Chapter 6(web)大连理工理论力学(英文)课件第六章

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Solution
10
11
Example A crane, which weighs P=520kN, is shown in the following figure. Provided the maximum heavy that the crane can sustain is W=250kN, determine: (1)the value of balance heavy Q to ensure the crane can not fall down when it is fully loaded or unloaded. (2) when the balance heavy Q= 355kN and the crane is fully loaded, the pressure that the crane acts on the rail.
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Solution
Fx 0
FAx 0
3 1 FB P qa 4 2
FAy P 3 qa 4 2
6
M
A
0 FB 4a M P 2a q 2a a 0
Fy 0
FAy q 2a P FB 0
Example Known P 100kN, M 20kN m, q 20kN/m,
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Solution 1. Plot FBD of the entire system
M
E
0
5 FA 2 2l P l 0 2
FA 5 2 P 8
0
F
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Solution plot FBD of the whole frame .
M
A
0
12 FBy 10 P 6 P 4 P2 2 P 5F 0 1
FBy 77.5kN
Fy 0
FAy FBy 2 P P P2 0 1
FAy 72.5kN
M
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Solution FBD of wheel I
MB 0
Pr F R 0
Pr F 10 P 1 R
Fr tan 20 0 F
Fr F tan 200 3.64 P 1
FBx 3,64 P 1
F 0 F 0
x
y
FBx Fr 0
Example Known AC=CB=l, P=10kN, Determine the forces at support A and inside rod DC. P
1
Solution
Fx 0
FAx Fc cos 450 0
FAy Fc sin 450 F 0
Fy 0
• A machine can be regarded as a mechanical device consisting one or more movable parts that serve to transmit and/or modify force. • A frame is defined as a structure that is built of bars. • If a frame is in equilibrium,each member of the frame must satisfy the conditions of equilibrium for a rigid body.(物体系的平衡 ：3n equations of equilibrium) • Simple frame (statically determinate frame静定结构) is defined as all the forces that act on the members of a frame and all the reactions of the supports can be determined by the equations of static equilibrium. Otherwise, it is called statically indeterminate frame(超静 定结构）. 20
FAy 300kN
MA 0
MA M F 1 l F cos 60 l F sin 60 3l 0
MA 1188kN m
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Example A pressure vessel which weighs 12kN is supported by three members. The wind load acting on the vessel is trapezoid (q1=1.2kN/m,q2=0.8kN/m). Determine the supportive forces of the three rods.
F
x
0
FAx F FBx 0
FAx 7.5kN
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Example 5 If DC=CE=CA=CB=2l, R=2r=l, 450 ,
and P is given, determine the reactions acting on supports A and E and the internal force of rod BD.
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Solution
∑MA（Fi）=0 m1 - FBCl1sinα=0 ∑MD（Fi）=0 （1） m1
B YA A
FBC
XA
FBCl2sinβ – m2 =0
（2） FBC YB
D XB
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Based on （1）（2）, yield
C m2
m1/m2= l1sinα/ l2sinβ
Chapter 6 Simple Structure and Machines
Fx 0
Fy 0
FAx FB 0
FAy P P1 0
M
A
0
FB 5 1.5 P 3.5P 0 1
Then
FAy 50kN
FB 31kN
FAx 31kN
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Example Known P, q, a, M pa;
Determine the reactions at point A and B
l 1m; F 400kN,
determine the reactions at fixed end A .
7q 3l 30kN 2
FAx F1 F sin 60 0
0
FAx 316.4kN
Fy 0
FAy P F cos 60 0
Foy FA cos 0
Fox
FR l 2 R2
F
y
Foy F
M FR
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M
o
0 FA cos R M 0
Example 2 The two-span beam is subjected the loads shown in figure. Known F =20kN, q=10kN/m, L=1m, M 20kN m, Determine the reactions of fixed end A and support B.
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Solution
FBD of CD
M
c
0
l FB sin 60 l ql F cos 300 2l 0 2
0
FB = 45.77kN
FBD of the entire beam
Fx 0
FAx FB cos 600 F sin 300 0
FAx 32.89kN
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Plot FBD of beam DE.
MD 0
8FE' 4P 2P2 0 1
FE' 12.5kN
Plot FBD of right frame.
M
C
0
6 FBy 10 FBx 4P FE 0
FBx 17.5kN
On the basis of the FBD of whole frame
Figure out the following structures are statically determinate or indeterminate and point out the degree of the statically indeterminate structure.
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Example 1 The slider-crank mechanism （曲柄滑块机 构）shown schematically in figure is in equilibrium in the case of action of F. Assume OA=R, AB=l. Determine the moment M and the reactions at pin O and block B when OA is horizontal.
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Solution 1. FBD of block B
F
y
0
F
x
0
F FB cos 0 F Fl FB cos l 2 R2 FN FB sin 0
FN F tan FR l 2 R2
2. FBD of wheel O
F
x
0
0
Fox FA sin 0
FBy P P2 F 0
FBy 32P 1
FAx 3.64 P1
FBD of wheel II
F F
x
0
0