Andyze asked in 科學工程學 · 9 years ago

# 求助靜力學高手

1.

Two planar pin-connected frames are supported and loaded asshow in the figures. For each structure, determine the following:

(a) The components of reactions at B and C.

(b) The axial force, shear force, and moment acting on the cross section at point D.

2.

A hollow transmission AB is supported at A and E by bearings and loaded as depicted in the figure. Calculate the following:

(a) The torque T required for equilibrium.

(b) The reaction at the bearings

Given: F1=4kN, F2=3kN,F3=5kN,F4=2kN

3.

Pin-connected members ADB and CD carry a load W applied by a cable-pulley arrangement, as shown in figure. Determine

(a)

The component of reactions at A and C.

(b)

The axial force, shear force, and moment acting on the cross-section at point G.

Given: The pulley at B has a radius of 150 mm. Load W=1.6kN.

4.

A bent rod is supported in the xz plane by bearings at B, C, D and loaded as shown in figure. Dimensions are in millimeters. Calculate moment and shear force in the rod on the cross section at point E, for P1=200N and P2=300N.

5.

A gear train is used to transmit a torque T=150 N.m from an electric motor to a drive machine. Determine the torque acting on driven machine shaft, Td, required for equilibrium.

6.

A frame AB and a simple beam CD are supported as shown in figure. A roller fits snugly between the two members at E. Determine the reactions at A and C in terms of load P.

Rating
• 阿銘
Lv 7
9 years ago

1

(1).左邊A至C元件對hinge處C點力矩合為0 A處水平反力Rax以向右為正

垂直反力Ray以向右為正 則Rax*3+Ray*4=0---------(1)

右邊元件B至C對hinge處C點力矩合為0 B處水平反力Rbx以向右為正

垂直反力Rby以向右為正 則10*6+1/2*8*6*2/3-Rbx*3-Rby*2=0---------(2)

水平力Fx合為0 則Rax=-Rbx--------(3)

垂直合力Fy為0 則10+1/2*8*6=Ray+Rby-------(4)

由(1)(2)(3)(4)可解得Rax=16/3T向左

Ray=4T向上

Rbx=16/3T向右

Rby=30T向上

(2)D點處之軸向力,剪力及力矩:

軸力=-6/3T(拉力)

剪力=10+1/2*4*3*-Rby=10+6-30=-14(斷面右邊向上)

力矩=30*3+16/3*2-10*3-1/2*4*3*1/3*3=61+2/3TM(往上彎)

2011-10-16 13:11:12 補充：

(1).左邊A至C元件對hinge處C點力矩合為0 , A處水平反力Rax以向右為正

垂直反力Ray以向右為正 ,則Rax*2+Ray*3=0---------(1)

對點A力矩合為0 , 則10*4-Rbx*2+Rby*1=0----------(2)

水平力Fx合為0 則Rax+Rbx=0--------(3)

垂直合力Fy為0 則10=Ray+Rby--------(4)

2011-10-16 13:11:59 補充：

由(1)(2)(3)(4)可解得Rax=-75/4KN向左

Rbx=75/4KN向右

Ray=25/2KN向上

Rby=-75/4KN向下

1.2.2D點斷面軸向力 剪力及力矩:設為Fda(拉力為正) Fds(向上為正) Md(逆時針為正)

取A點至D點部分分析

2011-10-16 13:12:05 補充：

水平合力為0; Rax+Fda*3/5+Fds*4/5=0--------(1)

垂直合力為0 Ray-Fda*4/5+Fds*3/5=0----------(2)

對點A合力矩為0 Fda*3/5*1-Fda*4/5*2.25+Fds*4/5*1-Fds*3/5*2.25+Md=0------(3)

將Rax RayRbx Rby值代入並由(1)(2)(3)即可解得Fda, Fds, Md

2011-10-16 13:41:28 補充：

2.

