同济大学 材料力学 习题解答1(练习册P61-P64)
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A
B
F2 920 N 500 N
＋
C F3
D
E
F1
F4
F5
640 N
＋ ＋
240 N
＋
FN 图
s=
FN
A
sAB = 1.25 MPa sCD = 1.6 MPa
sBC = 2.3 MPa sDE = 0.6 MPa
与点有关
A = 400 mm2
应力单位 ：MPa
P62 34-3
3 2
1
F1 F2 A
3
F3 D
1
B
2
C
F3 Fx = 0 FN1 FN1 = F3 = 30 kN 40kN FN1 – 2 2 30kN 30kN A1 = 4×10 m s1 = = 0.75 MPa F2 F ＋ ＋ N2＋
FN3
F1 F2
A1 F3 Fx = 0 FN2 = F3 + F2 = 40 kN FN2 FN图 – 2 2 A2 = 2×10 m s2 = = 2 MPa A2 F3 F = 0 FN3 = F3 + F2 – F1 = 30 kN x FN3 – 2 2 A3 = 4×10 m s3 = = 0.75 MPa A3
FN 2
C
SFx = 0
SFy = 0 FN1 = 0.897 F
s1 =
s2 =
FN1
FN2 = 0.732 F
≤ [s ]1
≤ [s ]2
A1 FN2
A2
F ≤ 107kN F ≤ 123kN
则 [ F ]= 107 kN
两杆一般不会同时达到容许应力！
P62 34-4
F
n
a
sa
FN = –F = –5 kN F 力的单位 ：kN FN s= = – 50 MPa A 正负 cos2a
s
s
s
sin2a 0
s
ta
sa = s
ta =
s
2
a = 0º a = 30º a = 45º a = 60º a = 90º
– 50 MPa – 37.5 MPa – 25 MPa – 12.5 MPa
Se ?
P63 35-2 h
A
Dl1 C dC
3 H
1
D
dH F = F N3 dH
l 2l
(1) FN1 = – 80 kN B Dl2 Dl = FN1 2h = – 0.4 mm 1 E1A1 2 FN2 = – 40 kN
E
2h
FN1
FN3 h D l3 = = 0.2 mm dH = dC + Dl3 = 0.6 mm E3A3 (2) Dl1 = – 0.2 mm Dl2 = – 0.4 mm Dl3 = 0.2 mm 2 1 dC = │Dl1│+ │Dl2│ = 0.267 mm 3 3 位移分析! dH = dC + Dl3 = 0.467 mm
dC =│Dl1│ = 0.4 mm
FN2 2h D l2 = = – 0.4 mm E2A2 FN2 FN3 = 120 kN
P64 35-3 设计截面尺寸 SMA = 0 FBC
B
A
FAx FAy
FN = FBC = 100 kN 3 p d2 (1) 斜杆用钢丝索 A = n 4 FN s= ≤ [s ] ∴ n ≥ 66.3 A 取 n = 67 或 66 C
– 21.65 MPa
– 25 MPa
– 21.65 MPa 0
0
P63 35-1
FA A
FB
FC
FD
a
B
a BC 1kN
C
a
D
AB FNi FNi a D li = EA – 4kN
CD 3kN
– 0.05mm 0.0125mm 0.0375mm 0
Dl = SDli
讨论：
e 呢?
e 与 Dl 区别?
六要素(阴影线方向)
F
FN1
F
＋ F
F FF F FN3
1 F
2 F
Fy = 0 Fy = 0 Fy = 0
FN1 = F FN2 = 0
3
FN3 F R
－
FN2 FN图
FN3 = － F
正向假设
P61 34-2 F1 = 500 N F2 = 420 N F3 = 280 N F4 = 400 N F5 = 240 N
(2) 斜杆用两根等边角钢 A = 2A1 FN s= ≤ [s ] ∴ A1 ≥ 1.044 cm2 A
取 ∟ 20×3 ( A1 = 1.13 cm2 )
用型号表示 结论!
