMPLIFIED METHOD OF EVALUATING STABILITYThis method is based on the tabulated values of the "LIMIT HEIGHT OF CENTER OF GRAVITY", which provides maximum allowable vertical center of gravity (KGmax), and minimum initial metacentric height (GMmin) against displacement of the ship. Following the procedure described below, we can easily evaluate whether the ship has sufficient stability or not.Maximum allowable vertical center of gravity is calculated in accordance with the intact stability criteria, but the condition for calculation of damage stability required by SOLAS 1974 Chapter II,Part B-1, Regulation are also incorporated. The conditions for calculation of damage stability are described on above page.The evaluating procedure is the following:1. Calculate displacement (DISP), vertical center of gravity of the ship, and total sum of initial free surface moments (IFSM) of all tanks that may have free surface during the voyage, in such a way as described in "Stability Calculation"2. Correct the vertical center of gravity KG by free surface effect as:KG'= KG + IFSM/DISP3. Find KGmax in the table "LIMIT HEIGHT OF CENTER OF GRAVITY" on page 1-24 according to displacement DISP.4. Compare KG' with KGmax. If KG' is less than KGmax, then, the ship has sufficient stability , and satisfies all intact stability criteria and the requirements of damage stability.If KG' is greater then KGmax, Master should alter the loading plan in order to modify stability. Then repeat step 1 thru 4, until KG' reaches the value less then KGmax.Alternately, by correcting initial metacentric height per formula:GMo = GM - IFSM/DISPwhereGMo is the corrected value of GM, the initial metacentric height uncorrected by the free surface effect of liquids in tanks, we can also use the table of "LIMIT HEIGHT OF CENTER OF GRAVITY" and find GMmin, the minimum allowable initial metacentric height. Then compare GMo and GMmin. If GMo is less then GMmin, the ship has bad stability, and the Master must alter the loading plan.8.CAPACITY TABLE OF TANKSTANK CAPACITYCEN. OF GRA.CEN. OF GRA FREENO. DESIGNATION FRAME VOLUMEWEIGHT Xg(LCG) Yg Zg(VCG)SURFACE项 名称 位置 容积 重量 重心纵向 重心横向 重心垂向自由液面目 (m3)t (m)(m)(m) t-m 货舱 CARGO HOLD1 NO.1 CARGO HOLD Fr110-Fr137 2806.49 33.050 0.000 7.0402 NO.2 CARGO HOLD Fr68-Fr110 3838.83 9.275 0.000 5.7033 NO.3 CARGO HOLD Fr28-Fr68 3588.43 -19.222 0.000 5.828合计 TOTAL 10233.75淡水舱 F.W.T4 F.W.T(P/S)Fr64-Fr70 41.98 41.98 -5.769 0.000 8.129 84.55合计 TOTAL 83.96 83.96压载舱 B.W.T5 F.P.T. Fr143-STEM 196.10 201.00 49.795 0.000 4.288 152.906 NO.1D.B.W.T(P) Fr122-Fr137 66.43 68.09 37.625 -3.022 0.704 260.867 NO.1D.B.W.T(S) Fr122-Fr137 72.42 74.23 37.301 2.865 0.699 240.258 NO.2D.B.W.T(P) Fr110-Fr122 62.60 64.17 28.265 -5.200 0.678 604.629 NO.2D.B.W.T(S) Fr110-Fr122 86.57 88.74 28.316 4.070 0.670 446.2810 NO.3D.B.W.T(P) Fr87-Fr110 135.81 139.21 16.171 -5.557 0.665 1366.9811 NO.3D.B.W.T(S) Fr87-Fr110 181.76 186.30 16.178 4.436 0.661 1011.2212 NO.4D.B.W.T(P) Fr68-Fr87 112.78 115.59 1.500 -5.574 0.663 1134.0113 NO.4D.B.W.T(S) Fr68-Fr87 150.73 154.50 1.500 4.452 0.660 838.9714 NO.1S.B.W.T(P/S) Fr122-Fr137 138.78 142.25 37.755 0.000 4.277 54.3215 NO.2S.B.W.T(P/S) Fr110-Fr122 76.80 78.72 28.843 0.000 4.093 5.5216 NO.3S.B.W.T(P/S) Fr87-Fr110 127.74 130.93 16.197 0.000 4.001 4.5517 NO.4S.B.W.T(P/S) Fr68-Fr87 105.58 108.21 1.500 0.000 4.000 3.7618 NO.5S.B.W.T(P/S) Fr46-Fr68 145.03 148.65 -12.756 0.000 3.488 4.3519 NO.6S.B.W.T(P/S) Fr28-Fr46 174.68 179.05 -27.966 0.000 4.033 49.5220 A.P.T(C) STERN-Fr9 62.44 64.00 -49.083 0.000 5.033 243.0221 A.B.W.T(P/S) STERN-Fr4 31.01 31.79 -51.604 0.000 7.914 20.37合计 TOTAL 2726.86 2795.03燃油舱 F.O.T22 NO.1F.O.T(P) Fr46-Fr68 102.90 98.78 -12.841 -4.917 0.651 185.9223 NO.1F.O.T(S) Fr46-Fr68 145.95 140.11 -12.844 3.797 0.650 530.0024 NO.2F.O.T(P) Fr28-Fr46 61.01 58.57 -25.819 -4.358 0.678 91.8325 NO.2F.O.T(S) Fr28-Fr46 96.23 92.38 -26.197 3.173 0.668 290.1926 F.O.OVERFLOW.T(P) Fr23-Fr28 9.29 8.92 -35.289 -2.645 0.862 5.1727 NO.1 F.O.SER.T(S) Fr21-Fr23 6.69 6.43 -37.350 6.355 7.400 1.70 28 NO.2 F.O.SER.T(S) Fr23-Fr2610.04 9.64 -35.600 6.355 7.400 2.54 29 NO.1F.O.SETTLING.T(S)Fr21-Fr23 7.93 7.61 -37.321 7.626 6.317 0.33 30 NO.2F.O.SETTLING.T(S)Fr23-Fr26 14.44 13.87 -35.565 7.709 6.160 0.54 合计 TOTAL 454.48 436.30 柴油舱 M.G.O31 NO.1 M.D.O(P) Fr16-Fr23 11.39 9.57 -38.651 -2.286 1.044 5.23 32 NO.1 M.D.O(S) Fr16-Fr25 17.23 14.47 -37.722 2.447 1.020 10.78 33NO.2 M.D.O(C)Fr12-Fr16 9.66 8.11 -42.666 0.000 0.700 12.18 34 M.D.O.SEV.T(S)Fr9-Fr13 8.98 7.54 -44.989 5.708 7.084 0.38 35 M.D.O.SETTLING.T(S)Fr9-Fr13 6.66 5.59 -44.897 7.042 7.570 1.13 合计 TOTAL53.91 45.28 滑油舱 L.O.T36 L.O. STORAGE TK(S)Fr5-Fr8 5.94 5.35 -47.997 6.020 7.566 1.93 37L.O.CIRCULATING TK(C)Fr17-Fr23 3.04 2.73 -38.750 0.000 0.900 0.57 合计 TOTAL8.98 8.08 杂项 MISCELLANEOUS38 S/T C.W.T （C ） STERN-Fr9 8.74 8.74 -47.265 0.013 2.285 0.96 39 BILGE WATER TK.(C) Fr9-Fr12 3.36 3.36 -45.263 0.000 0.753 2.67 40 SLUDGE TK.（S ）Fr25-Fr28 15.71 15.08 -34.168 2.384 0.915 29.94 41 PURIFY SLUDGE T(S) Fr15-Fr20 6.74 6.47 -39.732 6.044 5.220 3.18 42DIRTY L.O. TK (C)Fr17-Fr23 3.04 2.73 -38.750 0.000 0.300 0.57 43 SEWAGE TK(P) Fr4-Fr7 5.26 5.26 -48.594 -5.967 7.616 1.89 44 C.H. WASH TK.