混凝土结构设计原理 第四版 程文

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= 2 = 0.45 ⋅ = 0.0021 > 0.002
= = 0.218 2 = 0.45 ⋅ = 0.0021 > 0.002 第3章习题
3.1
解： f c = 14.3 N mm 2 ； f t = 1.43 N mm 2 ； f y = 300 N mm 2 ；
C30 混凝土，一类环境，c =25mm ，a s =35mm ，h 0=500-35=465mm
〈 s = M 90 ⋅106
〈1 f c bh 0 1.0 ⋅14.3⋅ 250 ⋅ 4652
= 0.116 查表得： ⎩ = 0.123 < ⎩b = 0.55
A s =
〈1 f c b ⎩ h 0 f y
= 1.0 ⋅14.3 ⋅ 250 ⋅ 0.123 ⋅ 465 300
= 681.6mm 2
选 2 16+2 14 钢筋， A s = 710mm 2 (满足构造要求
)。

〉min = 0.45
f t
f y
1.43 300 〉min bh = 0.0021⋅ 250 ⋅ 500 = 26
2.5mm
2 < A s = 710mm 2 取 A s = 710mm 2
3.2 解： f c = 19.1 N mm 2 ； f t = 1.71 N mm 2 ； f y = 360 N mm 2
C40 混凝土，二类 a 环境，c =30mm ，a s =40mm ，h 0=450-40=410mm
〈 s =
M 140 ⋅106
〈1 f c bh 0 1⋅19.1⋅ 200 ⋅ 4102
查表得： ⎩ = 0.249 < ⎩b = 0.518
A s =
〈1 f c b ⎩ h 0
f y = 1.0 ⋅19.1⋅ 200 ⋅ 0.249 ⋅ 410 360
= 1083.3mm 2 选 3 22 钢筋， A s = 1140mm 2 (满足构造要求)。

min = 0.45
f t f y
1.71 360
〉min bh = 0.0021⋅ 200 ⋅ 450 = 189mm 2 < A s = 1140mm 2
取 A s = 1140mm 2 3.4 解： f c = 14.3 N mm 2 ； f t = 1.43 N mm 2 ； f y = 360 N mm 2 ；
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C30 混凝土，一类环境，c=25mm，a s=25+16/2=33mm，h0=450-33=417mm
x =
f y A s
〈1 f c b
=
360 ⋅ 804
1.0 ⋅14.3⋅ 200
= 101.2mm < ⎩b h0 = 0.518 ⋅ 417 = 216mm
〉min = 0.45 f t
f y
1.43
360
〉min bh = 0.002 ⋅ 200 ⋅ 450 = 180mm2 < A s = 804mm2
x101.2
2 2
安全
) = 106kN ⊕ m > M = 84kN ⊕ m
3.5 解：f c = 11.9 N mm2；f y = f y = 300 N mm2；C25 混凝土，一类环境，c=25mm，设受拉钢筋放两层，a s=60mm，h0=h-a s=500-60=440mm
A s2 = M 〈1 f c bh0 ⎩b (1 0.5⎩b )
f y2(h0 a s2 )
=
260 ⋅106 1⋅11.9 ⋅ 200 ⋅ 4402 ⋅ 0.55⋅ (1 0.5⋅ 0.55)
300 ⋅ (440 35)
= 628mm2 A s =
〈1 f c b⎩b h0 + f y A2
f y
=
1⋅11.9 ⋅ 200 ⋅ 0.55 ⋅ 440 + 300 ⋅ 628
300
= 2548mm2
A s选3 25+3 22，A s＝2613mm2；A s2选2 20，A s2＝628mm2
3.6 解：1）f c = 19.1 N mm2；f t = 1.71 N mm2；f y = 360 N mm2，C40 混凝土，二类a 环境，c=30mm，设受拉钢筋放两层，a s=65mm，h0=750-65=685mm
〈1 f c b2 h2 (h0 ) = 1.0 ⋅19.1⋅ 550 ⋅100 ⋅ (685
2 2
)
= 667.1⋅106 N ⊕ mm > 500 ⋅106 N ⊕ mm
属于第一类T 形截面
〈 s =
M500 ⋅106
〈1 f c b2 h01⋅19.1⋅ 550 ⋅ 6852
查表得 ⎩ = 0.107 < ⎩b = 0.518（可不验算）
A s = 〈1 f c b2 ⎩ h0
f y
=
1.0 ⋅19.1⋅ 550 ⋅ 0.107 ⋅ 685
360
= 2139mm2
思路岛教育网整理提供选 6 22 钢筋，A s = 2281mm2 (满足构造要求)。

