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MECH 344 Problem Set 1-Chapter 2_study guide exam solution Concordia University
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MECH 344 Problem Set 1-Chapter 2_study guide exam solution Concordia University
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MECH 344 Problem Set 1-Chapter 2_study guide examsolution Concordia University
Known: The geometry and the loads acting on a pinned assembly are given.
Find: Draw a free-body diagram for the assembly and determine the magnitude of
theforces acting on each member of the assembly.
Schematic and Given Data:
1500 N 1500 N
Link 5
B C
45˚
45˚
Link 3
1000 mm
45˚
A 45˚ D
Link 1
Assumptions:
1. The links are rigid.
2. The pin joints are frictionless.
3. The weight of the links are negligible.
4. The links are two force members and are either in tension or compression.
Analysis:
1. We first draw a free-body diagram of the entire structure.
1500 N 1500 N
1
B C
45˚
45˚
2 1 2
45˚
A 45˚
D
Ax 1
Ay Dy
2. Taking moments about point A and assuming clockwise moments to be positive,
2-17
, ! MA = 0 = 1500(2) + 1500(1) - Dy(1)
3. Solving for Dy gives Dy = 4500 N.
4. Summation of forces in the y-direction and assuming vertical forces positive,
! Fy = 0 = Ay + Dy - 1500 - 1500 = Ay + Dy - 3000.
2-18
,5. Since, Dy = 4500 N, Ay = 3000 - Dy = -1500 N.
6. ! Fx = 0 gives, Ax = 0.
7. We now draw a free-body diagram for a section at C.
1500 N
CB
45˚
DC
8. ! Fx = 0 = - CB + DC sin 45˚ and
! Fy = 0 = DC sin 45˚ - 1500
9. Solving simultaneous equations gives DC = 2121 N, CB = 1500 N.
10. We now draw a free-body diagram for a section at A.
AB
45˚
DA
1500 N
11. ! Fy = 0 = AB sin 45˚- 1500
! Fx = 0 = AB cos 45˚- DA
12. Solving simultaneous equations gives AB = 2121 N, DA = 1500 N.
13. We now draw a free-body diagram for a section at D.
B
C
BD
45˚ 2121 N
D
A
1500 N
4500 N
2-19
, 14. ! Fy = 0 = 4500 - BD - 2121(sin 45˚)
Hence, BD = 3000 N.
15. The free-body diagrams for links DC, BC, AB, AD, and BD are:
B C
1500 1500
1500 2 = 2121 3000 1500 2 = 2121
B
B C
D D
A
3000 2121
2121
1500 1500
A D
16. We can now draw a free-body diagram of pin B:
1500 N
B
1500 N
45˚
2121 N
3000 N
17. Checking for static equilibrium at pin B gives:
! Fx = 2121 cos 45˚ - 1500 = 0
! Fy = 1500 + 2121 sin 45˚ - 3000 = 0
18. We can also draw a free-body diagram for pin C:
1500 N
2121 N
45˚
1500 N
C
2-20
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