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Top School in Ghaziabad By: school.edhole.com
Lecture 20
ENGR-1100 Introduction to Engineering Analysis
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Previous Lectures Outline
2D trusses analysis-
a) method of joints.
b) method of sections.
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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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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 neglected. school.edhole.com
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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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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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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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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Solution
x
y
d = 6/tan(70) + 6/tan(70) = 5.648 ft
MA = Ey (5.648) -500(2)- 400 (4)-
-300 (6) = 0
Fy = Ay + 779.0 = 0
Ay =- 779.0 lb=779 lb
Ey = 779.0 lb 779 lb
From a free-body diagram on the complete frame:
d
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From a free-body diagram on member CDE:
MC = 400 (2) + 779.0 (3.464) + Ex (6) = 0
Ex = -583.1 lb 583 lb
From a free-body diagram on member ABC: x
y
MC = 500 (4) + Ax (6) + 779.0 (2.184) = 0
Fx = Cx + 500 - 616.9 = 0
Ax = -616.9 lb 617 lb
Cx = 116.9 lb = 116.9 lbschool.edhole.com
Fy = Cy - 779.0 = 0
Cy = 779.0 lb = 779 lb
C = = = 787.7 lb 788 lb 22
x yC C 2 2116.9 799
c = tan -1 = 81.46 C 788 lb 81.5 779.0116.9
A = = = 993.7 lb 994 lb 22
x yA A 2 2616.9 799.0
A = tan -1 = -128.38A 994 lb 51.6 779.0616.9school.edhole.com
E = = = 973.1 lb 973 lb 22
x yE E 2 2583.1 799.0
E= tan -1 = 126.82 E 973 lb 53.2 779.0583.1
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Class Assignment: Exercise set P7-83please 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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C 85.73 lb 45
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:
MB = C cos 45 (2.0) - 40 cos 30 (3.5) = 0
C = 85.73 lb 85.7 lb
Fx = Bx + 40 cos 30 - 85.73 cos 45 = 0
Fy = By - 85.73 sin 45 + 40 sin 30 = 0
Bx = 25.98 lb 26.0 lb
By = 40.62 lb 40.6 lbschool.edhole.com
B = tan -1 = tan -1 = 57.440.6225.98
B 48.2 lb 57.4
B = ( Bx)2 + ( Bx)
2 = ( 25.98)2 + ( 40.62)2 = 48.22 lb
y
x
B
B
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Example 7-90
Determine all forces acting on member ABE of the frame shown in Fig. P7-90.
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Solution
x
y
From a free-body diagram on the complete frame:
MA = D (300) - 150 (300) = 0
D = 150.0 N = 150.0 N Fx = Ax + 150.0 = 0
Ax =- 150.0 N = 150.0 N Fy = Ay + 150.0 = 0
Ay =- 150.0 N = 150.0 N
A = tan -1 = 135.0 A = 212 N 45 150.0150.0
A = = = 212.1 N 212 N 22
x yA A 2 2150.0 150.0
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From a free-body diagram on member ABE:
MB = Ey (100) - 300 (100) - 150(100) = 0
Ey = 450 N = 450 N
E = = = 540.8 N 22
x yE E 2 2300 450
E = tan -1 = 56.31 E = 541 N 56.3 450300
MC = Ex (100) - 150 (200) = 0
Ex = 300 N = 300 N (on ABE)
From a free-body diagram on member CEF: 150N
Cx
Ex
Ey
Cy
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Fx = Bx + 300 - 150 = 0
Bx =- 150.0 N = 150.0 N Fy = By + 450 - 150 = 0
By =- 300 N = 300 N
B = = = 335.4 N 22
x yB B 2 2150 300
B = tan -1 = -116.56 B 335 N 63.4
300150
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Class Assignment: Exercise set P7-91please 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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Class Assignment: Exercise set P7-87please 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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Solution
From a free-body diagramfor the lever:
MF = 100 (30) - FDE (8) = 0
+ Fx = -FBD cos 67 - FCD cos 78 - 375=0
Fy = -FBD sin 67 + FCD sin 78 = 0
From a free-body diagramfor pin D:
FDE = 375 lb
FBD = -639.5 lb
FCD = -601.8 lbschool.edhole.com
From a free-body diagramfor the piston at B:
Fy = Ay - 639.5 sin 67 =0
Ay = 588.7 lb = 588.7 lb
Force on the can: F 589 lb
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Class Assignment: Exercise set P7-89please 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
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Solution
From a free-body diagramFor the complete system:
From a free-body diagramfor pin F:
+ Fx = Ax + Bx = Ax - 450 = 0
+ MA = - Bx (20) - 300 (30) = 0
Bx = -450 lb = 450 lb
Ax = 450 lb = 450 lb
+ Fy = TEF -300 sin 45 = 0
TEF =212.1 lb 212 lb (T)school.edhole.com
From a free-body diagramfor bar ACE:
+ MC = -Ay (10) + 450 (10)
- 212.1 (10/cos 45) = 0
Ay = 150.0 lb = 150.0 lb
+ Fy = Ay + Cy - 212.1 sin 45 = 150.0 + Cy - 212.1sin 45 = 0 Cy = 0
+ Fx = Cx+ 450+212.1 cos 45=0 Cx = - 600 lb
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From a free-body diagram for the complete system:
+ Fy = By + Ay - 300
= By + 150.0 - 300 = 0
By = 150.0 lb = 150.0 lb
A = = = 474.3 lb 474 lb 22
x yA A 2 2
450.0 150.0
A = tan-1 = 18.434 A 474 lb 18.43
150.0450.0
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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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SolutionFrom a free-body diagramfor member CDE:
+ Fx = Cx = 0
Cx = 0
+ MD = Cy (3) - E(2) = 0
23
Cy = E
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From a free-body diagramFor handle ABC:
+ MB = Cy (1) - 50 (20) = 0
Cy =1000 lb = 1000 lb
Fx = Bx = 0
Bx = 0
C = 1000 lb
Fy = By + 50 + 1000 = 0
By = -1050 lbB = 1000 lb
E = 1.5Cy = 1.5 (1000) = 1500 lb
Force on the bolt: E = 1500 lb school.edhole.com
Class Assignment: Exercise set P7-110please 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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SolutionFrom a free-body diagram for the cylinder:
W = mg = 150 (9.807) = 1471.1 N + Fx = D sin 45 - E sin 45 = 0
+ Fy = 2D cos 45 - 1471.1 = 0
E 1040 lb 45.0 (on member ACE)
From a free-body diagramfor the complete frame:
MB = A (2) - 1471.1 (1) = 0
A =735.6 N = 735.6 N
A 736 N
D = E = 1040.2 lb
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From a free-body diagram for ACE:
MC = T (1) - 735.6 (1) - 1040.2 (0.8) = 0
T = 1567.8 N 1568 N
T 1568 N
Fx = Cx + T + 1040.2 cos 45 = Cx + 1567.8 + 1040.5 cos 45 = 0
Cx = -2303 2300 N
Fy = Cy + 735.6 - 1040.2 sins 45 = 0
Cy = 0C 2300 N school.edhole.com
