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1. Top School in Ghaziabad
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2. ENGR-1100 Introduction to
Engineering Analysis
Lecture 20
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3. Previous Lectures Outline
2D trusses analysis-a)
method of joints.
b) method of sections.
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4. Today’s Lecture Outline
• Frames
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5. Two Important Structures Types
• Trusses: Structures composed
entirely of two force
members.
• Frames: Structures containing at least one
member acted on by forces at three or more
points.
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6. Trusses Assumptions
1) Truss members are connected together at
their ends only.
2) Truss members are connected together by
frictionless pins.
3) The truss structure is loaded only at the
joints.
4) The weight of the member may be
school.endehgolelcet.ceodm.

7. Frames vs. Machines
• Frames
- Rigid structure
- Overall equilibrium is sufficient to determine
support reaction.
• Machines
- Not a rigid structure
- Overall equilibrium is not sufficient to
determine support reaction.
additional support reaction is needed for
equilibrium
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8. Beware
• Members of a frame are not necessarily a
two force member.
• The direction of the force applied by the
members on the pins are not necessarily
known.
F
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9. Method of solving frames
• Draw a free body diagram for each component
• Not all members can be treated as two-force
members.
• Write the equilibrium equations for each free
body diagram.
• Solve the equilibrium equations of the system
of rigid bodies.
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10. Example 7-85
A two-bar frame is loaded and supported as shown in Fig.
F7-85. Determine the reactions at supports A and E and the
force exerted on member ABC by the pin at C.
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11. Solution
x
y
From a free-body diagram on the
d = 6/tan(70°) + 6/tan(70°) = 5.648 ft
åMA = Ey (5.648) -500(2)- 400 (4)-
-300 (6) = 0
Ey = 779.0 lb @ 779 lb
åFy = Ay + 779.0 = 0
Ay =- 779.0 lb=779 lb
complete frame:
d
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12. From a free-body diagram on member CDE:
åMC = 400 (2) + 779.0 (3.464) + Ex (6) = 0
Ex = -583.1 lb @ 583 lb
y
From a free-body diagram on member ABC: x
åMC = 500 (4) + Ax (6) + 779.0 (2.184) = 0
Ax = -616.9 lb @ 617 lb
åFx = Cx + 500 - 616.9 = 0
school.edhole.cCoxm = 116.9 lb = 116.9 lb

13. åFy = Cy - 779.0 = 0
Cy = 779.0 lb = 779 lb
C = = ( ) ( ) = 787.7 lb @ 788 lb 2 2
x y C + C ( ) ( ) 2 2 116.9 + 799
qc = tan -1 7 7 9 . 0 = 81.46° C @ 788 lb 81.5°
116.9
A = ( ) ( ) = = 993.7 lb @ 994 lb 2 2
x y A + A ( ) ( ) 2 2 616.9 + 799.0
qA = tan -1 7 7 9 . 0 = -128.38° A @ 994 lb 51.6°
school.edhol-e6.1c6o.9m

14. E = ( E ) 2 + ( E ) 2
= 2 x y ( 583.1 ) + ( 799.0
) 2 = 973.1 lb @ 973 lb 779.0
-583.1
qE= tan -1 = 126.82° E @ 973 lb 53.2°
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15. Class Assignment: Exercise set P7-83
please submit to TA at the end of the lecture
Determine all forces acting on member BCD of the linkage
shown in Fig. P7-83.
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16. Member AC is a two-force member;
Therefore, the line of action of force C is
known as shown on the free-body diagram
for member BCD:
åMB = C cos 45° (2.0) - 40 cos 30° (3.5) = 0
C = 85.73 lb @ 85.7 lb
C @ 85.73 lb @ 45°
åFx = Bx + 40 cos 30° - 85.73 cos 45° = 0
Bx = 25.98 lb @ 26.0 lb
åFy = By - 85.73 sin 45° + 40 sin 30° = 0
By = 40.62 lb @ 40.6 lb
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17. B = Ö( Bx)2 + ( Bx)2 = Ö( 25.98)2 + ( 40.62)2 = 48.22 lb
40.62
25.98
y
x
B
B
qB = tan -1 = tan -1 = 57.4°
B @ 48.2 lb 57.4°
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18. Example 7-90
Determine all forces acting on member ABE of the
frame shown in Fig. P7-90.
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19. Solution
x
y
From a free-body diagram on the complete frame:
åMA = D (300) - 150 (300) = 0
D = 150.0 N = 150.0 N ­
åFx = Ax + 150.0 = 0
Ax =- 150.0 N = 150.0 N ¬
åFy = Ay + 150.0 = 0
Ay =- 150.0 N = 150.0 N ¯
A = ( A ) 2 + ( A )2
2 2 x y = ( 150.0 ) + ( -150.0
) = 212.1 N @ 212 N qA = tan -1 1 5 0 . 0 = 135.0° A = 212 N 45°
150.0
-
-
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20. From a free-body diagram on member CEF: 150N
åMC = Ex (100) - 150 (200) = 0
Ex = 300 N = 300 N ® (on ABE)
Ex
Cx
From a free-body diagram on member ABE:
åMB = Ey (100) - 300 (100) - 150(100) = 0
Ey = 450 N = 450 N ­
x y E + E ( ) ( ) 2 2 300 + 450
Ey
Cy
E = = = ( ) ( ) 540.8 N 2 2
450
300
qE = tan -1 = 56.31° E = 541 N 56.3°
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21. åFx = Bx + 300 - 150 = 0
Bx =- 150.0 N = 150.0 N ¬
åFy = By + 450 - 150 = 0
By =- 300 N = 300 N ¯
B = ( B ) 2 + ( B ) 2
= ( -150 ) 2 + ( -300
) 2 = 335.4 N x y 300
-150
qB = tan -1 = -116.56° B @ 335 N 63.4°
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22. Class Assignment: Exercise set P7-91
please submit to TA at the end of the lecture
Determine all forces acting on member ABCD of the Frame
shown in Fig. P7-91.
Solution:
A=167.7 lb 63.4o
B=424 lb 45o
C=335 lb 26.6o
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23. Class Assignment: Exercise set P7-87
please submit to TA at the end of the lecture
A pin-connected system of leaves and bars is used as a toggle
for a press as shown in Fig. P7-87. Determine the force F
exerted on the can at A when a force P = 100 lb is applied to
the lever at G.
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24. Solution
From a free-body diagram
for the lever:
åMF = 100 (30) - FDE (8) = 0
FDE = 375 lb
From a free-body diagram
for pin D:
+® åFx = -FBD cos 67° - FCD cos 78° - 375=0
­ åFy = -FBD sin 67° + FCD sin 78° = 0
FBD = -639.5 lb
FCD = -601.8 lb
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25. From a free-body diagram
for the piston at B:
­åFy = Ay - 639.5 sin 67° =0
Ay = 588.7 lb = 588.7 lb ­
Force on the can: F @ 589 lb ¯
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26. Class Assignment: Exercise set P7-89
please submit to TA at the end of the lecture
A pin-connected system of bars supports a 300 lb
load as shown in Fig. P7-87. Determine the
reactions at supports A and B and the force exerted
by the pin at C on member ACE.
By = 150.0 lb = 150.0 lb ­
Bx = -450 lb = 450 lb ¬
Ax = 450 lb = 450 lb ®
Ay = 150.0 lb = 150.0 lb ­
Cy = 0 Cx = - 600 lb school.edhole.com