(C) Fr106-Fr110 16.1016.10 22.899 0.000 8.134 17.06 45合计 TOTAL58.9557.749. FREE SURFACE EFFECT CAPSIZING MOMENTLENGTH BREADTH HEIGHT VOLUMEDENSITYFREE SURFACE EFF. CAPSIZING MOMENTCOMPARTMENTl b h V ρMfs( m ) ( m ) ( m ) ( m 3)( t / m 3) 10 20 30 40 50 60* F.P.T. 8.360 10.400 8.100 196.10 1.025 20.8 43.0 68.8 101.6 122.4 129.4 * NO.1D.B.W.T(S) 10.500 7.863 1.300 72.42 1.025 42.4 54.8 54.5 50.8 45.2 37.9 * NO.2 D.B.W.T(S) 8.400 9.100 1.300 86.57 1.025 73.6 87.4 85.5 79.1 69.8 58.0 * NO.3 D.B.W.T(S) 16.100 9.100 1.300 181.76 1.025 161.6 192.0 188.0 173.9 153.3 127.5 * NO.4 D.B.W.T(S) 13.3009.1001.300150.731.025 134.3159.5156.2144.5127.4105.9NO.1S.B.W.T(P/S) 10.500 4.439 5.400 138.78 1.025 5.7 11.7 18.7 27.9 42.1 57.3NO.2S.B.W.T(P/S) 8.400 2.750 5.400 76.80 1.025 1.3 2.6 4.2 6.3 9.4 15.6 NO.3S.B.W.T(P/S) 16.100 1.500 5.400 127.74 1.025 0.8 1.6 2.6 3.9 5.9 9.7 NO.4S.B.W.T(P/S) 13.300 1.500 5.400 105.58 1.025 0.7 1.4 2.2 3.2 4.9 8.1 NO.5S.B.W.T(P/S) 15.400 1.500 6.700 145.03 1.025 0.7 1.5 2.3 3.5 5.3 8.7 NO.6S.B.W.T(P/S)12.600 5.000 6.700 174.68 1.025 6.3 13.1 20.9 31.1 46.9 69.3 * A.P.T(C) 7.60010.0002.40062.441.025 22.939.842.340.737.132.1A.B.W.T(P/S) 4.500 4.500 2.700 31.01 1.025 2.6 5.5 8.7 11.3 12.0 11.8 F.W.T(P/S) 4.200 6.270 2.300 41.98 1.000 8.8 18.2 23.4 24.2 23.2 21.0 * NO.1 F.O.T(S) 15.400 7.600 1.300 145.95 0.960 89.5 118.7 118.5 110.9 98.7 83.0 * NO.2 F.O.T(S) 12.6007.6001.30096.230.960 53.070.370.265.658.449.1NO.1 M.D.O(S) 6.300 3.800 1.660 17.23 0.840 1.2 2.5 3.7 4.0 3.9 3.6NO.2 M.D.O(C)2.800 4.728 1.200 9.660.8401.7 3.1 3.4 3.3 3.02.6JZ MARINE TRIM & STABILITY CALCULATION JZ403.101.005JS Page 161711.MINIMUM INITIAL METACENTRIC HEIGHTJZ MARINE TRIM& STABILITY CALCULATION JZ403.101.005JS Page 1812.MISCELLANEOUS CONSUMABLES ON DEPARTUREMISCELLANEOUS CONSUMABLES ON DEPARTUREVERTICAL LONGITUDINAL FREEDESIGNATION FRAME WEIGHT(t)Zg(VCG) MOMENT Xg(LCG)MOMENTSURFACE REMARK燃油溢油舱(左) F.O.OVERFLOW.T(P) Fr23-Fr28 8.74 0.86 7.53 -35.29 -308.43 0.00 98% NO.1燃油日用舱(右) NO.1 F.O.SER.T(S) Fr21-Fr23 6.30 7.40 46.60 -37.35 -235.18 0.00 98% NO.2燃油日用舱(右) NO.2 F.O.SER.T(S) Fr23-Fr26 9.45 7.40 69.90 -35.60 -336.26 0.00 98%NO.1燃油沉淀舱(右)NO.1F.O.SETTLING.T(S) Fr21-Fr23 7.46 6.32 47.12 -37.32 -278.40 0.00 98%NO.2燃油沉淀舱(右)NO.2F.O.SETTLING.T(S) Fr23-Fr26 13.59 6.16 83.70 -35.57 -483.26 0.00 98%柴油日用舱(右) M.D.O.SEV.T(S) Fr9-Fr13 7.39 7.08 52.36 -44.99 -332.54 0.00 98% 柴油沉淀舱(右) M.D.O.SETTLING.T(S)Fr9-Fr13 5.48 7.57 41.49 -44.90 -246.07 0.00 98% 滑油储存舱(右) L.O. STORAGE TK(S) Fr5-Fr8 5.24 7.57 39.65 -48.00 -251.54 1.93 98%滑油循环舱(中) L.O.CIRCULATINGTK(C) Fr17-Fr23 2.68 0.90 2.41 -38.75 -103.83 0.57 98%艉轴冷却水舱(中) S/T C.W.T(C) STERN-Fr98.74 2.29 19.97 -47.27 -413.00 0.96 100% 舱底水舱(中) BILGE WATER TK.(C) Fr9-Fr12 0.00 0.75 0.00 -45.26 0.00 2.67 0% 油渣舱(右) SLUDGE TK.(S) Fr25-Fr28 0.00 0.92 0.00 -34.17 0.00 29.94 0%分油机油渣舱(右) PURIFY SLUDGE T(S)Fr15-Fr20 0.00 5.22 0.00 -39.73 0.00 3.18 0%污滑油舱(中) DIRTY L.O. TK (C) Fr17-Fr23 0.00 0.30 0.00 -38.75 0.00 0.57 0% 污水舱(左) SEWAGE TK(P) Fr4-Fr7 0.00 7.62 0.00 -48.59 0.00 1.89 0%TOTAL 75.061 5.472 410.734 -39.815 -2988.52 41.710 JZ MARINE TRIM& STABILITY CALCULATION JZ403.101.005JS Page 1913.MISCELLANEOUS CONSUMABLES ON ARRIVALMISCELLANEOUS CONSUMABLES ON ARRIVALVERTICAL LONGITUDINAL FREEDESIGNATION FRAME WEIGHT(t)Zg(VCG) MOMENT Xg(LCG)MOMENTSURFACE REMARK燃油溢油舱(左) F.O.OVERFLOW.T(P) Fr23-Fr28 8.74 0.86 7.53 -35.29 -308.43 0.00 98% NO.1燃油日用舱(右) NO.1 F.O.SER.T(S) Fr21-Fr23 6.30 7.40 46.60 -37.35 -235.18 0.00 98% NO.2燃油日用舱(右) NO.2 F.O.SER.T(S) Fr23-Fr26 9.45 7.40 69.90 -35.60 -336.26 0.00 98%NO.1燃油沉淀舱(右)NO.1F.O.SETTLING.T(S) Fr21-Fr23 7.46 6.32 47.12 -37.32 -278.40 0.00 98%NO.2燃油沉淀舱(右)NO.2F.O.SETTLING.T(S) Fr23-Fr26 13.59 6.16 83.70 -35.57 -483.26 0.00 98%柴油日用舱(右) M.D.O.SEV.T(S) Fr9-Fr13 7.39 7.08 52.36 -44.99 -332.54 0.00 98% 柴油沉淀舱(右) M.D.O.SETTLING.T(S)Fr9-Fr13 5.48 7.57 41.49 -44.90 -246.07 0.00 98% 滑油储存舱(右) L.O. STORAGE TK(S) Fr5-Fr8 0.53 7.57 4.05 -48.00 -25.67 1.93 10%滑油循环舱(中) L.O.CIRCULATINGTK(C) Fr17-Fr23 0.27 0.90 0.25 -38.75 -10.60 0.57 10%艉轴冷却水舱(中) S/T C.W.T(C) STERN-Fr98.74 2.29 19.97 -47.27 -413.00 0.96 100% 舱底水舱(中) BILGE WATER TK.(C) Fr9-Fr12 3.36 0.75 2.53 -45.26 -151.90 2.67 100% 油渣舱(右) SLUDGE TK.(S) Fr25-Fr28 15.08 0.92 13.80 -34.17 -515.25 29.94 100% 分油机油渣舱(右) PURIFY SLUDGE T(S)Fr15-Fr20 6.47 5.22 33.79 -39.73 -257.19 3.18 100% 污滑油舱(中) DIRTY L.O. TK (C) Fr17-Fr23 2.73 0.30 0.82 -38.75 -105.94 0.57 100% 污水舱(左) SEWAGE TK(P) Fr4-Fr7 5.26 7.62 40.08 -48.59 -255.70 1.89 100%TOTAL 100.854 4.600 463.972 -39.219 -3955.39 41.710 JZ MARINE TRIM& STABILITY CALCULATION JZ403.101.005JS Page 20JZ MARINE TRIM AND STABILITY CALCULATION JZ403.101.005JS Page 21General Loading Case No.1LIGHT SHIPWEIGHTVCGMOMENT LCGMOMENT FSMI T E Mabt.BLabt.MS( t ) ( m )( t-m )( m ) ( t-m )( t-m )Crew, Luggage 0.0017.0000.0-45.0000.0Provisions(P&S)0.008.1290.0-5.7690.00.00No.1 F.O.T(P&S)0.000.6500.0-12.8430.00.00No.2 F.O.T(P&S)0.000.6720.0-26.0500.00.00NO.1 M.D.O(P&S)0.00 1.0300.0-38.0920.00.00NO.2 M.D.O(C)0.000.7000.0-42.6660.00.00Miscellaneous Consumables 0.00 5.470.0-39.820.00.000.00.00.00Cargo00.00.00.00 No.1 Cargo Hold 0.007.0400.033.0500.00.00 No.2 Cargo Hold 0.00 5.7030.09.2750.00.00 No.3 Cargo Hold 0.00 5.8280.0-19.2220.00.00BALLAST WATER (C)0.00 4.2880.049.7950.00.00No.1 D.B Tk(P&S)0.000.7010.037.4560.00.00No.2 D.B Tk(P&S)0.000.6730.028.2950.00.00No.3 D.B Tk(P&S)0.000.6630.016.1750.00.00No.4 D.B Tk(P&S)0.000.6610.0 1.5000.00.00No.1 S.B.W.T(P&S)0.00 4.2770.037.7550.00.00No.2 S.B.W.T(P&S)0.00 4.0930.028.8430.00.00No.3 S.B.W.T(P&S)0.00 4.0010.016.1970.00.00No.4 S.B.W.T(P&S)0.00 4.0000.0 1.5000.00.00No.5 S.B.W.T(P&S)0.00 3.4880.0-12.7560.00.00No.6 S.B.W.T(P&S)0.00 4.0330.0-27.9660.00.00APTk(C)0.00 5.0330.0-49.0830.00.00Aft (P&S)0.007.9140.0-51.6040.00.00DEADWEIGHT 0.000.00.0LIGHT SHIP 3050.007.04021472.0-6.280-19154.0 DISPLACEMENT3050.07.04021472.0-6.280-19154.00.00Lpp =104.52 ( m ) TRIM AND INITIAL STABILITYDp =3.400 ( m ) h =2.000( m )Draft Moulded ( m ) 2.147 3.9497KG ( m )7.040LCG *( m )-6.280TKM 8.649( m )12.963LCB *( m ) 3.300 3.0401IFSM ( t-m ) 0LCF *( m ) 3.168 2.0743IFSM/DISP ( m ) 0.000MTC ( t-m/cm )88.17397.849GM ( m ) 5.923Trim **( m ) 3.314GMo ( m ) 5.923Draft at F.P.( m )0.591KG'=KG+IFSM/DISP ( m ) 7.040Draft at A.P.( m ) 3.904Immersion RateDraft at M.S.( m )2.248of Propeller( % )106.0GMo5.9230.150Note:* Minus sign "-" stands for position after midship ** Trim = draft at A.P. - draft at F.P.STABILITY REQUIREMENTALLOWABLE GMoJUDGEMENTYES。
as受弯构件正截面计算PPT学习教案
M/ Mu
1.0 Mu My
0.8
Ⅱa Ⅲ Ⅲa
2. 第II阶段：带裂缝工作阶段
0.6 Ⅱ
0.4
→IIa，M→My。正常使用状态 是变形和裂缝宽度验算依据
Mcr ⅠⅠa
0
f
3. 第III阶段：破坏阶段
→IIIa，M→Mu。 是正截面抗弯计算依据
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4.2 钢筋混凝土受弯构件（2/5）
4.2.2 受弯构件正截面计算1（5/14）
