Andyze asked in 科學工程學 · 9 years ago

求助靜力學高手

誠心發問

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.

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

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

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

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.

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

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.

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

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.

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

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.

圖片參考:http://imgcld.yimg.com/8/n/AB00167288/o/1611101103...

1 Answer

Rating
  • 阿銘
    Lv 7
    9 years ago
    Favorite Answer

    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)

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