27. Solution
From a free-body diagram
For the complete system:
+ åMA = - Bx (20) - 300 (30) = 0
Bx = -450 lb = 450 lb ¬
+ ® åFx = Ax + Bx = Ax - 450 = 0
Ax = 450 lb = 450 lb ®
From a free-body diagram
for pin F:
+ ® åFy = TEF -300 sin 45° = 0
school.edhToElFe =.2c1o2.m1 lb @ 212 lb (T)

28. From a free-body diagram
for bar ACE:
+ åMC = -Ay (10) + 450 (10)
- 212.1 (10/cos 45°) = 0
Ay = 150.0 lb = 150.0 lb ­
+ ­ åFy = Ay + Cy - 212.1 sin 45°
= 150.0 + Cy - 212.1sin 45° = 0 Cy = 0
+ ® åFx = Cx+ 450+212.1 cos 45°=0 Cx = - 600 lb
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29. From a free-body diagram for the complete system:
+ ­ åFy = By + Ay - 300
= By + 150.0 - 300 = 0
By = 150.0 lb = 150.0 lb ­
A = ( A ) 2 + ( A ) 2
= ( 450.0 ) 2 + ( 150.0
) 2 = 474.3 lb @ 474 lb x y 150.0
450.0
qA = tan-1 = 18.434° A @ 474 lb 18.43°
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30. Example 7-101
Forces of 50 lb are applied to the handles of the bolt
cutter of Fig. P7-101. Determine the force exerted by on
the bolt at E and all forces acting on the handle ABC.
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31. Solution
From a free-body diagram
for member CDE:
+ ® åFx = Cx = 0
Cx = 0
+ åMD = Cy (3) - E(2) = 0
2
3
Cy = E
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32. From a free-body diagram
For handle ABC:
+ åMB = Cy (1) - 50 (20) = 0
Cy =1000 lb = 1000 lb ­
C = 1000 lb ­
åFx = Bx = 0
Bx = 0
åFy = By + 50 + 1000 = 0
By = -1050 lb
B = 1000 lb ¯
E = 1.5Cy = 1.5 (1000) = 1500 lb
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Force on the bolt: E = 1500 lb ­

33. Class Assignment: Exercise set P7-110
please submit to TA at the end of the lecture
A cylinder with a mass of 150 kg is supported by a
two-bar frame as shown in Fig. P7-110. Determine
all forces acting on member ACE.
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34. Solution
From a free-body diagram for
the cylinder:
W = mg = 150 (9.807) = 1471.1 N
+ ® åF= D sin 45° - E sin 45° = 0
x + ­ åF= 2D cos 45° - 1471.1 = 0
y D = E = 1040.2 lb
E @ 1040 lb 45.0° (on member ACE)
From a free-body diagram
for the complete frame:
åM= A (2) - 1471.1 (1) = 0
B A =735.6 N = 735.6 N ­
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A @ 736 N ­

35. From a free-body diagram for ACE:
åMC = T (1) - 735.6 (1) - 1040.2 (0.8) = 0
T = 1567.8 N @ 1568 N
T @ 1568 N ®
åFx = Cx + T + 1040.2 cos 45°
= Cx + 1567.8 + 1040.5 cos 45° = 0
Cx = -2303 @ 2300 N ¬
åFy = Cy + 735.6 - 1040.2 sins 45° = 0
Cy = 0
school.edhole.coCm @ 2300 N ¬