2. 界限相对受压区高度和最大配筋率
a.相对受压区高度 x
h0
M xbx=bbxh0n
aa1 fcc CC=aa1fcfbcbxxb
b.界限破坏
Ts=fsAy As s
当梁的配筋率达到最大配筋率 ma时x ，受拉钢筋屈服的同时 ，受压区边缘的混凝土也达到极限压应变被压碎破坏，这
种破坏称为界限破坏。
a sba1 fcbh02
M
Mu
f
y
As
h0
x 2
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s f y Ash0
4.2 钢筋混凝土受弯构件（2/5）
4.2.2 受弯构件正截面计算1（*/14）
【习题1】
如图所示简支梁，跨中有集中力作用，试设计跨中截面配筋 。混凝土选用C20，钢筋为II级。
C 20 : fc 9.6N / mm 2；II级钢筋：f y 300N / mm 2 跨中弯矩为 Pl 4
3. 超筋梁
过多 , 出现许多小裂缝，但 s<fy, 当 c=cu, 压区砼被压碎，梁破坏。 “脆性破坏”
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4.2 钢筋混凝土受弯构件（2/5）
桁架模板计算书
2.3 连续3跨2.3.1 设计数据1 基本数据混凝土强度 C30施工阶段结构重要性系数 γ01＝0.9使用阶段结构重要性系数 γ02＝1.1永久荷载分项系数 γG ＝ 1.2可变荷载分项系数 γQ ＝ 1.4次梁间距l 1＝2250mm l 2＝2250mm l 3＝2250mm 楼板厚度 h ＝100mm 假设支承梁上翼缘宽度b 1均为200mm 模板在梁上的支承长度 a ＝100mm 单榀桁架计算宽度 b ＝188mm 钢筋桁架节点间距 l s ＝200mm混凝土抗压强度设计值 f c ＝14.3混凝土抗拉强度设计值 f t ＝ 1.43混凝土抗拉强度标准值 f tk ＝2.01混凝土弹性模量 E c ＝30000构件受力特征系数2.1受拉区纵向钢筋的相对粘结特性系数 v i ＝1mm 相对受压区高度0.3732m m N 2m m N =b ξ2m m N 2m m N 2m m N 2m m N 2m m N 2m m N ρσρψ=ψξξρρρρσψαωωωωσρψ=ψσψψπρρρραρπ=cr α钢筋抗压强度设计值 f y '＝340钢筋抗拉强度设计值 f y ＝340钢筋强度标准值 f yk ＝480钢筋弹性模量 E s ＝190000连接钢筋抗拉强度设计值 f y ＝340混凝土上保护层厚度 c'＝15mm 混凝土下保护层厚度 c ＝15mm 桁架高度 h t =70mm2 荷载施工阶段：模板自重+湿混凝土重量2.5施工荷载 1.5使用阶段：楼板2.5附加恒载1楼面活荷载102.3.2 使用阶段计算一、荷载计算楼板净跨l 1n =l 1-b 1＝2050mm l 2n =l 2-b 1＝2050mm l 3n =l 3-b 1＝2050mm 计算跨度l 10=l 1n +a/2+b 1/2＝2200mm 2m kN 2m kN2m kN=b ξ2m m N 2m m N2m m N2m kN 2m kN2m m N2m m N 2m m Nl 20=l 2n +b 1＝2250mm l 30=l 3n +a/2+b 1/2＝2200mm 楼板自重g 1=0.470施工荷载p 1=0.282除楼板自重外的永久荷载 g 2=0.188楼面活荷载 p 2=1.8801 施工阶段荷载计算(2) 三等跨连续板施工阶段计算跨度 l 3左＝l 3n +a/2+b 1/2＝2200mm l 3中＝l 30＝2050mm l 3右＝l 3左＝2200mm ∵ (l 3中－l 3左)/l 3中×100%＝-0.073<10%∴ 可按三等跨连续板计算内力。
单向板计算书
实训题目一2010、12 付丽文一、设计题目某工业用现浇钢筋混凝土单向板助形楼盖，楼盖平面如图1所示，试设计该楼盖。
图3-1 楼盖平面图二、设计资料1、楼面可变荷载标准值按表1设计题号对应荷载值选择，中柱断面尺寸400m×400m。
表1设计题号对应荷载值1）20mm厚水泥砂浆面层，重度γ=20KN/m32）钢筋混凝土现板，重度γ=25KN/ m3。
3)15mm厚混合砂浆板底抹灰，重度γ=17KN/ m33、材料选用(1)混凝土采用C20(可以调整)(f C=9.6N/mm2, f t=1.1N/mm2)(2)钢筋梁中受力钢筋采用HRB335级(f y=300N/mm2,)其余钢筋一律采用HPB235级(f y =210N/mm2,)(可以调整)三、设计内容1)结构平面布置(柱网布置，主梁、次梁及板的布置。
)_2)板的内力计算、配筋计算(按塑性理论计算内力)3)次梁的内力计算、配筋计算(按塑性理论计算内力)4）主梁的内力计算、配筋计算(按弹性理论计算内力)5)绘制结构施工图(图幅比例可以自行掌握)①结构平面布置(比例1：100，可以在此图中画出板的配筋图)。
②次梁配筋图(比例1：50、1：25)。
③主梁配图及M、V包络图(比例1：40、1：20)④钢筋明细表及必要的说明。
四、设计要求(1)计算书要求书写工整、数字准确、画出必要的计算简图。
(2)制图要求所有图线、图例尺寸和标注方法均应符合国家现行的建筑制图标准，图样上所有汉字和数字均应书写端正、排列整齐、笔画清晰，中文书写为长仿宋字。
解：一、选择题目选设计任务书中的第4组题目，l1=30m,l2=20.7m二、结构平面布置结构平面布置如图1所示。
三、单向板的设计1、板厚及主次粱截面尺寸多跨连续板的厚度按不进行挠度验算条件不小于l0/40,同时对于现浇工业建筑单向楼板厚度不小于70mm.222222000000/26.0/17015.0153/00.2/2508.080)2(/40.0/2002.02015594%,10%95.02100/21002120.2120,2140)2/1202/2001202300(2/2120)2/802/2001202300(2/)2/2/(2100)2002300(32250,325~217650)2/1~3/1()2/1~3/1(650,863~4936900)8/1~14/1()8/1~14/1(200,225~150450)2/1~3/1()2/1~3/1(.450,5003336000)12/118/1()12/118/1(.80,5.57)40/2300(40/m KN m KN mm m KN m KN mm m KN m KN mm mm l mm mm a l mm mm h l a h l l mm mm l l mm b mm mm h b mm h mm mm l h mm b mm mm h b mm h mm mm l h mm h mm mm l n n n n =⨯=⨯=⨯<=-==+--=+=+--=++==-====⨯====⨯====⨯===-=⨯-=-====厚混合沙浆板底抹灰）（钢筋混凝土现浇板厚水泥沙浆面层）（、荷载计算跨连续板计算。
PKPM结构计算书
1工程概况1.1 结构设计条件本工程采用框架结构。
设计使用年限为50年，结构安全等级为二级。
1.1.1 气象条件基本风压0.35 KN/m2，基本雪压0.35KN/m2，地面粗糙程度为C类，全年主导风向北偏南。
1.1.2 抗震设防设防烈度为7度，设计地震分组为第二组，设计基本地震加速度值为0.10g，II类场地。
1.1.3 工程地质条件场地地形平坦，地质总体状况为上覆盖新生界(代)2第四系,下伏太古界(代)。
勘测期间，勘测范围内未见地下水。
土层及其主要物理力学指标见表1.1。
表1.1 土层及其主要物理力学指标1．2 工程设计概况工程设计概况见表1.2。
表1.2 工程概况表1.2(续)注：结构高度指室外地坪至檐口或大屋面（斜屋面至屋面中间高）1.3 设计依据建筑地基基础设计规范（GB50007-2002）建筑结构荷载规范（GB50009-2001）建筑抗震设计规程（GB50011-2001）混凝土结构设计规范（GB50010-2002）高层建筑混凝土结构技术规程（JGJ3-2002）建筑抗震设计规范（GB50011-2001）建筑地基基础设计规范（GB50007-2002）1.4 可变荷载标准值选用（kN/㎡）可变荷载标准值选用见表1.3。
表1.3 可变荷载标准值选用1.5 上部永久荷载标准值及构件计算1.5.1 楼面荷载1. 首层卧室、起居室、书房：150厚砼板 3.75kN/m2板面装修荷载 1.0kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 5.25kN/m2厨房、普通卫生间：150厚砼板 3.75kN/m2板面装修荷载 1.1kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 5.35kN/m2带采暖卫生间：150厚砼板 3.75kN/m2板面装修荷载 20x0.13=2.6kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 6.85kN/m2门厅、电梯间：150厚砼板 3.75kN/m2板面装修荷载 20x0.07=1.4kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 5.65kN/m22. 标准层:卧室、起居室、书房：110厚砼板 2.75kN/m2板面装修荷载 1.0kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 4.25kN/m2对于120厚楼板：4.5kN/m2对于130厚楼板：4.75kN/m2对于140厚楼板：5.0kN/m2对于150厚楼板：5.25kN/m2对于200厚楼板：8.0kN/m2(包括大跨度楼板上轻质隔墙的折算荷载1.5kN/m2) 厨房、普通卫生间：100厚砼板 2.5kN/m2板面装修荷载 1.1kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 4.10kN/m2带采暖卫生间：100厚砼板 2.5kN/m2板面装修荷载 20x0.13=2.6kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 5.60kN/m2楼电梯间：120厚砼板 3.0kN/m2板面装修荷载 20x0.07=1.4kN/m2板底粉刷或吊顶 0.50kN/m2恒载合计 4.90kN/m2 1.5.2 屋面荷载屋面板120厚：4cm细石混凝土面层 1.00kN/m2三元乙丙橡胶卷材防水层 0.3kN/m2=40) 0.5kN/m2聚苯乙烯板隔热层2%找坡(hmin防水卷材 0.30kN/m22cm找平 0.40kN/m2120厚砼板 3.0kN/m22cm底粉刷 0.4kN/m2恒载合计 6.2kN/m2对于130厚楼板：6.5kN/m2对于140厚楼板：6.7kN/m2露台:水泥砂浆+10厚防滑地砖 0.7kN/m24cm细石混凝土面层 1.00kN/m2三元乙丙橡胶卷材防水层 0.3kN/m2=40) 0.5kN/m2聚苯乙烯板隔热层2%找坡(hmin防水卷材 0.30kN/m22cm找平 0.40kN/m2110厚砼板 2.75kN/m22cm底粉刷 0.4kN/m2恒载合计 6.4kN/m2对于120厚楼板：6.7kN/m2对于130厚楼板：6.9kN/m21.5.3墙体荷载可变荷载标准值选用见表1.3。
(完整版)篱笆计算书
(完整版)篱笆计算书引言篱笆计算书是一种传统的数学研究工具，用于帮助学生进行数学运算和解决问题。
它由一组小木球和编织的篱笆构成，可用于展示和可视化数学概念和操作。
本文档将介绍篱笆计算书的基本原理、用法以及一些与之相关的应用场景。
篱笆计算书的构成篱笆计算书由一个长方形的篱笆构成，篱笆上有等距排列的小木球。
每根篱笆通常由10或20个小木球组成，根据需要可以增加或减少。
篱笆计算书以及木球通常以不同的颜色来表示不同的数值或数学概念。
篱笆计算书的原理篱笆计算书的原理基于进（进位）和出（退位）的概念，用于辅助进行加法、减法、乘法和除法等数学运算。
- 进位：当某个小球的数量达到篱笆上一定的数值时，它会进位，即向上移动到上一根篱笆的相应位置。
- 退位：当某个小球的数量不足以维持在当前位置时，它会退位，即向下移动到下一根篱笆的相应位置。
通过进位和退位的操作，篱笆计算书可以帮助学生以一种可视化的方式理解和运用数学运算的概念。
篱笆计算书的用法篱笆计算书可以用于进行多种数学运算，下面以加法为例进行说明：1. 将两个数值分别表示为篱笆上的小球数量。
2. 从右到左逐位进行加法运算。
如果有进位则向上移动小球，没有则保持不动。
3. 重复步骤2，直到所有位数都计算完毕。
4. 最终结果即为篱笆上每个位数上的小球数量。
篱笆计算书还可以用于减法、乘法和除法等数学运算，具体的方法和步骤类似。
篱笆计算书的应用场景篱笆计算书具有直观、可视化的特点，适用于多种数学教学和研究场景，如：- 小学数学教学：篱笆计算书可以帮助学生直观理解加法、减法等运算概念，提高计算准确性和速度。
- 心算训练：通过使用篱笆计算书进行心算训练，学生可以提高计算技巧和速度。
- 数学问题解决：篱笆计算书可以用于解决一些数学问题，如排列组合、概率和统计等。
结论篱笆计算书是一种有趣且实用的数学研究工具，可以帮助学生理解和运用多种数学运算概念。
它的直观性和可视化特点适用于不同年龄段的学生，为他们提供了一种新的研究和思考数学的方式。
42m钢箱梁计算书
