同济大学材料力学习题解答6(练习册P86-P94)

50 MPa
sx = 30 MPa sy = 50 MPa
20 MPa
52.32
30 MPa 18.66
tx = - 20 MPa
a = 30º
30º
sa =
sx + sy
2
+
sx - sy
2
cos2a –txsin2a
= 52.32 MPa
ta =
sx - sy
2
sin2a +txcos2a
=
Ma 2EI
(2l
-
a)
P87 46-4 q
mA = 0
FB =
9 4
qa
Fy = 0
FA =
3 4
qa
A xD
FAa
B a
C弯矩方程 AB FBaM1(x)= 43qax- 21qx2
BC q M2(x)= - 2(3a-x)2
y边条1(E界件xEqI)yx1I=xE(1d=d=x=I2qy2x)4201Ad=ad1qE4,=,2=xyy2yI(q261141xq16qx==EE=4a4-I03-qI0(123：2x4：3a43x挠qDCxq-3qx311B-a曲2+==839=-x4qqa03线0a463a+ax3x23qCx近+a2)连条1y4xx似+Da续件+yC=3微2D)11(xxqx12分a==)EE=4Eq22I方2aaII(2y,,Eddx程42qyyqx)I=E222===2dId=[q4q2y6x(-y(B31E2q23=a6q=Ia=[q-0ya(-xB2q3C：3x)=a-=4D)(0+(4-3：8324q+xa=EaaCa)C-3--34I2xx2=+83xx-))C162q+93a2qa]D4a42]3
连 续
x = a,
q2= q1= qC
=
Ma EI
C2= Ma
条 件
x= 0, y1= 0：D1= 0
条 件
x = a,
y2=
y1
=
yC
=
Ma2 2EI
D2
=
-
1 2
Ma2
q1(x)=
Mx EI
y1(x)=
Mx2 2EI
q2(x)=
Ma EI
y2(x)=
2MEaI(2x - a)
qB
=
Ma EI
yB
P86 46-2(b)
弯矩方程
x
M
AC M1(x)= -M BC M2(x)= 0
挠曲线近似微分方程
A
C
a
l
B
AC
EI
d2y1 dx2
=
M
BC
EI
d2y2 dx2
=
0
积分
EI
dy1 dx
= Mx
+ C1
EI
dy2 dx
=
C2
EIy1=
1 2
Mx2 + C1x +D1
EIy2= C2x+D2
边 界
x= 0, q1= 0：C1= 0
qCF = qBF
Fa2 qa3 = 2EI = 2EI
P224 表19-1 序号5
yCF = yBF + qBF·a
=
Fa3 3EI
+ qBF·a
=
5qa4 6EI
qCq = qBq
qa3 =
6EI
yCq = yBq +qBq·a
=
qa4
8EI +qBq·a
=
7qa4 24EI
P88 47-2(a) F
习题解答(六)
挠转AByqP(C(处处8xx曲角6))F挠转1==线方M4-C6度角x6方程6-ylE3M1EMm程IIaxlxl[[积xlA挠(3==x(分曲0x-l2l.法5：-l线：)ll3)近-q2y坐-BAl0F似2Bl标.==x20B]6+微-12弯Ml65边63分EM]lE矩m2I界方lI积方误=条程差分程0x件2=讨.6EM(Exx％E论1IFII==(-qyBxd：d0l)x2===,,y12=3令6M2MMyyl=)lll==M((ql=Mx(xl00l=y--F：(：-mxll0DCxCa))：-x)230==+=l+.0)C-C61694xMM61135+Mll2E2DlI
P88 47-1(b)
q
A a
M(qa2) F(qa)
B
C
百度文库
a
qC
=
qCM
+
qCF
+
qCq
=-
4qa3 3EI
yC
=
yCM
+
yCF +
yCq
=-
7qa4 8EI
P224 表19-1 序号1
qCM =-
M(2a) EI
=-
2qa3 EI
yCM = -
M(2a)2 2EI
=-
2qa4 EI
P224 表19-1 序号3
Fl3
yC = yCM + yCF
= 48EI
P89 47-3(c)
q
F(qa)
A
B C
a
a
P225 表19-1 序号11
qCq
=
qBq
=
-
qa3 24EI
P226 表19-1 序号12
qCF =
5qa3 6EI
qC
=
qCq
+
qCF
=
19qa3 24EI
yC =
yCq +
yCF
=
5qa4 8EI
yCq
=qBq·a
=-
qa4 24EI
2qa4 yCF = 3EI
P90 48-1 q
A
B
l b
1 ql2
2 smax =
M│max Wz
≤ [s ]
h
M│max
=
1 2
ql
2
= 45 kN·m
Wz =
bh2 = h3 6 12
h ≥ 3 12 M│max = 16.5 cm
[s ]
M图
查表：
ymax
=
P92 49-1 危险截面(内力分布) 危险点(应力分布) F
B
A
2T0 3T0
T0
C T0
tB
T0
F
F
tmax =
Mn WP
Mn = 2T0 pd3
WP = 16
t
C
A
s
s = FN
A
FN = F
横截面位置 应力方向
s= F
A
tmax =
Mn WP
s
Mn = T0 pd3
WP = 16
P92 49-2(a) 应力表示(方向 数值)
M(Fl) F
A
C
B
0.5l
0.5l
P225 表19-1 序号8
qAM
=-
Ml 24EI
=-
Fl2 24EI
= qBM
yCM =0
P225 表19-1 序号9
qAF =
Fl2 16EI
=- qBF
Fl3 yCF = 48EI
qA = qAM + qAF
Fl2 =
48EI
qB = qBM + qBF
=- 5Fl2 48EI
ql4 8EI
其中：I =
bh3 = h4 12 24
ymax l
=
ql3 8EI
≤[
f]
l
h≥
4 3ql3
E[
f l
]
= 17.8 cm
取 h = 17.8 cm b = 8.9 cm
P91 48-3
q
静定基 多余约束力： FC
A
C
FA l FC l
3 8
ql
5 8
ql
B
FB 变形协几调何条方件程： yCq=+0yCF = 0
查表： yCq =
5q(2l)4 384EI
yCF
=-
FC(2l)3 48EI
FS 图 补充方程： 5q(2l)4 - FC(2l)3 = 0
384EI 48EI
5 ql 8
83（ql 简便方法）
FC =
5 ql 4
1 ql2 8
9 ql2 1 ql2
128
16
M图
平衡： FA = FB =
3 ql 8