〉 min = 0.45 f t
f y 1.71 360
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〉min bh = 0.0021⋅ 250 ⋅ 750 = 393.75mm2 < A s = 2281mm2
取A s = 2281mm2
2）f c = 27.5 N mm2；f t = 2.04 N mm2；f y = 360 N mm2，C60 混凝土，二类a 环境，c=30mm，设受拉钢筋放两层，a s=65mm，h0=750-65=685mm， 〈1 = 0.98
〈1 f c b2 h2 (h0 ) = 0.98 ⋅ 27.5 ⋅ 550 ⋅100 ⋅ (685
2 2
)
= 941.2 ⋅106 N ⊕ mm > 500 ⋅106 N ⊕ mm
属于第一类T 形截面
〈 s =
M500 ⋅106
〈1 f c b2 h00.98 ⋅ 27.5 ⋅ 550 ⋅ 6852
= 0.072
查表得 ⎩ = 0.075 < ⎩b = 0.499 （可不验算）
A s =
〈1 f c b2 ⎩ h0
f y
0.98 ⋅ 27.5 ⋅ 550 ⋅ 0.075 ⋅ 685
360
= 2115mm2 选8 18 钢筋，A s = 2036mm2 (满足构造要求)。

〉 min = 0.45 f t
f y
2.04
360
〉min bh = 0.00255 ⋅ 250 ⋅ 750 = 478.125mm2 < A s = 2036mm2
取A s = 2036mm2
3.7 解：f c = 1
4.3 N mm2；f t = 1.43 N mm2；f y = 360 N mm2；C30 混凝土，一类环境，c=25mm，设受拉钢筋放两层，a s=60mm，h0=h-a s=500-60=440mm
〈1 f c b2 h2 (h0 ) = 1.0 ⋅14.3 ⋅ 400 ⋅ 80 ⋅ (440
2 2
)
= 183.04 ⋅106 N ⊕ mm < 300 ⋅106 N ⊕ mm
属于第二类T 形截面
M u1 = 〈1 f c (b2 b)h2 (h0
2 ) = 1.0⋅14.3⋅ (400 200) ⋅80⋅ (440
80
2
) = 91.52⋅106 N ⊕ mm
M u 2 = M M u1 = 300 ⋅106 91.52 ⋅106 =208.48×106N﹒mm
= =
0.377 2
〈 s =
M u 2 208.48 ⋅106
〈1 f c bh 0 1⋅14.3 ⋅ 200 ⋅ 4402
查表得 ⎩ = 0.504 < ⎩b = 0.518
A s 2 =
〈1 f c b ⎩ h 0
f y
=
1.0 ⋅14.3 ⋅ 200 ⋅ 0.504 ⋅ 440 360 = 1762mm 2
A s 1 =
〈1 f c (b 2f b )h 2f
f y
=
1⋅14.3 ⋅ (400 200) ⋅ 80 360
= 636mm 2
A s = A s 1 + A s 2 = 1762 + 636 = 2398mm 2
选 5 25 钢筋， A s = 2454mm 2 (满足构造要求)。

斜截面受剪承载力计算例题
1 1
2 2
2）复合截面尺寸
h w=h0=h-c-25/2=500-30-12.5=457.5 h w
b
457.5
200
= 2.3 < 4
0.25® c c f bh0 = 0.25 ⋅1.0 ⋅ 9.6 ⋅ 200 ⋅ 457.5 = 219.6kN > V = 124.6kN
满足。