ES匝道钢箱梁上部结构计算书2017.11目录一、概述 (1)1.1桥梁简介 (1)1.2 模型概况 (1)1 设计规范 (1)2 参考规范 (1)3 主要材料及性能指标 (1)4 荷载 (2)二、模型概述 (3)2.1 第一体系建模 (3)2.2 第二体系建模 (4)三、结果验算 (5)3.1顶底板强度验算 (5)1 计算结果 (5)2 强度验算 (6)3.2 腹板验算 (7)1 厚度验算 (7)2 腹板强度验算 (7)3 腹板纵向加劲肋构造验算 (8)4 腹板横向加劲肋构造验算 (8)3.3 构件设计验算 (9)1 加劲肋构造验算 (9)2 受压板加劲肋刚度验算 (10)3 闭口肋几何尺寸验算 (10)4 支承加劲肋验算 (11)3.4刚度验算 (11)1 车道荷载挠度值 (11)2 正交异形板桥面顶板挠跨比 (12)3 横隔板刚度验算 (13)3.5 整体稳定验算 (13)3.6 疲劳验算 (13)四、结论 (14)一、概述1.1桥梁简介ES匝道桥为一单跨42m简支钢箱梁桥。
截面采用等截面形式，梁宽10.2m，梁高2m。
主梁线型为圆曲线，中心线位于半径R=682m的圆弧上。
顶板厚18mm，腹板和底板厚20mm，顶板U肋厚8mm，开口肋厚20mm。
材料采用Q345C材质。
图1.1典型钢箱梁横断面(mm)1.2 模型概况1 设计规范《公路工程结构可靠度设计统一标准》（GB/T 50283-1999）;《公路工程技术标准》（JTG B01-2014）《公路桥涵设计通用规范》（JTG D60-2015）《公路钢结构桥梁设计规范》（JTG D64-2015）《钢结构设计规范》(GB50017-2014)《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62-2004)2 参考规范《道路桥示方书·同解说》（日本道路协会，平成8年12月）3 主要材料及性能指标主梁采用Q345C钢材，其主要力学性能见下表。
板的pkpm计算书
楼板计算书日期： 7/02/2012时间： 4:18:44:92 pm一、基本资料：1、房间编号： 12、边界条件（左端/下端/右端/上端）：固定/铰支/固定/固定/3、荷载：永久荷载标准值：g ＝ 2.86 kN/m2可变荷载标准值：q ＝ 3.00 kN/m2计算跨度Lx = 2200 mm ；计算跨度Ly = 6065 mm板厚H = 80 mm；砼强度等级：C30；钢筋强度等级：HPB2354、计算方法：弹性算法。
5、泊松比：μ＝1/5.6、考虑活荷载不利组合。
二、计算结果：Mx =(0.04255+0.00790/5)*(1.20* 2.86+1.40* 1.50)* 2.2^2 = 1.18kN·m 考虑活载不利布置跨中X向应增加的弯矩：Mxa =(0.11430+0.00857/5)*(1.4* 1.50)* 2.2^2 = 1.18kN·mMx= 1.18 + 1.18 = 2.36kN·mAsx= 245.64mm2，实配φ 8@200 (As ＝ 251.mm2)ρmin ＝ 0.307% ，ρ＝ 0.314%My =(0.00790+0.04255/5)*(1.20* 2.86+1.40* 1.50)* 2.2^2= 0.44kN·m考虑活载不利布置跨中Y向应增加的弯矩：Mya =(0.00857+0.11430/5)*(1.4* 1.50)* 2.2^2 = 0.32kN·mMy= 0.44 + 0.32 = 0.76kN·mAsy= 245.64mm2，实配φ 8@200 (As ＝ 251.mm2)ρmin ＝ 0.307% ，ρ＝ 0.314%Mx' =0.08330*(1.20* 2.86+1.40* 3.00)* 2.2^2 = 3.07kN·mAsx'= 276.37mm2，实配φ 8@100 (As ＝ 503.mm2,可能与邻跨有关系)ρmin ＝ 0.307% ，ρ＝ 0.628%My' =0.05583*(1.20* 2.86+1.40* 3.00)* 2.2^2 = 2.06kN·mAsy'= 245.64mm2，实配φ 8@200 (As ＝ 251.mm2,可能与邻跨有关系)ρmin ＝ 0.307% ，ρ＝ 0.314%三、跨中挠度验算：Mq -------- 按荷载效应的准永久组合计算的弯矩值(1)、挠度和裂缝验算参数：Mq =(0.00790+0.04255/5)*(1.0* 2.86+0.5* 3.00 )* 2.2^2 = 0.35kN·mEs ＝ 210000.N/mm2 Ec ＝ 29791.N/mm2Ftk ＝ 2.01N/mm2 Fy ＝ 210.N/mm2(2)、在荷载效应的准永久组合作用下，受弯构件的短期刚度 Bs：①、裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 0.35/(0.87* 48.* 251.) = 32.958N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 251./ 40000.=0.00628ψ＝ 1.1 - 0.65* 2.01/(0.00628* 32.96) = -5.197当ψ＜0.2 时，取ψ＝ 0.2②、钢筋弹性模量与混凝土模量的比值αE：αE ＝Es / Ec ＝210000.0/ 29791.5 = 7.049③、受压翼缘面积与腹板有效面积的比值γf'：矩形截面，γf' ＝ 0④、纵向受拉钢筋配筋率ρ＝ As / b / ho ＝ 251./1000/ 48.=0.00524⑤、钢筋混凝土受弯构件的 Bs 按公式（混凝土规范式 7.2.3－1）计算:Bs=Es*As*ho^2/[1.15ψ+0.2+6*αE*ρ/(1+ 3.5γf')]Bs＝ 210000.* 251.* 48.^2/[1.15*0.200+0.2+6*7.049*0.00524/(1+3.5*0.00)]= 186.66kN·m2 (3)、考虑荷载长期效应组合对挠度影响增大影响系数θ：按混凝土规范第 7.2.5 条，当ρ' ＝ 0时，θ＝ 2.0(4)、受弯构件的长期刚度 B，可按下列公式计算：B ＝ Bs / θ（混凝土规范式 7.2.2）B＝ 186.66/2 = 93.332kN·m2(5)、挠度 f ＝κ * Qq * L ^ 4 / Bf ＝0.00260* 4.4* 2.20^4/ 93.332= 2.842mmf / L ＝ 2.842/2200.= 1/ 774.，满足规范要求！四、裂缝宽度验算：①、X方向板带跨中裂缝：裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 0.93*10^6/(0.87* 56.* 251.) = 75.971N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 251./ 40000.= 0.006当ρte ＜0.01 时，取ρte ＝ 0.01ψ＝ 1.1 - 0.65* 2.01/( 0.01* 75.97) = -0.616当ψ＜0.2 时，取ψ＝ 0.2ωmax ＝αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) （混凝土规范式 7.1.2－1）ωmax ＝1.9*0.200* 75.971/210000.*(1.9*20.+0.08*11.43/0.01000) = 0.018mm，满足规范要求！②、Y方向板带跨中裂缝：裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 0.35*10^6/(0.87* 48.* 251.) = 32.958N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 251./ 40000.= 0.006当ρte ＜0.01 时，取ρte ＝ 0.01ψ＝ 1.1 - 0.65* 2.01/( 0.01* 32.96) = -2.856当ψ＜0.2 时，取ψ＝ 0.2ωmax ＝αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) （混凝土规范式 7.1.2－1）ωmax ＝1.9*0.200* 32.958/210000.*(1.9*20.+0.08*11.43/0.01000) = 0.008mm，满足规范要求！③、左端支座跨中裂缝：裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 1.76*10^6/(0.87* 56.* 503.) = 71.697N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 503./ 40000.= 0.013ψ＝ 1.1 - 0.65* 2.01/( 0.01* 71.70) = -0.347当ψ＜0.2 时，取ψ＝ 0.2ωmax ＝αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) （混凝土规范式 7.1.2－1）ωmax ＝1.9*0.200* 71.697/210000.*(1.9*20.+0.08*11.43/0.01257) = 0.014mm，满足规范要求！⑤、右端支座跨中裂缝：裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 1.76*10^6/(0.87* 56.* 279.) = 129.055N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 279./ 40000.= 0.007当ρte ＜0.01 时，取ρte ＝ 0.01ψ＝ 1.1 - 0.65* 2.01/( 0.01* 129.05) = 0.090当ψ＜0.2 时，取ψ＝ 0.2ωmax ＝αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) （混凝土规范式 7.1.2－1）ωmax ＝1.9*0.200*129.055/210000.*(1.9*20.+0.08*11.43/0.01000) = 0.030mm，满足规范要求！⑥、上端支座跨中裂缝：裂缝间纵向受拉钢筋应变不均匀系数ψ，按下列公式计算：ψ＝ 1.1 - 0.65 * ftk / (ρte * σsq) （混凝土规范式 7.1.2－2）σsq ＝ Mq / (0.87 * ho * As) （混凝土规范式 7.1.4－3）σsq ＝ 1.18*10^6/(0.87* 56.* 251.) = 96.102N/mm2矩形截面，Ate＝0.5*b*h＝0.5*1000* 80.= 40000.mm2ρte ＝ As / Ate （混凝土规范式 7.1.2－4）ρte ＝ 251./ 40000.= 0.006当ρte ＜0.01 时，取ρte ＝ 0.01ψ＝ 1.1 - 0.65* 2.01/( 0.01* 96.10) = -0.257当ψ＜0.2 时，取ψ＝ 0.2ωmax ＝αcr*ψ*σsq/Es*(1.9c+0.08*Deq/ρte) （混凝土规范式 7.1.2－1）ωmax ＝1.9*0.200* 96.102/210000.*(1.9*20.+0.08*11.43/0.01000) = 0.023mm，满足规范要求！---------------------------------------------------一、基本资料：1、房间编号： 22、边界条件（左端/下端/右端/上端）：铰支/铰支/固定/固定/3、荷载：永久荷载标准值：g ＝ 2.86 kN/m2可变荷载标准值：q ＝ 3.00 kN/m2计算跨度Lx = 2265 mm ；计算跨度Ly = 6065 mm板厚H = 80 mm；砼强度等级：C30；钢筋强度等级：HPB2354、计算方法：弹性算法。
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DESIGN CALCULATIONIn Accordance with ASME Section VIII Division 1ASME Code Edition : 2007,up to 2009 AddendaDrawing No. : YX09RW-015-00 Rev.0Calculation Rev. : Rev.0Analysis Performed by : ZHANGJIAGANG HUALING CHEMICAL MACHINERY CO.LTD.Date of Analysis : Jul 26,2010Job File : VANPHONG BONDED PETROLEUM TERMINAL PROJECT-V-9601&V-9602PREPARED BY: ____________REVIEWED BY: ____________APPROVED BY: ____________Statement:The used Modules in this calculation of computer program (PVElite Edition 2010) have been verified by hand calculation,and found meet the requirements of ASME Code Section VIII-1,2007 edition, up to 2009 addenda.Verified by DateApproved by Date1Table of Contents:1. Shell Input Data (3)2. Internal Pressure Calculations (9)3. Nozzle Flange MAWP (13)4. Stress Calcs. for saddles (Operating case) (14)5. Stress Calcs. for saddles (Test case) (28)6. Nozzle Calcs. N1 (42)7. Nozzle Calcs. N2 (45)8. Nozzle Calcs. N3 (50)9. Nozzle Calcs. N4 (53)10. Nozzle Calcs. N5 (56)11. Nozzle Calcs. K1A (59)12. Nozzle Calcs. K1B (62)13. Nozzle Calcs. K2 (66)14. Nozzle Calcs. M1 (69)15. Nozzle Schedule and nozzle summary (75)2Vessel Analysis: Input DataDesign Internal Pressure (for Hydrotest) 0.4000 MPaDesign Internal Temperature 60 CType of Hydrotest UG99-bHydrotest Position HorizontalProjection of Nozzle from Vessel Top 0.0000 mmProjection of Nozzle from Vessel Bottom 0.0000 mmMinimum Design Metal Temperature -196 CType of Construction WeldedSpecial Service NoneDegree of Radiography RT 1UG-22 LoadingsComplete Listing of Vessel Elements and Details:Element From Node 10Element To Node 20Element Type EllipticalDescriptionDistance "FROM" to "TO" 25.000 mm3Inside Diameter 1850.0 mmElement Thickness 7.2000 mmInternal Corrosion Allowance 0.0000 mmNominal Thickness 8.0000 mmExternal Corrosion Allowance 0.0000 mmDesign Internal Pressure 0.4000 MPaDesign Temperature Internal Pressure 60 CDesign External Pressure 0.0000 MPaDesign Temperature External Pressure 20 CEffective Diameter Multiplier 1.2Material Name SA-240 316Allowable Stress, Ambient 137.90 MPaAllowable Stress, Operating 137.90 MPaAllowable Stress, Hydrotest 186.16 MPaMaterial Density 0.007750 kg/cm^3P Number Thickness 0.0000 mmYield Stress, Operating 192.51 MPaExternal Pressure Chart Name HA-2UNS Number S31600Product Form PlateEfficiency, Longitudinal Seam 1.Efficiency, Circumferential Seam 1.Elliptical Head Factor 2.Element From Node 20Element To Node 30Element Type CylinderDescriptionDistance "FROM" to "TO" 4150.0 mmInside Diameter 1850.0 mmElement Thickness 8.0000 mmInternal Corrosion Allowance 0.0000 mmNominal Thickness 8.0000 mmExternal Corrosion Allowance 0.0000 mmDesign Internal Pressure 0.4000 MPaDesign Temperature Internal Pressure 60 CDesign External Pressure 0.0000 MPaDesign Temperature External Pressure 20 CEffective Diameter Multiplier 1.2Material Name SA-240 316Efficiency, Longitudinal Seam 1.Efficiency, Circumferential Seam 1.Element From Node 204Detail Type SaddleDetail ID Lft SdlDist. from "FROM" Node / Offset dist 500.00 mmWidth of Saddle 272.00 mmHeight of Saddle at Bottom 1183.0 mmSaddle Contact Angle 120.Height of Composite Ring Stiffener 0.0000 mmWidth of Wear Plate 460.00 mmThickness of Wear Plate 10.000 mmContact Angle, Wear Plate (degrees) 132.Element From Node 20Detail Type SaddleDetail ID Rht SdlDist. from "FROM" Node / Offset dist 3650.0 mmWidth of Saddle 272.00 mmHeight of Saddle at Bottom 1183.0 mmSaddle Contact Angle 120.Height of Composite Ring Stiffener 0.0000 mmWidth of Wear Plate 460.00 mmThickness of Wear Plate 10.000 mmContact Angle, Wear Plate (degrees) 132.Element From Node 20Detail Type NozzleDetail ID N1Dist. from "FROM" Node / Offset dist 1040.0 mmNozzle Diameter 50. mmNozzle Schedule 40Nozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID N2Dist. from "FROM" Node / Offset dist 3150.0 mmNozzle Diameter 100. mmNozzle Schedule 40Nozzle Class 150Layout Angle 270.5Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID N3Dist. from "FROM" Node / Offset dist 150.00 mmNozzle Diameter 50. mmNozzle Schedule 40Nozzle Class 150Layout Angle 270.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID N4Dist. from "FROM" Node / Offset dist 3640.0 mmNozzle Diameter 50. mmNozzle Schedule 40Nozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID N5Dist. from "FROM" Node / Offset dist 640.00 mmNozzle Diameter 50. mmNozzle Schedule 40Nozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP3166Element From Node 20Detail Type NozzleDetail ID K1ADist. from "FROM" Node / Offset dist 4090.0 mmNozzle Diameter 20. mmNozzle Schedule 40Nozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID K2Dist. from "FROM" Node / Offset dist 140.00 mmNozzle Diameter 50. mmNozzle Schedule 40Nozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316Element From Node 20Detail Type NozzleDetail ID M1Dist. from "FROM" Node / Offset dist 2340.0 mmNozzle Diameter 508. mmNozzle Schedule NoneNozzle Class 150Layout Angle 90.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-240 316Element From Node 30Element To Node 40Element Type EllipticalDescriptionDistance "FROM" to "TO" 25.000 mm7Inside Diameter 1850.0 mmElement Thickness 7.2000 mmInternal Corrosion Allowance 0.0000 mmNominal Thickness 8.0000 mmExternal Corrosion Allowance 0.0000 mmDesign Internal Pressure 0.4000 MPaDesign Temperature Internal Pressure 60 CDesign External Pressure 0.0000 MPaDesign Temperature External Pressure 20 CEffective Diameter Multiplier 1.2Material Name SA-240 316Efficiency, Longitudinal Seam 1.Efficiency, Circumferential Seam 1.Elliptical Head Factor 2.Element From Node 30Detail Type NozzleDetail ID K1BDist. from "FROM" Node / Offset dist 600.00 mmNozzle Diameter 20. mmNozzle Schedule 40Nozzle Class 150Layout Angle 270.Blind Flange (Y/N) NWeight of Nozzle ( Used if > 0 ) 0.0000 NGrade of Attached Flange GR 2.2Nozzle Matl SA-312 TP316PVElite is a registered trademark of COADE, Inc. [2010]8Element Thickness, Pressure, Diameter and Allowable Stress :| | Int. Press | Nominal | Total Corr| Element | Allowable |From| To | + Liq. Hd | Thickness | Allowance | Diameter | Stress(SE)|| | MPa | mm | mm | mm | MPa | ---------------------------------------------------------------------------10| 20| 0.40000 | 8.00000 | 0.00000 | 1850.00 | 137.900 |20| 30| 0.40000 | 8.00000 | 0.00000 | 1850.00 | 137.900 |30| 40| 0.40000 | 8.00000 | 0.00000 | 1850.00 | 137.900 |Element Required Thickness and MAWP :| | Design | M.A.W.P. | M.A.P. | Minimum | Required |From| To | Pressure | Corroded | New & Cold | Thickness | Thickness || | MPa | MPa | MPa | mm | mm | ----------------------------------------------------------------------------10| 20| 0.40000 | 1.07255 | 1.07255 | 7.20000 | 2.68388 |20| 30| 0.40000 | 1.18649 | 1.18649 | 8.00000 | 2.68778 |30| 40| 0.40000 | 1.07255 | 1.07255 | 7.20000 | 2.68388 |Minimum 1.073 1.073MAWP: 1.073 MPa, limited by: Elliptical Head.Internal Pressure Calculation Results :ASME Code, Section VIII, Division 1, 2007 A-09Elliptical Head From 10 To 20 SA-240 316 at 60 CRequired Thickness due to Internal Pressure [tr]:= (P*D*Kcor)/(2*S*E-0.2*P) Appendix 1-4(c)= (0.400*1850.0000*1.000)/(2*137.90*1.00-0.2*0.400)= 2.6839 + 0.0000 = 2.6839 mmMax. Allowable Working Pressure at given Thickness, corroded [MAWP]:= (2*S*E*t)/(Kcor*D+0.2*t) per Appendix 1-4 (c)= (2*137.90*1.00*7.2000)/(1.000*1850.0000+0.2*7.2000)= 1.073 MPaMaximum Allowable Pressure, New and Cold [MAPNC]:= (2*S*E*t)/(K*D+0.2*t) per Appendix 1-4 (c)= (2*137.90*1.00*7.2000)/(1.000*1850.0000+0.2*7.2000)= 1.073 MPa9Actual stress at given pressure and thickness, corroded [Sact]:= (P*(Kcor*D+0.2*t))/(2*E*t)= (0.400*(1.000*1850.0000+0.2*7.2000))/(2*1.00*7.2000)= 51.429 MPaStraight Flange Required Thickness:= (P*R)/(S*E-0.6*P) + c per UG-27 (c)(1)= (0.400*925.0000)/(137.90*1.00-0.6*0.400)+0.000= 2.688 mmStraight Flange Maximum Allowable Working Pressure:= (S*E*t)/(R+0.6*t) per UG-27 (c)(1)= (137.90 * 1.00 * 8.0000 ) / (925.0000 + 0.6 * 8.0000 )= 1.186 MPaPercent Elongation per UHA-44 (75*tnom/Rf)*(1-Rf/Ro) 1.884 %Note: Please Check Requirements of Table UHA-44 for Elongation limits.Cylindrical Shell From 20 To 30 SA-240 316 at 60 CRequired Thickness due to Internal Pressure [tr]:= (P*R)/(S*E-0.6*P) per UG-27 (c)(1)= (0.400*925.0000)/(137.90*1.00-0.6*0.400)= 2.6878 + 0.0000 = 2.6878 mmMax. Allowable Working Pressure at given Thickness, corroded [MAWP]:= (S*E*t)/(R+0.6*t) per UG-27 (c)(1)= (137.90*1.00*8.0000)/(925.0000+0.6*8.0000)= 1.186 MPaMaximum Allowable Pressure, New and Cold [MAPNC]:= (S*E*t)/(R+0.6*t) per UG-27 (c)(1)= (137.90*1.00*8.0000)/(925.0000+0.6*8.0000)= 1.186 MPaActual stress at given pressure and thickness, corroded [Sact]:= (P*(R+0.6*t))/(E*t)= (0.400*(925.0000+0.6*8.0000))/(1.00*8.0000)= 46.490 MPaPercent Elongation per UHA-44 (50*tnom/Rf)*(1-Rf/Ro) 0.431 %Note: Please Check Requirements of Table UHA-44 for Elongation limits.Elliptical Head From 30 To 40 SA-240 316 at 60 C10Required Thickness due to Internal Pressure [tr]:= (P*D*Kcor)/(2*S*E-0.2*P) Appendix 1-4(c)= (0.400*1850.0000*1.000)/(2*137.90*1.00-0.2*0.400)= 2.6839 + 0.0000 = 2.6839 mmMax. Allowable Working Pressure at given Thickness, corroded [MAWP]:= (2*S*E*t)/(Kcor*D+0.2*t) per Appendix 1-4 (c)= (2*137.90*1.00*7.2000)/(1.000*1850.0000+0.2*7.2000)= 1.073 MPaMaximum Allowable Pressure, New and Cold [MAPNC]:= (2*S*E*t)/(K*D+0.2*t) per Appendix 1-4 (c)= (2*137.90*1.00*7.2000)/(1.000*1850.0000+0.2*7.2000)= 1.073 MPaActual stress at given pressure and thickness, corroded [Sact]:= (P*(Kcor*D+0.2*t))/(2*E*t)= (0.400*(1.000*1850.0000+0.2*7.2000))/(2*1.00*7.2000)= 51.429 MPaStraight Flange Required Thickness:= (P*R)/(S*E-0.6*P) + c per UG-27 (c)(1)= (0.400*925.0000)/(137.90*1.00-0.6*0.400)+0.000= 2.688 mmStraight Flange Maximum Allowable Working Pressure:= (S*E*t)/(R+0.6*t) per UG-27 (c)(1)= (137.90 * 1.00 * 8.0000 ) / (925.0000 + 0.6 * 8.0000 )= 1.186 MPaPercent Elongation per UHA-44 (75*tnom/Rf)*(1-Rf/Ro) 1.884 %Note: Please Check Requirements of Table UHA-44 for Elongation limits.Hydrostatic Test Pressure Results:Pressure per UG99b = 1.3 * M.A.W.P. * Sa/S 1.394 MPa Pressure per UG99b[34] = 1.3 * Design Pres * Sa/S 0.520 MPa Pressure per UG99c = 1.3 * M.A.P. - Head(Hyd) 1.376 MPa Pressure per UG100 = 1.1 * M.A.W.P. * Sa/S 1.180 MPa Pressure per PED = 1.43 * MAWP 1.534 MPa User Defined Hydrostatic Test Pressure at High Point 0.520 MPaHorizontal Test performed per: UG-99bHorizontal Test performed per: UG-99bStresses on Elements due to Hydrostatic Test Pressure:From To Stress Allowable Ratio Pressure 10 20 69.2 186.2 0.372 0.54 20 30 62.5 186.2 0.336 0.54 30 40 69.2 186.2 0.372 0.54Elements Suitable for Internal Pressure.PVElite is a registered trademark of COADE, Inc. [2010]Nozzle Flange MAWP Results :Nozzle ----- Flange RatingDescription Operating Ambient Temperature Class Grade|GroupMPa MPa C----------------------------------------------------------------------------N1 1.8 1.9 60 150 GR 2.2 N2 1.8 1.9 60 150 GR 2.2 N3 1.8 1.9 60 150 GR 2.2 N4 1.8 1.9 60 150 GR 2.2 N5 1.8 1.9 60 150 GR 2.2 K1A 1.8 1.9 60 150 GR 2.2 K2 1.8 1.9 60 150 GR 2.2 M1 1.8 1.9 60 150 GR 2.2 K1B 1.8 1.9 60 150 GR 2.2 ----------------------------------------------------------------------------Minimum Rating 1.786 1.896 MPaNote: ANSI Ratings are per ANSI/ASME B16.5 2003 EditionPVElite is a registered trademark of COADE, Inc. [2010]ASME Horizontal Vessel Analysis: Stresses for the Left SaddleHorizontal Vessel Stress Calculations : Operating CaseInput and Calculated Values:Vessel Mean Radius Rm 929.00 mm Tangent to Tangent Length L 4200.00 mm Distance from Saddle to Vessel tangent a 525.00 mmSaddle Width b 272.00 mm Saddle Bearing Angle theta 120.00 degreesWear Plate Width b1 460.00 mm Wear Plate Bearing Angle theta1 132.00 degrees Wear Plate Thickness tr 8.0 mmWear Plate Allowable Stress Sr 137.90 MPaInside Depth of Head h2 462.50 mmShell Allowable Stress used in Calculation 137.90 MPaHead Allowable Stress used in Calculation 137.90 MPaSaddle Force Q, Operating Case 10533.17 NHorizontal Vessel Analysis Results: Actual Allowable-------------------------------------------------------------------Long. Stress at Top of Midspan 23.00 137.90 MPa Long. Stress at Bottom of Midspan 23.45 137.90 MPaLong. Stress at Top of Saddles 23.48 137.90 MPa Long. Stress at Bottom of Saddles 23.08 137.90 MPaTangential Shear in Shell 1.09 82.74 MPaCirc. Stress at Horn of Saddle 2.41 172.37 MPaCirc. Compressive Stress in Shell 0.25 137.90 MPaLoad Combination Results for Q + Wind or Seismic [Q]:= Saddle Load + Max( Fwl, Fwt, Fsl, Fst )= 10533 + Max( 0 , 0 , 0 , 0 )= 10533.2 NSummary of Loads at the base of this Saddle:Vertical Load (including saddle weight) 12554.07 NTransverse Shear Load Saddle 0.00 NLongitudinal Shear Load Saddle 0.00 NFormulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plates are assumed to be Welded k = 0.1The Computed K values from Table 4.15.1:K1 = 0.1066 K2 = 1.1707 K3 = 0.8799 K4 = 0.4011 K5 = 0.7603 K6 = 0.0529 K7 = 0.0184 K8 = 0.3405 K9 = 0.2711 K10 = 0.0581 K1* = 0.1923 K6p = 0.0434 K7P = 0.0151The suffix 'p' denotes the values for a wear plate if it exists.Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]:= -Q*a [1 - (1- a/L + (R^(2)-h2^(2))/(2a*L))/(1+(4h2)/3L)]= -10533*525.00[1-(1-525.00/4200.00+(929.000^(2)-462.500^(2))/(2*525.00*4200.00))/(1+(4*462.50)/(3*4200.00))]= -601197.4 N-mmMoment per Equation 4.15.4 [M2]:= Q*L/4(1+2(R^(2)-h2^(2))/(L^(2)))/(1+(4h2)/( 3L))-4a/L= 10533*4200/4(1+2(929^(2)-462^(2))/(4200^(2)))/(1+(4*462)/(3*4200))-4*525/4200= 4825672.5 N-mmLongitudinal Stress at Top of Shell (4.15.6) [Sigma1]:= P * Rm/(2t) - M2/(pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) - 4825672 /(pi*929.0^(2)*8.00 )= 23.00 MPaLongitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]:= P * Rm/(2t) + M2/(pi * Rm^(2) * t)= 0.4 * 929.000 /(2 * 8.00 ) + 4825672 /(pi * 929.0^(2) * 8.00 )= 23.45 MPaLongitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]:= P * Rm/(2t) - M1/(K1*pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) - -601197.4 /(0.1066 *pi*929.0^(2)*8.00 )= 23.48 MPaLongitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]:= P * Rm/(2t) + M1/(K1*pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) + -601197.4 /(0.1066 *pi*929.0^(2)*8.00 )= 23.08 MPaMaximum Shear Force in the Saddle (4.15.5) [T]:= Q(L-2a)/(L+(4*h2/3))= 10533 ( 4200.00 - 2 * 525.00 )/(4200.00 + ( 4 * 462.50 /3))= 6888.5 NShear Stress in the shell no rings, not stiffened (4.15.14) [tau2]:= K2 * T / ( Rm * t )= 1.1707 * 6888.47 / ( 929.0001 * 8.0000 )= 1.09 MPaDecay Length (4.15.22) [x1,x2]:= 0.78 * sqrt( Rm * t )= 0.78 * sqrt( 929.000 * 8.000 )= 67.243 mmCircumferential Stress in shell, no rings (4.15.23) [sigma6]:= -K5 * Q * k / ( t * ( b + X1 + X2 ) )= -0.7603 * 10533 * 0.1 / ( 8.000 * ( 272.00 + 67.24 + 67.24 ) )= -0.25 MPaEffective reinforcing plate width (4.15.1) [B1]:= min( b + 1.56 * sqrt( Rm * t ), 2a )= min( 272.00 + 1.56 * sqrt( 929.000 * 8.000 ), 2 * 525.000 )= 406.49 mmWear Plate/Shell Stress ratio (4.15.29) [eta]:= min( Sr/S, 1 )= min( 137.900 / 137.900 , 1 )= 1.0000Circumferential Stress at wear plate (4.15.26) [sigma6,r]:= -K5 * Q * k / ( B1( t + eta * tr ) )= -0.7603 * 10533 * 0.1 / ( 406.486 ( 8.000 + 1.000 * 8.000 ) )= -0.12 MPaCirc. Comp. Stress at Horn of Saddle, L<8Rm (4.15.28) [sigma7,r*]: = -Q/(4(t+eta*tr)b1) - 12*K7*Q*Rm/(L(t+eta*tr)^(2))= -10533 /(4(8.000 + 1.000 * 8.000 )406.486 ) -12*0.018*10533*929.000/(4200.00(8.000+1.000*8.000)^(2))= -2.41 MPaFree Un-Restrained Thermal Expansion between the Saddles [Exp]:= Alpha * Ls * ( Design Temperature - Ambient Temperature )= 0.158E-04 * 3150.000 * ( 60.0 - 21.1 )= 1.931 mmResults for Vessel Ribs, Web and Base:Baseplate Length Bplen 1360.0000 mm Baseplate Thickness Bpthk 16.0000 mm Baseplate Width Bpwid 220.0000 mm Number of Ribs ( inc. outside ribs ) Nribs 4Rib Thickness Ribtk 12.0000 mmWeb Thickness Webtk 12.0000 mm Web Location Webloc SideMoment of Inertia of Saddle - Lateral DirectionY A AY Ay^(2) Io Shell 4. 4754. 19014. 76057. 25352. Wearplate 12. 3680. 44160. 529920. 19627. Web 121. 2520. 304920. 36895284. 9260990. BasePlate 234. 3520. 823680. 192740896. 75093. Totals 371. 14474. 1191774. 230242144. 9381063.Value C1 = Sumof(Ay)/Sumof(A) = 82. mm Value I = Sumof(Ay^(2) ) + Sumof(Io) - C1*Sumof(Ay) = 141490944. mm^4 Value As = Sumof(A) - Ashell = 9720. mm^2K1 = (1+Cos(beta)-.5*Sin(beta)^(2) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2035Fh = K1 * Q = 0.2035 * 10533.165 = 2143.7278 NTension Stress, St = ( Fh/As ) = 0.2206 MPaAllowed Stress, Sa = 0.6 * Yield Str = 111.6990 MPad = B - R*Sin(theta) / theta = 402.0312 mmBending Moment, M = Fh * d = 862194.8750 N-mmBending Stress, Sb = ( M * C1 / I ) = 0.5016 MPaAllowed Stress, Sa = 2/3 * Yield Str = 124.1100 MPaMinimum Thickness of Baseplate per Moss := ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 2 * BasePlateLength *AllStress ))^(.5)= ( 3 * (10533 + 2020 ) * 220.00 / ( 2 * 1360.000 * 124.110 ))^(.5)= 4.954 mmCalculation of Axial Load, Intermediate Values and Compressive StressEffective Baseplate Length [e]:= ( Bplen - Clearance ) / ( Nribs - 1)= ( 1360.0000 - 25.4 ) / ( 4 - 1 ) = 444.8667 mmBaseplate Pressure Area [Ap]:= e * Bpwid / 2= 444.8667 * 220.0000 / 2 = 48935.3320 mm^2Axial Load [P]:= Ap * Bp= 48935.3 * 0.04 = 1722.7 NArea of the Rib and Web [Ar]:= ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk= ( 220.000 - 25.4 - 12.000 ) * 12.000 + 444.8667 /2 * 12.000= 4860.400 mm^2Compressive Stress [Sc]:= P/Ar= 1722.7 / 4860.3999 = 0.3545 MPaCheck of Outside Ribs:Inertia of Saddle, Outer Ribs - Longitudinal DirectionY A AY Ay^(2) Io Rib 103.3 2655.6 274323.5 6317474.0 12698247.0 Web 6.0 2669.2 16015.2 6285286.5 64060.7 Values 54.5 5324.8 290338.7 12602760.0 12762308.0Bending Moment [Rm]:= Fl /( 2 * Bplen ) * e * rl / 2= 0.0 /( 2 * 1360.00 ) * 444.867 * 700.50 / 2= 0.000 N-mmKL/R < Cc ( 9.6967 < 145.6069 ) per AISC E2-1Sca = (1-(Klr)^(2)/(2*Cc^(2)))*Fy/(5/3+3*(Klr)/(8*Cc)-(Klr^(3))/(8*Cc^(3))Sca = ( 1-( 9.70 )^(2)/(2 * 145.61^(2) )) * 186 /( 5/3+3*(9.70 )/(8* 145.61 )-( 9.70^(3))/(8*145.61^(3))Sca = 109.81 MPaAISC Unity Check on Outside Ribs ( must be <= 1.0 )Check = Sc/Sca + (Rm/Z)/SbaCheck = 0.35 / 109.81 + (0.00 /181083.016) / 124.11Check = 0.00Check of Inside RibsInertia of Saddle, Inner Ribs - Axial DirectionY A AY Ay^(2) Io Rib 103.3 2191.2 226350.9 10427632.0 7369330.0 Web 6.0 5338.4 32030.4 4280125.0 64060.7 Values 34.3 7529.6 258381.3 14707757.0 7433390.5KL/R < Cc ( 4.6595 < 145.6069 ) per AISC E2-1Sca = (1-(Klr)^(2)/(2*Cc^(2)))*Fy/(5/3+3*(Klr)/(8*Cc)-(Klr^(3))/(8*Cc^(3))Sca = ( 1-( 4.66 )^(2)/(2 * 145.61^(2) )) * 186 /( 5/3+3*(4.66 )/(8* 145.61 )-( 4.66^(3))/(8*145.61^(3))Sca = 110.84 MPaAISC Unity