A
2m
Example #1:
■ Find the reactions at supports.
2m
B
■ Find the force in all members.
75 kg
2m
C
Solution:
Ay
Ax
∑MA=0
−75 × 9.8 × 2 sin 60 + 𝐶𝑥 × 2 = 0
−1273.06 + 𝐶𝑥 × 2 = 0  𝐶𝑥 = 636.5 𝑁
∑Fy=0
−𝐴𝑦 − 75 × 9.8 = 0  𝐴𝑦 = −735 𝑁 = 735 𝑁 ↑
∑Fx=0
75 x 9.8
Cx
𝐴𝑥 + 636.5 = 0  𝐴𝑥 = −636.5 𝑁 = 636.5 𝑁 ←
FCA
For joint C:
FCB
∑Fx=0
636.5 N
636.5 + 𝐹𝐶𝐵 sin 60 = 0
C
𝐹𝐶𝐵 = −735 𝑁 = 735 𝑁 (𝑐𝑜𝑚𝑝)
∑Fy=0
𝐹𝐶𝐴 + −735 cos 60 = 0
𝐹𝐶𝐴 = 367.5 𝑁 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛)
Eng. Ibrahim Al Mohanna, Civil Engineering, 2 A 92
Tutorial # 1
FBA
For joint B:
B
∑Fx=0
−𝐹𝐵𝐴 cos 30 + 735 cos 30 = 0
735 N
75 x 9.8
𝐹𝐵𝐴 = 735 𝑁 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛)
Eng. Ibrahim Al Mohanna, Civil Engineering, 2 A 92
Tutorial # 1
H
3m
G
3m
F
3m
E
Example # 2:
Calculate the force in the
member CG
3m
A
B
C
D
20 KN
20 KN
20 KN
Solution:
∑MG=0
−20 × 3 − 20 × 6 − 𝐹𝐶𝐵 × 3 = 0
G
F
𝐹𝐶𝐵 = −60 𝐾𝑁 = 60 𝐾𝑁 (𝑐𝑜𝑚𝑝)
B
∑MF=0
C
20 KN
D
20 KN
−20 × 3 + 60 × 3 − 𝐹𝐶𝐵 × cos 45 × 3 = 0
𝐹𝐶𝐺 = 56.6 𝐾𝑁 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛)
Eng. Ibrahim Al Mohanna, Civil Engineering, 2 A 92
Tutorial # 1
4 KN
Example #3:
Draw the shear and moment
diagram for the loaded beam.
2m
2 KN
2m
2m
A
B
Solution:
∑MA=0
−4 × 2 + 𝐵𝑦 × 4 − 2 × 6 = 0  𝐵𝑦 = 5 𝐾𝑁
∑Fy=0
−4 + 5 − 2 + 𝐴𝑦 = 0  𝐴𝑦 = 1 𝐾𝑁
First Segment:
∑M=0  −1 × 𝑥 + 𝑀 = 0  𝑀 = 𝑥 𝐾𝑁. 𝑚
∑Fy=0  1 − 𝑉 = 0  𝑉 = 1 𝐾𝑁
V
M
V
M
x
1 KN
Second Segment:
4 KN
∑M=0  −𝑥 + 4 × 𝑥 − 2 + 𝑀 = 0
2m
𝑀 = 8 − 3𝑥
x
1 KN
∑Fy=0  1 − 4 − 𝑉 = 0  𝑉 = −3 𝐾𝑁
Third Segment:
∑M=0  −𝑥 + 4 × 𝑥 − 2 − 5 × (𝑥 − 4) + 𝑀 = 0
𝑀 = 2𝑥 − 12
∑Fy=0  1 − 4 + 5 − 𝑉 = 0  𝑉 = 2 𝐾𝑁
4 KN
2m
2m
V
1 KN
Eng. Ibrahim Al Mohanna, Civil Engineering, 2 A 92
x
M
5 KN
Tutorial # 1
2 KN
4 KN
Load Diagram
1 KN
5 KN
2 KN
1 KN
Shear Force Diagram
-3 KN
2 KN.m
Bending Moment Diagram
-4 KN.m
Eng. Ibrahim Al Mohanna, Civil Engineering, 2 A 92
Tutorial # 1