3）验算是否按计算配置腹筋
0.7 f t bh0 = 0.7 ⋅1.1⋅ 200 ⋅ 457.5 = 70.5kN < V = 124.6kN
应按计算配置腹筋
4）计算腹筋数量
①只配箍筋
由V δ 0.7 f t bh0 + 1.25 f yv A sv
s
h0得：
ε = = 0.451 mm2/mm s 1.25 f yv h0 1.25 ⋅ 210 ⋅ 457.5
选双肢 ⎫ 8 箍筋s δ nA sv12 ⋅ 50.3 0.451 0.451
取s=200mm 验算最小配箍率
〉 sv = nA sv1
bs
2 ⋅ 50.3f 1.1 200 ⋅ 200f yv210
满足
仅配箍筋时的用量为双肢 ⎫ 8@200
②即配箍筋又配弯筋
a. 先选弯筋，再算箍筋
根据已配的 2 25+1 22 纵向钢筋，将 1 22 的纵筋以45°角弯起，则弯筋承担的剪力：
V sb = 0.8 f y A sb sin 〈 s = 0.8 ⋅ 380.1⋅ 300 ⋅
2
= 64.5kN
nA sv1
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ε = =负值
1.25 f yv h0 1.25 ⋅ 210 ⋅ 457.5
按构造要求配置箍筋并满足最小配箍率要求
思路岛教育网整理提供选双肢⎫6@200 的箍筋，
〉 sv = nA sv1
bs
2 ⋅ 28.3f 1.1 200 ⋅ 200f yv210
b. 先选箍筋，再算弯筋
先按构造要求选双肢⎫6@200 的箍筋，
〉 sv = nA sv1
bs
2 ⋅ 28.3f 1.1 200 ⋅ 200f yv210
满足要求。

A sb ε V 0.7 f t bh0 1.25 f yv
0.8 f y sin 〈 s
A sv
s
h0
=
2
2
2 ⋅ 28.3
200 = 118.5mm
2
2
5）验算弯起钢筋弯起点处的斜截面抗剪承载力
弯起钢筋弯起点距支座边缘500-30-22/2-30-20/2+50=469mm，该处剪力
V1=124.6-0.469×70=91.77kN
V cs = 0.7 f t bh0 + 1.25 f yv A sv
s
2 ⋅ 28.3
200
3
故不需要弯起第二排钢筋或加大箍筋用量。

4－2 解：1）剪力图见书
2）复合截面尺寸
h w=h0=h-c-20/2=600-30-10=560 h w
b
560
200
= 2.8 < 4
0.25®c f c bh0 = 0.25 ⋅1.0 ⋅ 9.6 ⋅ 200 ⋅ 560 = 268.8kN > V max = V A = 180kN
满足。

3）验算是否按计算配置腹筋
A支座：V集
V总
=
160
180
V 140
V总160
梁的左右区段均应按集中荷载作用下的独立梁计算
将梁分为AC、CD、DE、EB段来计算斜截面受剪承载力AC段： ⎣ =
a h0
= = 0.00142 > 〉 sv ,min = 0.24 t = 0.24 ⋅ = 0.0013
= = 0.00142 > 〉 sv ,min = 0.24 t = 0.24 ⋅ = 0.0013
124.6 ⋅103 70.5 ⋅103 1.25 ⋅ 210 ⋅ ⋅
457.5 弯起 1 22 的纵筋，A sb =380.1mm
h 0 = 70.5 ⋅103 + 1.25 ⋅ 210 ⋅ ⋅ 457.5 = 104.5 ⋅103 N
>91.77×10 N
= 88% ；B 支座： 集 = = 87.5%
= = 1.79
f t bh 0 = ⋅1.1⋅ 200 ⋅ 560 = 77.3kN <V A =180kN
1000 560
1.75 1.75 ⎣ + 1.0 1.79 + 1
应按计算配置腹筋
CD段： ⎣ = a
h0
2000
560
1.75 1.75
⎣ + 1.0 3 + 1
按构造要求配置箍筋，选用 ⎫ 6@350 的箍筋。