Check on Inside Ribs ( must be <= 1.0 )Check = Sc/Sca + (Rm/Z)/SbaCheck = 0.46 / 110.84 + ( 0.00 /227555.484) / 124.11Check = 0.00Input Data for Base Plate Bolting Calculations:Total Number of Bolts per BasePlate Nbolts 2Total Number of Bolts in Tension/Baseplate Nbt 1Bolt Material Specification SA-193 B7Bolt Allowable Stress Stba 172.00 MPa Bolt Corrosion Allowance Bca 0.0000 mm Distance from Bolts to Edge Edgedis 80.0100 mm Nominal Bolt Diameter Bnd 20.0000 mm Thread Series Series TEMA Metric BasePlate Allowable Stress S 98.60 MPa Area Available in a Single Bolt BltArea 217.0510 mm^2 Saddle Load QO (Weight) QO 12554.1 N Saddle Load QL (Wind/Seismic contribution) QL 0.0 N Maximum Transverse Force Ft 0.0 N Maximum Longitudinal Force Fl 0.0 N Saddle Bolted to Steel Foundation YesBolt Area Calculation per Dennis R. MossBolt Area Requirement Due to Longitudinal Load [Bltarearl]:= 0.0 (QO > QL --> No Uplift in Longitudinal direction)Bolt Area due to Shear Load [Bltarears]:= Fl / (Stba * Nbolts)= 0.00 / (172.00 * 2.00 )= 0.0000 mm^2Bolt Area due to Transverse LoadMoment on Baseplate Due to Transverse Load [Rmom]:= B * Ft + Sum of X Moments= 1183.00 * 0.00 + 0.00= 0.00 N-mmEccentricity (e):= Rmom / QO= 0.00 / 12554.07= 0.00 mm < Bplen/6 --> No Uplift in Transverse directionBolt Area due to Transverse Load [Bltareart]:= 0 (No Uplift)Required of a Single Bolt [Bltarear]= max[Bltarearl, Bltarears, Bltareart]= max[0.0000 , 0.0000 , 0.0000 ]= 0.0000 mm^2ASME Horizontal Vessel Analysis: Stresses for the Right SaddleInput and Calculated Values:Vessel Mean Radius Rm 929.00 mm Tangent to Tangent Length L 4200.00 mm Distance from Saddle to Vessel tangent a 525.00 mmSaddle Width b 272.00 mm Saddle Bearing Angle theta 120.00 degreesWear Plate Width b1 460.00 mm Wear Plate Bearing Angle theta1 132.00 degrees Wear Plate Thickness tr 8.0 mmWear Plate Allowable Stress Sr 137.90 MPaInside Depth of Head h2 462.50 mmShell Allowable Stress used in Calculation 137.90 MPa Head Allowable Stress used in Calculation 137.90 MPaSaddle Force Q, Operating Case 10720.04 NHorizontal Vessel Analysis Results: Actual Allowable-------------------------------------------------------------------Long. Stress at Top of Midspan 23.00 137.90 MPa Long. Stress at Bottom of Midspan 23.45 137.90 MPa Long. Stress at Top of Saddles 23.49 137.90 MPa Long. Stress at Bottom of Saddles 23.08 137.90 MPaTangential Shear in Shell 1.10 82.74 MPa Circ. Stress at Horn of Saddle 2.45 172.37 MPa Circ. Compressive Stress in Shell 0.25 137.90 MPaLoad Combination Results for Q + Wind or Seismic [Q]:= Saddle Load + Max( Fwl, Fwt, Fsl, Fst )= 10720 + Max( 0 , 0 , 0 , 0 )= 10720.0 NSummary of Loads at the base of this Saddle:Vertical Load (including saddle weight) 12740.95 N Transverse Shear Load Saddle 0.00 N Longitudinal Shear Load Saddle 0.00 NFormulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plates are assumed to be Welded k = 0.1The Computed K values from Table 4.15.1:K1 = 0.1066 K2 = 1.1707 K3 = 0.8799 K4 = 0.4011 K5 = 0.7603 K6 = 0.0529 K7 = 0.0184 K8 = 0.3405 K9 = 0.2711 K10 = 0.0581 K1* = 0.1923 K6p = 0.0434 K7P = 0.0151The suffix 'p' denotes the values for a wear plate if it exists.Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]:= -Q*a [1 - (1- a/L + (R^(2)-h2^(2))/(2a*L))/(1+(4h2)/3L)]= -10720*525.00[1-(1-525.00/4200.00+(929.000^(2)-462.500^(2))/(2*525.00*4200.00))/(1+(4*462.50)/(3*4200.00))]= -611863.6 N-mmMoment per Equation 4.15.4 [M2]:= Q*L/4(1+2(R^(2)-h2^(2))/(L^(2)))/(1+(4h2)/( 3L))-4a/L= 10720*4200/4(1+2(929^(2)-462^(2))/(4200^(2)))/(1+(4*462)/(3*4200))-4*525/4200= 4911287.5 N-mmLongitudinal Stress at Top of Shell (4.15.6) [Sigma1]:= P * Rm/(2t) - M2/(pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) - 4911287 /(pi*929.0^(2)*8.00 )= 23.00 MPaLongitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]:= P * Rm/(2t) + M2/(pi * Rm^(2) * t)= 0.4 * 929.000 /(2 * 8.00 ) + 4911287 /(pi * 929.0^(2) * 8.00 )= 23.45 MPaLongitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]:= P * Rm/(2t) - M1/(K1*pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) - -611863.6 /(0.1066 *pi*929.0^(2)*8.00 )= 23.49 MPaLongitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]: = P * Rm/(2t) + M1/(K1*pi*Rm^(2)t)= 0.4 * 929.000 /(2*8.00 ) + -611863.6 /(0.1066 *pi*929.0^(2)*8.00 )= 23.08 MPaMaximum Shear Force in the Saddle (4.15.5) [T]:= Q(L-2a)/(L+(4*h2/3))= 10720 ( 4200.00 - 2 * 525.00 )/(4200.00 + ( 4 * 462.50 /3))= 7010.7 NShear Stress in the shell no rings, not stiffened (4.15.14) [tau2]:= K2 * T / ( Rm * t )= 1.1707 * 7010.68 / ( 929.0001 * 8.0000 )= 1.10 MPaDecay Length (4.15.22) [x1,x2]:= 0.78 * sqrt( Rm * t )= 0.78 * sqrt( 929.000 * 8.000 )= 67.243 mmCircumferential Stress in shell, no rings (4.15.23) [sigma6]:= -K5 * Q * k / ( t * ( b + X1 + X2 ) )= -0.7603 * 10720 * 0.1 / ( 8.000 * ( 272.00 + 67.24 + 67.24 ) )= -0.25 MPaEffective reinforcing plate width (4.15.1) [B1]:= min( b + 1.56 * sqrt( Rm * t ), 2a )= min( 272.00 + 1.56 * sqrt( 929.000 * 8.000 ), 2 * 525.000 )= 406.49 mmWear Plate/Shell Stress ratio (4.15.29) [eta]:= min( Sr/S, 1 )= min( 137.900 / 137.900 , 1 )= 1.0000Circumferential Stress at wear plate (4.15.26) [sigma6,r]:= -K5 * Q * k / ( B1( t + eta * tr ) )= -0.7603 * 10720 * 0.1 / ( 406.486 ( 8.000 + 1.000 * 8.000 ) )= -0.13 MPaCirc. Comp. Stress at Horn of Saddle, L<8Rm (4.15.28) [sigma7,r*]:= -Q/(4(t+eta*tr)b1) - 12*K7*Q*Rm/(L(t+eta*tr)^(2))= -10720 /(4(8.000 + 1.000 * 8.000 )406.486 ) -12*0.018*10720*929.000/(4200.00(8.000+1.000*8.000)^(2))= -2.45 MPaResults for Vessel Ribs, Web and BaseBaseplate Length Bplen 1360.0000 mm Baseplate Thickness Bpthk 16.0000 mm Baseplate Width Bpwid 220.0000 mm Number of Ribs ( inc. outside ribs ) Nribs 4Rib Thickness Ribtk 12.0000 mm Web Thickness Webtk 12.0000 mm Web Location Webloc SideMoment of Inertia of Saddle - Lateral DirectionY A AY Ay^(2) Io Shell 4. 4754. 19014. 76057. 25352. Wearplate 12. 3680. 44160. 529920. 19627.。