同济大学材料力学习题解答(练习册PP)

20 MPa 30 MPa
sx = 30 MPa
sy = 50 MPa
30º
18.66
tx = - 20 MPa
2
a = 30º
cos2a – tx sin2a
= 52.32 MPa
sa =
ta =
sx + sy sx - sy
2
+
sx - sy
2
sin6 MPa
h
P91 48-3
A FA l 3 ql 8
q
C FC l B FB
静定基
多余约束力： FC
+0yCF = 0 变形协调条件： yCq 变形几何方程 C= FC(2l)3 yCF = 48EI
5 ql 8
5 q ( 2 l) 4 查表： yCq = 384EI
5 ql 8 1 ql2 8 9 ql2 128
a
s1 = s2 = 0 s3 = - 100 MPa s1 = 30 MPa = - s3 s2 = 0 t = 22.5 MPa s = 100 MPa d s = 50 MPa c s1 = 100 MPa s2 = s3 = 0 s1 = 58.6 MPa s2 = 0 sx - sy 2 sx + sy ± ( ) + tx 2 s max = s3 = - 8.6 MPa 2 2 min
P90 48-1
q
A
l B b ≤ [s ]
1 ql2 2
smax =
M│max
Wz
1 2 M│max = ql 2 = 45 kN· m bh2 h3 Wz = = 6 12 3 12 M│max h≥ = 16.5 cm [s ]
ql4 bh3 h4 M 图 查表： ymax = 8EI 其中： I = 12 = 24 4 ymax f ql3 3ql3 = h≥ = 17.8 cm ≤[ ] f l l 8EI E[ ] l 取 h = 17.8 cm b = 8.9 cm
2qa3 M(2a) =qCM = EI EI
P224 表19-1 序号3
qCF = qBF
P224 表19-1 序号5
Fa2 qa3 = = 2EI 2EI
qa3 = 6EI
qCq = qBq
P88 47-2(a)
A
0.5l
F M(Fl) F
C
B
0.5l
Ml Fl2 == qBM yCM = 0 P225 表19-1 序号8 qAM = 24EI 24EI Fl2 Fl3 P225 表19-1 序号9 qAF = = - qBF yCF = 16EI 48EI Fl2 qA = qAM + qAF = Fl3 48EI y = y + y C CM CF = 2 48EI 5Fl qB = qBM + qBF = 48EI
P87 46-4 q
9 3 mA = 0 FB = qa Fy = 0 FA = qa 4 4 C 弯矩方程 AB BC A xD B q 1 3 FB M (x)= qax - qx2 M (x)= - (3a - x)2 FA 2 a a a 1 2 2 4 d2y1 1 2 3 d2y2 q EI 2 = 挠曲线近似微分方程 qx - qax EI 2 = (3a - x)2 4 2 dx dy2 dx q 2 3 dy1 1 EI = qx3 - qax 2 + C1 EI dx = - 6 (3a - x)3 + C2 dx 1 6 3 8 q EIy1= qx4 - qax3 + C1x +D1 EIy2= (3a - x)4 + C2x + D2 24 24 24 1 3 C = qa = 0 ： x =2a, q = q = q D = 0 2 2 1 B 连续 边界 x = 0, y1= 0： 1 6 qa 3 3 条件 x =2a, y2= yB = 0： C1= 条件 x =2a, y1= 0： D2 = - qa4 6 8 q q q1(x)= (4x3 -9ax2+ 4a3) q2(x)= [a3 - (3a - x)3] 6EI q 24EI q y1(x)= (x4-3ax3+4a 3x) y2(x)= [(3a-x)4+4a3x-9a4] 24EI 3 24EI 4 qa qa qa4 qA = qB = 0 yD = yC = 6EI 12EI 8EI
P88 47-1(b) M(qa2) q F(qa)
A a B a C
P224 表19-1 序号1
4qa3 qC = qCM + qCF + qCq = 3EI 7qa4 yC = yCM + yCF + yCq = 8EI
