最新同济大学 材料力学 习题解答2(练习册P65-P70)
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同济大学 材料力学 习 题解答2(练习册P65-
P70)
P65 36-1
F
AB
F
FN = 5 kN
s = FN = 25 MPa
A
s = Ee1
s E = e1 = 208 GPa
n =│ee2│1
=4 15
= 0.267
P65 36-2
FN
SMC = 0 FN = 2F
A
s = FN = 8F
FN2 = 60 kN
P66 36-4 受力图! FN3
FN2
SFx = 0
2
3
SFy = 0
FN1
1 30º Dl1
a F Dl2
位移分析! Dl3
代入：
3 2
Dl1 +
1 2
Dl3 = Dl2
∴ 3FN1 + FN3 = 4FN2
联立 求解
FN1 = – 0.122 F FN2 = 0.141 F
]
满足强度条件
WP
0.4kN·m |Mn|max = 1.0 kN·m
六要素
Mn图
0.6kN·m |Mn|max = 0.6 kN·m
P69 38-2 T0
tl Mn
t
x
t
530.5N·m
Nk = 53 kW
T0 = 9549
Nk n
= 530.5 N·m
m = 0 tl = T0
t = 132.6 N·m/m
m = 0 Mn(x) = - tx |Mn|max = tl = 530.5 N·m
F
剪切面
F
FC FS
a
挤压面
ll
剪切 FS = F AS = bl
挤压 FC = F AC = ba
sC =
FC AC
=
FC ba
≤ [sC ]
t = FS = FS ≤ [t ]
AS bl ∴ l ≥ 20 cm
∴ a ≥ 2 cm
取 l = 20 cm
取 a = 2 cm
结论
P68 37-4 双剪切 FC 铆钉均匀受力 计算内容？
= =
F
3 4
A1 F
=( b–d )t
A2=( b–2d F ≤ 224 kN
s1
)t
=
FN1 A1
s2 =
则
≤[s ] F≤180kN FN2 ≤[s ]
A2 [ F ]= 62.83 kN
P69 38-1
T4
T3
A
B
T2
T1
C
D
1.0kN·m
0.2kN·m 基线
0.2kN·m
0.4kN·m
Mn图
①
②
l
Dl1
=
FN1l EA
FAx A
a
FAy
B
Dl1 a
C
D
Dl2 a F
∵ FN2 ＞ FN1 且 A 相同
∴ s2 ＞ s1
s2 =
FN2 A
≤ [s ]
∴ A ≥ 6 cm2
Dl2
=
FN2l EA
代入 Dl2 = 2Dl1
∴ FN2 = 2FN1 补充方程
联立 求解
FN1 = 30 kN
取 A = 6 cm2 结论!
FS = 4FS AS pd2
= 132 MPa
挤压强度：
＜ [t ]
sC =
FC = FC = 176 MPa AC dt
＜ [sC ]
钢板 FN1 = F = 2FN2 = 60 kN A相同
1-1截面：smax =
FN1 A1
=
FN1 (b–d)t
= 140 MPa ＜ [s ]
满足强度要求
P67 37-2
= 40.75 MPa
P70 38-4
TA
TB
TC
TD
0.5kN·m 0.25kN·m
|Mn|max = 0.5 kN·m
Mn图 0.5kN·m
WP =
tmax =
pd3 = 24.54×10-6 m3 16 |Mn|max = 20.37 MPa
|tmax [t
≈ [t ]
[t ]| ×100％＜ 5％
Mn图
P70 38-3 T0
T0
A
符号 Mn = T0 = 1 kN·m 单位
公式
IP =
pd4 32
= 61.36×10-8 m4（整个Biblioteka Baidu面）
tr
=
Mn·r
IP
=
20.37
MPa
tmax = 2tr = 40.74 MPa
或
WP =
pd3 16
= 24.54×10-6 m3
tmax =
Mn WP
FC1 FC1 ∵ FC = 2FC1
t ＜ 2t1
d
3
3 挤压
∴剪切校核主板
t1 t
2 1
b
2
1
FC
FC =
F 4
AC = dt
FS =
FC 2
=
F 8
AS =
pd2 4
F
F
FsC =
FC AC
≤
[sC
]
t = FS ≤ [t ]
AS
主板： 拉压
F ≤ 100 kN
F ≤ 62.83 kN
1-1截面：FN1 2-2截面：FN2
FN3 = 0.930 F
3 FN1 + 2 FN2 = 0
1 2 FN2 + FN3 = F
Dl1
=
FN1a EA
Dl2 =
FN2
2 3
EA
a
Dl3 =
FN3
1 3
EA
a
P67 37-1 单剪切
F
F
12
F FC FC
拉压强度： 1 2
F
FC FS
FS = FC
=
F 2
= 30 kN
剪切强度：
F
t=
A pd2
a
C
FCx
a
FCy
B
D
a
F
(2) 令 s = sb
F=
pd2 8
sb
= 62.8 kN
(1) 令 s = sS
F=
pd2 8
sS
=
37.7
kN
若 s = sP ( ≈ se )
F=
pd2 8