1.所需平衡扭距=(F1-F2)*0.3/2+(F3-F4)*0.3/2=0.15+0.45=0.6(KNM)

2.支承反力:(均以圖上所定坐標定正負)

Raz+Rcz=F3+F4=5+2=7---------------------(1)[Z方向合力為0]

Rcz*2.5=(F3+F4)*0.5=7*0.5=3.5-----------(2)[對Y軸繞A點合力矩為0]

2011-10-16 13:41:34 補充：

Ray+Rcy=-(F1+F2)=-7-------------------------(3)[Y方向合力為0]

Rcy*2.5=-(F1+F2)*1=-7--------------------------(4)[對Z軸繞A點合力矩為0]

由(1)(2)(3)(4)解得Raz=28/5KN, Rcz=7/5KN,Ray=-21/5KN ,Rcy=-14/5KNR

2011-10-17 11:07:22 補充：

3.Y方向合力為0 則Ray+Rcy=W=1.6KN----------(1)

X方向合力為0 則Rax+Rcx=0-----------------------(2)

對點C點合力距為0 則Rax*0.4-Ray*0.5=W*(1.0+0.075)=1.6*1.075=1.72KNM----------(3)

取CD及拉繩部分 對點D合力矩為0 則 -Rcx*0.4=W*0.075=1.6*0.075=0.12KNM------(4)

2011-10-17 11:08:31 補充：

由(1)(2)(3)(4)解得:

Rax=0.3KN(向右) Rcx=-0.3KN(向左) Ray=-3.2KN(向下) Rcy=4.8KN(向上)

3.2點 G之各項所求(取下段A至滑輪拉繩處分析)

2011-10-17 11:08:39 補充：

(1)軸力:Rax+Rcx+1.6=1.6kKN(壓力)

(2)剪力:Ray+Rcy=1.6KN(AG斷面往下)

(3)力矩:Ray*0.9+Rcy*0.4=-0.96KNM(AG斷面往順時針)

2011-10-20 11:30:18 補充：

6.取元件AB 設A點水平反力為Rax 垂直反力為Ray 滾輪反力為Rry

X方向平衡:Rax+Pcos45度=0----------------------------(1)

Y方向平衡:Ray+Rry=Psin45度-------------------------(2)

對A點力矩平衡:Pcos45*1+Psin45度*6=Rry*4-----------------(3)

由(1)(2)(3)解得

Rax=-P*(2)^-1 (向上) Ray=P*-3/4*2^-1/2(向下) Rry=P*7/4*2^-1/2(向上)

2011-10-20 11:56:51 補充：

6.2 取C至D基座元件

因D為HINGE 故基座至點 D之間桿件傳至點D之內力必為只有軸向力 設為Rd

X方向平衡:Rcx+Rd*cos30度=0-------------------------------(1)

Y方向平衡:Rcy+Rd*sin30度=Rry--------------------------(2)

對C點力矩平衡:Rry*2=Rd*sin30度*6------------------------(3)

由(1)(2)(3)解得

Rcx=-P*7/12*(3/2)^1/2(向左) Rcy=P*7/6*2^-1/2(向上)

2011-10-20 14:14:09 補充：

4.設BCD點反力為分別為Rby.,Rcy,Rdy

Y向合力為0 則Rby+Rcy+Rdy=P1+P2=200+300=500N-------(1)

對AC連線X軸 之扭距應為0 故Rdy*2.5=P*1 =300NM ---------(2)

對D點與Z軸平行方向扭距為0 故 Rby*4.5+Rcy*2=P1*5.5=1100NM--(3)

解(1)(2)(3)得Rby=136N(向上) Rcy=244N(向上)

故E點之剪力Se=P1-Rby=200N-136N=64N

E點之力矩Me=P1*2.75-Rby*1.75=550-238=312NM

2011-10-20 14:15:57 補充：

漏個P2對AC線X軸 之扭距應為0 故Rdy*2.5=P2*1=300NM ---------(2)