h = 3m
q = 10kN/m
l = 4m
平衡计算
P64 35-4 确定许用荷载
FN 1 A ① 45º60º B F ②
习题解答(一)
P61 34-1(1)
1 1 2 3
1 1 F 2
截面法
FN1
＋wenku.baidu.com
分段相等
4F
Fy = 0 Fy = 0 Fy = 0
FN1 = 4F
F＋ F ＋ 2F
F 3 2F
FN2 3F FN3 F 2F FN图 2F 2F
FN2 = 3F
FN3 = 2F
P61 34-1(4)
F
1 2 3
B
F2 920 N 500 N
＋
C F3
D
E
F1
F4
F5
640 N
＋ ＋
240 N
＋
FN 图
s=
FN
A
sAB = 1.25 MPa sCD = 1.6 MPa
sBC = 2.3 MPa sDE = 0.6 MPa
与点有关
A = 400 mm2
应力单位 ：MPa
P62 34-3
3 2
1
F1 F2 A
3
F3 D
1
B
2
C
F3 Fx = 0 FN1 FN1 = F3 = 30 kN 40kN FN1 – 2 2 30kN 30kN A1 = 4×10 m s1 = = 0.75 MPa F2 F ＋ ＋ N2＋
FN3
F1 F2
A1 F3 Fx = 0 FN2 = F3 + F2 = 40 kN FN2 FN图 – 2 2 A2 = 2×10 m s2 = = 2 MPa A2 F3 F = 0 FN3 = F3 + F2 – F1 = 30 kN x FN3 – 2 2 A3 = 4×10 m s3 = = 0.75 MPa A3
FN 2
C
SFx = 0
SFy = 0 FN1 = 0.897 F
s1 =
s2 =
FN1
FN2 = 0.732 F
≤ [s ]1
≤ [s ]2
A1 FN2
A2
F ≤ 107kN F ≤ 123kN
则 [ F ]= 107 kN
两杆一般不会同时达到容许应力！
P62 34-4
F
n
a
sa
FN = –F = –5 kN F 力的单位 ：kN FN s= = – 50 MPa A 正负 cos2a
s
s
s
sin2a 0
s
ta
sa = s
ta =
s
2
a = 0º a = 30º a = 45º a = 60º a = 90º
– 50 MPa – 37.5 MPa – 25 MPa – 12.5 MPa
Se ?
P63 35-2 h
A
Dl1 C dC
3 H
1
D
dH F = F N3 dH
l 2l
(1) FN1 = – 80 kN B Dl2 Dl = FN1 2h = – 0.4 mm 1 E1A1 2 FN2 = – 40 kN
E
2h
FN1
FN3 h D l3 = = 0.2 mm dH = dC + Dl3 = 0.6 mm E3A3 (2) Dl1 = – 0.2 mm Dl2 = – 0.4 mm Dl3 = 0.2 mm 2 1 dC = │Dl1│+ │Dl2│ = 0.267 mm 3 3 位移分析! dH = dC + Dl3 = 0.467 mm
dC =│Dl1│ = 0.4 mm
FN2 2h D l2 = = – 0.4 mm E2A2 FN2 FN3 = 120 kN
P64 35-3 设计截面尺寸 SMA = 0 FBC
B
A
FAx FAy
FN = FBC = 100 kN 3 p d2 (1) 斜杆用钢丝索 A = n 4 FN s= ≤ [s ] ∴ n ≥ 66.3 A 取 n = 67 或 66 C
– 21.65 MPa
– 25 MPa
– 21.65 MPa 0
0
P63 35-1
FA A
FB
FC
FD
a
B
a BC 1kN
C
a
D
AB FNi FNi a D li = EA – 4kN
CD 3kN
– 0.05mm 0.0125mm 0.0375mm 0
Dl = SDli
讨论：
e 呢?
e 与 Dl 区别?
六要素(阴影线方向)
F
FN1
F
＋ F
F FF F FN3
1 F
2 F
Fy = 0 Fy = 0 Fy = 0
FN1 = F FN2 = 0
3
FN3 F R
－
FN2 FN图
FN3 = － F
正向假设
P61 34-2 F1 = 500 N F2 = 420 N F3 = 280 N F4 = 400 N F5 = 240 N
(2) 斜杆用两根等边角钢 A = 2A1 FN s= ≤ [s ] ∴ A1 ≥ 1.044 cm2 A
取 ∟ 20×3 ( A1 = 1.13 cm2 )
用型号表示 结论!
h = 3m
q = 10kN/m
l = 4m
平衡计算
P64 35-4 确定许用荷载
FN 1 A ① 45º60º B F ②
习题解答(一)
P61 34-1(1)
1 1 2 3
1 1 F 2
截面法
FN1
＋wenku.baidu.com
分段相等
4F
Fy = 0 Fy = 0 Fy = 0
FN1 = 4F
F＋ F ＋ 2F
F 3 2F
FN2 3F FN3 F 2F FN图 2F 2F
FN2 = 3F
FN3 = 2F
P61 34-1(4)
F
1 2 3