sP
= 31.4 kN
P66 36-3
FN1
FN2 SMA = 0 FN1 + 2FN2 = 3F
P70)
P65 36-1
F
AB
F
FN = 5 kN
s = FN = 25 MPa
A
s = Ee1
s E = e1 = 208 GPa
n =│ee2│1
=4 15
= 0.267
P65 36-2
FN
SMC = 0 FN = 2F
A
s = FN = 8F
FN2 = 60 kN
P66 36-4 受力图! FN3
FN2
SFx = 0
2
3
SFy = 0
FN1
1 30º Dl1
a F Dl2
位移分析! Dl3
代入：
3 2
Dl1 +
1 2
Dl3 = Dl2
∴ 3FN1 + FN3 = 4FN2
联立 求解
FN1 = – 0.122 F FN2 = 0.141 F
]
满足强度条件
WP
0.4kN·m |Mn|max = 1.0 kN·m
六要素
Mn图
0.6kN·m |Mn|max = 0.6 kN·m
P69 38-2 T0
tl Mn
t
x
t
530.5N·m
Nk = 53 kW
T0 = 9549
Nk n
= 530.5 N·m
m = 0 tl = T0
t = 132.6 N·m/m
m = 0 Mn(x) = - tx |Mn|max = tl = 530.5 N·m
F
剪切面
F
FC FS
a
挤压面
ll
剪切 FS = F AS = bl
挤压 FC = F AC = ba
sC =
FC AC
=
FC ba
≤ [sC ]
t = FS = FS ≤ [t ]
AS bl ∴ l ≥ 20 cm
∴ a ≥ 2 cm
取 l = 20 cm
取 a = 2 cm
结论
P68 37-4 双剪切 FC 铆钉均匀受力 计算内容？
= =
F
3 4
A1 F
=( b–d )t
A2=( b–2d F ≤ 224 kN
s1
)t
=
FN1 A1
s2 =
则
≤[s ] F≤180kN FN2 ≤[s ]
A2 [ F ]= 62.83 kN
P69 38-1
T4
T3
A
B
T2
T1
C
D
1.0kN·m
0.2kN·m 基线
0.2kN·m
0.4kN·m
Mn图
①
②
l
Dl1
=
FN1l EA
FAx A
a
FAy
B
Dl1 a
C
D
Dl2 a F
∵ FN2 ＞ FN1 且 A 相同
∴ s2 ＞ s1
s2 =
FN2 A
≤ [s ]
∴ A ≥ 6 cm2
Dl2
=
FN2l EA
代入 Dl2 = 2Dl1
∴ FN2 = 2FN1 补充方程
联立 求解
FN1 = 30 kN
取 A = 6 cm2 结论!
FS = 4FS AS pd2
= 132 MPa
挤压强度：
＜ [t ]
sC =
FC = FC = 176 MPa AC dt
＜ [sC ]
钢板 FN1 = F = 2FN2 = 60 kN A相同
1-1截面：smax =
FN1 A1
=
FN1 (b–d)t
= 140 MPa ＜ [s ]
满足强度要求
P67 37-2
= 40.75 MPa
P70 38-4
TA
TB
TC
TD
0.5kN·m 0.25kN·m
|Mn|max = 0.5 kN·m
Mn图 0.5kN·m
WP =
tmax =
pd3 = 24.54×10-6 m3 16 |Mn|max = 20.37 MPa
|tmax [t
≈ [t ]
[t ]| ×100％＜ 5％
Mn图
P70 38-3 T0
T0
A
符号 Mn = T0 = 1 kN·m 单位
公式
IP =
pd4 32
= 61.36×10-8 m4（整个Biblioteka Baidu面）
tr
=
Mn·r
IP
=
20.37
MPa
tmax = 2tr = 40.74 MPa
或
WP =
pd3 16
= 24.54×10-6 m3
tmax =
Mn WP
FC1 FC1 ∵ FC = 2FC1
t ＜ 2t1
d
3
3 挤压
∴剪切校核主板
t1 t
2 1
b
2
1
FC
FC =
F 4
AC = dt
FS =
FC 2
=
F 8
AS =
pd2 4
F
F
FsC =
FC AC
≤
[sC
]
t = FS ≤ [t ]
AS
主板： 拉压
F ≤ 100 kN
F ≤ 62.83 kN
1-1截面：FN1 2-2截面：FN2
FN3 = 0.930 F
3 FN1 + 2 FN2 = 0
1 2 FN2 + FN3 = F
Dl1
=
FN1a EA
Dl2 =
FN2
2 3
EA
a
Dl3 =
FN3
1 3
EA
a
P67 37-1 单剪切
F
F
12
F FC FC
拉压强度： 1 2
F
FC FS
FS = FC
=
F 2
= 30 kN
剪切强度：
F
t=
A pd2
a
C
FCx
a
FCy
B
D
a
F
(2) 令 s = sb
F=
pd2 8
sb
= 62.8 kN
(1) 令 s = sS
F=
pd2 8
sS
=
37.7
kN
若 s = sP ( ≈ se )
F=
pd2 8
sP
= 31.4 kN
P66 36-3
FN1
FN2 SMA = 0 FN1 + 2FN2 = 3F