2qa4 M(2a)2 =yCM = EI 2EI yCF = yBF + qBF· a Fa3 5qa4 = + qBF· a= 3EI 6EI yCq = yBq + qBq· a qa4 7qa4 = + qBq· a= 8EI 24EI
- 2t x tan2a0 = sx - sy 57.02 MPa sx - sy 2 sx + sy ± ( ) + tx 2 = s max = - 7.02 MPa 2 2 min s1 = 57.02 MPa s2 = 0 s3 = - 7.02 MPa 57.02 MPa 32.02 MPa sx - sy 2 ) + tx 2 = t max = ± ( 19.33º - 32.02 MPa 2 min 7.02 MPa sx - sy tan2a1 = = 1.25 a1 = 64.33º 2t x a1' = - 25.67º 64.33º sx + sy s a1 = = 25 MPa 25 MPa 32.02 MPa 2
3T0
T0 T0
t
F
B
F
A
Mn Mn = 2T0 tmax = 3 p d WP W = P 16
横截面位置 应力方向
t
C
s
FN s= FN = F A
F s= A Mn tmax = WP
s
Mn = T0 pd3 WP = 16
P92 49-2(a) 应力表示(方向 数值)
50 MPa 52.32
P92 49-2(b)
20 MPa 27.32 40 MPa 27.32
sx = - 40 MPa tx = 20 MPa
2
+
sy = 0 a = 60º
30º
sa = ta =
sx + sy sx - sy
2
sx - sy
2
cos2a – tx sin2a = - 27.32 MPa
sin2a + tx cos2a
= - 27.32 MPa
P93 49-3(a) 解析法 sx = 50 MPa
20 MPa
11.65 54.82
50 MPa
sy = 0 tx = - 20 MPa a = 30º sx + sy sx - sy sa = + cos2a – tx sin2a
2 2 = 54.82 MPa sx - sy ta = sin2a + tx cos2a = 11.65 MPa 2 = 0.8 a0 = 19.33º a0' = 109.33º
2 bh P93 49-4 M = 10 kN· m Wz = = 100 cm3 6 h/ 2 a M M smax = = 100 MPa = sd = - sa = 2sc Wz FS 3 bh b h/4 FS = 120 kN Iz = = 500 cm4 c 12 d Sz*max = 75 cm3 Sz*3 = 56.25 cm3 * * F · S FS· Sz max z S = 22.5 MPa = 30 MPa = tb tc = 3 tmax = Iz· b Iz· b s = 100 MPa t = 30 MPa b
1 ql2 16
3 4 F ( 2 l ) 5 q ( 2 l ) C FS 图 补充方程： =0 48EI 384EI 3 ql 5 8 FC = ql （简便方法） 4 3 平衡： FA = FB = ql 8 M图
P92 49-1 危险截面(内力分布) 危险点(应力分布)
A
B C
F
T0
2T0
30º
P93 49-3(e)sx = - 20 MPa sy = 30 MPa tx = 20 MPa a = - 30º 30 MPa sx + sy sx - sy + cos2a – tx sin2a 20 MPa sa = 2 2 30º 20 MPa = 9.82 MPa s s x y 9.82 ta = sin2a + tx cos2a = 31.65 MPa 31.65 2 - 2t x tan2a0 = = 0.8 a0 = 19.33º a0' = 109.33º sx - sy 37.02 MPa sx - sy 2 sx + sy 2 ± ( ) + tx = s max = - 27.02 MPa 2 2 min s1 = 37.02 MPa s2 = 0 s3 = - 27.02 MPa 27.02 MPa 32.02 MPa smax + smin )= 19.33ºt max = ±( - 32.02 MPa 2 min 37.02 MPa 5 MPa a1 = a0 + 45º= 64.33º a1' = - 25.61º sx + sy 64.33º s a1 = = 5 MPa 2 32.02 MPa