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同济大学 理论力学 孙杰 习题解答2 (练习册P4-P10)
A1
对称性: yC = 0 负面积法: A1 = pR2
A2 = - pr2 x1 = 0 x2 = R/2
R O ·
A2 r
x
xC =
A1x1 + A2x2
A1 + A2
=–
r2 R 2(R2 - r2)
R/2
P6 习题:2
对称性: xC = xE = l /2 负面积法:
A1 = l 2 A2 = - l yE /2
cosb = — — √6 cosg = — — √6
1
F1
F5 y
x
AF
B MOx = 0 MOy = - F1· + M34 = 0 OA MOz = F2· - F5· = 0 OA OC
FR' = 5 i + 10 j + 5 k (kN)
MO = 0
FR = FR' 最终结果
P5 习题:1
∴
FAD = FBD = - 26.4 kN
FCD = 33.5 kN
P7 习题:2
F
O
[ 几何法 ] FB P
FB P F = = 5 4 3
∴ F = 15 kN
q F
5 3 FB = F ∴ F = 15 kN Fx = 0 5 3 sinj = ⑵ Fy = 0 cos(j -q )F = sinj P 5 当q = j Fmin = 12 kN
2 cosq = 5 sinq = 21 5
P
P1
q = 66.4º
FB
Fy = 0
FB – P1 + P2sinq = 0
FB = 14.2 kN
P8 习题:2
B C M1 A
力偶系 力偶系平衡特性 B C M2 M1 A FC
FC C l
FD M2
j
j
D
j
FA
2l
j
D
S mi = 0
S mi = 0
O
MO ' FR FR
x bR a
x
cosb = - 0.9512
MO = - F · - P1· - P2· = - 622200 kN· 3h b 5b m 主矩等于零处: d = |MO|/FR = 18.97 m 最终结果 或
xR = |MO|/|FRy| = 19.94 m
P6 习题:1 y
P10 习题:5
qC F1 F2 q B M
四个方程?
F1 =( qC – qB )×3l/2 = 6 kN
A
C
FBy
B F Bx
q
FA
F2 = qB×3l = 12 kN
MB = 0
Fx = 0
FAcos30º 4l + F1×2l + F2×1.5l - M = 0 ×
FA = 0
FBx - FAsin30º 0 =
—
力系的平衡: 指定方法 ! 结果(正确) 方法(力系 途径 表示):空间问题 平面问题 简单 汇交力系 力偶系 任意力系 平行力系
受力图 ! (约束 公理 力系 准确 完整 单独) 固定端 二力杆 研究对象 (除非 取整体) 整体
计算: 力的投影平衡方程 力矩平衡方程 (力系) 平衡方程 平衡方程
FBy + FAcos30º F1 - F2 = 0 -
FBx = 0
Fy = 0
FBy = 18 kN
P9 习题:2 空间任意力系 受力图(不考虑重力) Mx = 0 × × z - F1×Rsinq - Fsin30º a - Fcos30º 2a - FAy×3a = 0 1 FAy FAy = - (1 + )P 3 A 3 My = 0 F1×Rcos30º FAx×3a = 0 + FAx F 3 60º FAx = – P 6 3 FBx = P Fx = 0 FAx + FBx = 0 6 F1 q Fy = 0 FAy + FBy + Fcos30º 0 = 1 FBy y FBy = (2 - )P 3 B 6 x F Fz = 0 FBz – F1 – Fsin30º= 0 FBx Bz 3 Mz ≡ 0 FBz = P 2
选择 (投影轴 矩轴与矩心 方程形式研究对象) 物体系统 方程 (与受力图对应 直接列 代数式 独立方程个数)
P7 习题:1 z
A
空间汇交力系 FAD
力的投影平衡方程
j FCD
2q
O B
q
C
D P FBD
FBDcos45º FADcos45º 0 = y
j
FAD = FBD
x
- 2FADsin45º cos30º FCDcos15º 0 = - 2FADsin45º sin30º FCDsin15º P = 0 -
2 力螺旋
P4 习题:1
空间任意力系的简化
m) M34 = F3· j = 6 j (kN· OA
z
E G O
' 向O点简化: FRx = F5 = 5 kN
— ' FR = 5 6 kN √ 1
cosa = — — √6
2
D
F'
Ry
= F2 = 10 kN
g a
2
M H F4 34 ' F b FR3 C FRz = F1 = 5 kN
⑴ Fy = 0
B 5 4P j P F 3 = P FB F 5 3 A B 45 P FB FA [ 解析法 ] 3 ∴ F = 12 kN = Fmin 4 F
F F
FB = P
P7 习题:3 P1 = 60 kN P2 = 50 kN
P2 A q
Fx = 0
P – P2cosq = 0
P10 习题:3(b)
q ql D A FA M B FB F C
平面平行力系的平衡
平面平行力系的平衡特性
MB = 0
ql×2.5l - FA×2l - F×l - M = 0 5ql F M FA = 2 2l 4
Fiy = 0
FA + FB = F + ql 3F M FB = + + 2 2l 4
平面任意力系的简化
' FRx = F = 10120 kN ' FRy = - P1 - P2 = - 31200 kN ' F 'F 2 945kN F F22 Rx ix Ry 2iy = 32800 kN
y
向O点简化:
' = FR
F
P1 d
P2
cosa = 0.3085
a = 72.03° b = 162.03°
静力学 2
习题解答
练习册
P4 习题:2
向O点简化:
' FRx = F2 = 100 N ' FRy = F1 = 100 N ' FRz = 0
FR' = 100 i + 100 j (N)
— ' FR = 100 2 N √
a = b = 45° g = 90°
MO = 20 j + 10 k (N· m)
y
y1 = l /2
y2 = yE /3 3l 2 – yE2 令 3(2l – yE )
· E A1 A2 O
l x
yC =
A1 y1 + A2 y2
A1 + A2
=–
= yE
2yE2 - 6yEl + 3l 2 = 0 则
yE =
— 3 ±√3
2
l
l = yC
E( 0.5l,0.634l )
取
3 -√3 yE = 2
FC×2l - M1 = 0
M2 - FC×l = 0
∴ M1 = 2M2
P10 习题:3(a) 平面力系 第一小题
FAx A FA M B FB F C
平面平行力系的平衡
平面平行力系的平衡特性
MB = 0
- FA×2l - F×l - M = 0 F M FA = 2 2l
Fiy = 0
FA + FB - F = 0 3F M FB = + 2 2l
ql
P10 习题:4
M
F j B
平面任意力系的平衡
FQ = hq/2 = 6 kN
2h/3 FQ q FAx A FAy
Fx = 0
FAx + FQ - Fcos45º 0 = FAx = 0
Fy = 0
FAy - Fsin45º 0 = FAy = 6 kN
MA = 0
MA
4 MA + Fcos45º 4 - Fsin45º 3 - FQ× - M = 0 × × 3 MA = 12 kN· m
F2
z
MO
M⊥
y
MOy O
MOz M∥
h
x F1
' FR
MOx = 0 M
m MOy = F2· - M = 20 N· a ' = 90° h m MOz = F1·- F2· = 10 N· b ' = 26.57° l b
MO = 10 5 √
— N· m
g ' = 63.43°
— M⊥ √6— √2— M = 10 2— N· 最终结果 a = — = — m √ M∥ = m Oy ' 2 20 FR — 2 — 2 √2 = 12.25 cm 2 M )2 10 3 M = √ N· m FR ⊥ Fix (— FOy = 945kN ( MOz ) iy