Mechanics of Materials (DME516126)
Homework 1
Due : N/A
1. Determine the average normal and average shear force in the member at section
b-b. The load P is applied along the centroidal axis of the member.
b
P
P
θ
b
Sol :
FBD
EQM
+
∑ Fx = 0
↗
⇒ -V + P cos θ = 0
⇒ V = P cos θ
+
∑ Fy = 0
↖
⇒ N - P sin θ = 0
⇒ N = P sin θ
Ans : V = P cos θ
N = P sin θ
2. A steel pad supporting heavy machinery rests on four short, hollow, cast iron piers.
The failure stress of the cast iron in compression is  fail . The outer diameter of the
piers is d, and the wall thickness is t. Determine the total force F that may be
supported by the pad.
F
t
d
Sol :
A=
π 2
[d - ( d - 2t )2 ]
4
F = σfail ∙A∙ 4 = σfail ∙ π[d 2 - ( d - 2t )2 ] = 4πσfail ( dt - t2 )
Ans : F = 4πσfail ( dt - t2 )
3. The rod AB and CD are made of steel. Determine their smallest diameter so that
they can support these loads as shown. The beam is assumed to be pin connected at
B and D. Use a factor of safety against failure of 2 and the failure stress is
 fail = 345 MPa.
A
C
4 kN 5 kN
6 kN
B
D
3m
3m
4m
4m
Sol :
FBD
EQM
+
∑ MB = 0
↻
⇒4∙3 + 5∙( 3 + 3 ) + 6∙( 3 + 3 + 4 ) - FCD ∙( 3 + 3 + 4 + 4 ) = 0
⇒FCD =
51
kN
7
+
∑ Fy = 0
↑
⇒ FAB - 4 - 5 - 6 + FCD =0
⇒ FAB =
54
kN
7
The factor of safety n =
so we can get σallow =
σfail
σallow
σfail 345
=
=172.5 Mpa
n
2
For the left rod AB
FAB
FAB
σAB =
=π
2
AAB
4 𝑑AB
54 3
∙10
6
⇒ 172.5∙10 = π7
𝑑 2
4 AB
⇒ 𝑑AB = 7.545 ∙ 10−3 m = 7.545 mm
For the left rod CD
FCD
FCD
σCD =
=π
2
ACD
4 𝑑𝐶𝐷
51 3
∙10
6
⇒ 172.5∙10 = π7
2
4 𝑑𝐶𝐷
⇒ 𝑑𝐶𝐷 = 7.333 ∙ 10−3 m = 7.333 mm
Ans : 𝑑𝐶𝐷 = 7.333 mm
𝑑AB = 7.545 mm
4. Determine the maximum magnitude P of the load the beam will support if the
average shear stress in each pin is not to allowed to exceed 60 Mpa. All pins are
subjected to double shear as shown, and each has a diameter of 18 mm.
Sol :
FBD
EQM
+
∑ MA = 0
↺
⇒ FBC sin 30°∙( 2 + 2 + 2 ) - P ∙( 2 + 2 ) - P ∙2 =0
⇒ FBC = 2P
+
∑ Fx = 0
→
⇒ Ax - FBC cos 30° = 0
⇒ Ax = √3P
+
∑ Fy = 0
↑
⇒ Ay - P - P + FBC sin 30° = 0
⇒ Ay = P
For the pin A
2
FA = √Ax 2 +Ay 2 = √(√3P) +P 2 = 2P = 2V
( FA = FBC = 2P)
⇒ V=P
𝜏𝑎𝑙𝑙𝑜𝑤 ≥
V
𝑃
=π
2
𝐴
4𝑑
π
π
2
⇒ P = 4 d 2 ⋅τallow = 4 (18⋅10-3 ) ⋅60⋅106 = 4860π N = 15.3 kN
Ans : P = 15.3 kN
5. A square piece of material is deformed into the dashed parallelogram.
(a) Determine the shear strain γ at corners A and D.
(b) Determine the average normal strain that occurs along the diagonals AC and BD.
y
15.18 mm
B
C
15.24 mm
15 mm
89.7 ﾟ
A
15 mm
D
x
15.18 mm
Sol:
𝜋
89.7°
𝜋
90.3°
(a) (𝛾𝐴 )𝑥𝑦 = 2 − ( 180° ) 𝜋 = 5.24 × 10−3 rad
(𝛾𝐷 )𝑥𝑦 = 2 − ( 180° ) 𝜋 = −5.24 × 10−3 rad
(b) 𝐴𝐶 = 𝐵𝐷 = √152 + 152 = 21.2132 𝑚𝑚
𝐴𝐶 ′ = √15.182 + 15.242 − 2 × 15.18 × 15.24 cos 90.3° = 21.5665 mm
𝐵 ′ 𝐷′ = √15.182 + 15.242 − 2 × 15.18 × 15.24 cos 89.7° = 21.4538 mm
𝜀𝐴𝐶 =
𝜀𝐵𝐷 =
A𝐶 ′ −𝐴𝐶
𝐴𝐶
= 0.01665 = 16.7 × 10−3
𝐵′ 𝐷 ′ −𝐵𝐷
𝐵𝐷
= 0.01134 = 11.3 × 10−3
Ans : (a) (𝛾𝐴 )𝑥𝑦 = 5.24 × 10−3 rad
(𝛾𝐷 )𝑥𝑦 = −5.24 × 10−3 rad
(b) 𝜀𝐴𝐶 = 16.7 × 10−3
𝜀𝐵𝐷 = 11.3 × 10−3