DE段： ⎣ = a
h0
2000
560
1.75 1.75
⎣ + 1.0 3 + 1
应按计算配置腹筋
EB段： ⎣ = a
h0
1000
560
1.75 1.75
⎣ + 1.0 1.79 + 1
应按计算配置腹筋4）计算腹筋数量
AC段：nA sv1
s ε
V A
1.75
f yv h0210 ⋅ 560
选双肢 ⎫ 8 箍筋s δ nA sv12 ⋅ 50.3 0.873 0.873
取s=110mm 验算最小配箍率
〉 sv = nA sv1
bs
2 ⋅50.
3 f
1.1
200 ⋅110 f
210
满足
箍筋用量为双肢 ⎫ 8@110
DE段：nA sv1
s ε
V E
1.75
h
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= = 3.57 > 3 ，取 ⎣ = 3
f t bh 0 = ⋅1.1⋅ 200 ⋅ 560 = 53.9kN >V C =50kN
= = 3.57 > 3 ，取 ⎣ = 3
f t bh 0 = ⋅1.1⋅ 200 ⋅ 560 = 53.9kN <V E =70Kn = = 1.79 f t bh 0 = ⋅1.1⋅ 200 ⋅ 560 = 77.3kN <V B =160Kn
(180 77.3) ⋅103
= = 115.2mm
= = 0.00457 > 〉 sv ,min = 0.24 t = 0.24 ⋅ = 0.00126
(70 53.9) ⋅103
= = 734.3mm
f yv h 0 10 ⋅ 560
选双肢 ⎫ 8 箍筋 s δ
nA sv 1 2 ⋅ 50.3
0.137 0.137 取 s=250mm
验算最小配箍率
〉 sv = nA sv1
bs
2 ⋅50.
3 f
1.1
200 ⋅250 f
210
满足
箍筋用量为双肢 ⎫ 8@250
EB段：nA sv1
s ε
V B
1.75
f yv h0210 ⋅ 560
选双肢 ⎫ 8 箍筋s δ nA sv12 ⋅ 50.3 0.703 0.703
取s=140mm 验算最小配箍率
〉 sv = nA sv1
bs
2 ⋅ 50.3f 1.1 200 ⋅140f yv210
满足
箍筋用量为双肢 ⎫ 8@140
4－3 解：1）剪力图见书
2）复合截面尺寸
h w=h0-h f′=h-60-200=700-60-200=440 h w
b
440
250
= 1.76 < 4
0.25®c f c bh0 = 0.25 ⋅1.0 ⋅14.3⋅ 250 ⋅ 640 = 572kN > V max = V A = 343.75kN
满足。

3）验算是否按计算配置腹筋
AC段： ⎣ = a
h0
1500
640
1.75 1.75
⎣ + 1.0 2.34 + 1
应按计算配置腹筋
CB段： ⎣ = a
h0
2500
640
1.75 1.75
⎣ + 1.0 3 + 1
应按计算配置腹筋
4）计算腹筋数量
= = 0.00201 > 〉 sv,min = 0.24t = 0.24 ⋅ = 0.00126
(160 77.3) ⋅103
= = 143.1mm
= = 0.00359 > 〉 sv,min = 0.24t = 0.24 ⋅ = 0.00126
= = 2.34
f t bh0 = ⋅1.43 ⋅ 250 ⋅ 640 = 119.88kN <V A=343.75kN
= = 3.9 >3，取 ⎣ = 3
f t bh0 = ⋅1.43⋅ 250 ⋅ 640 = 100.10kN <V B=206.25kN
AC段：在上排2 根
25 纵筋中先弯起一根（A sb=490.9mm2）
nA sv1 s
ε
V A
1.75
⎣ + 1.0 t0
f yv h0
=
343.75 ⋅103
1.75
2.34 + 1
⋅1.43⋅ 250 ⋅ 640 0.8 ⋅ 360 ⋅ 490.9 ⋅
300 ⋅ 640
2
2 = 0.645mm2 / mm
选双肢 ⎫ 8 箍筋s δ
nA sv12 ⋅ 50.3
0.645 0.645
取s=150mm
验算最小配箍率
〉 sv = nA sv1
bs
2 ⋅ 50.3f 1.43
250 ⋅150f yv300
由于AC段剪力值均为343.75kN，所以只要配弯起钢筋就必须在此段弯满，弯起两排即可弯满，所以需要再弯起一排（一根25 的）如图。

CB段：取箍筋配置与AC段相同，即配置双肢 ⎫ 8@150 的箍筋
V cs =
1.75
⎣ + 1.0
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