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The rigid bar is supported by the pin-connected rod CB
that has a cross-sectional area of 20 m m...
Question:
The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 20mm2 and is made
from 6061-T6 aluminium.
Part A: Determine the vertical deflection of the bar at D when the load is applied.
Strain:
Deformation is the increase or decrease in the dimension of the member as the result of the load applied to the
member. Strain in the member is the ratio of deformation divided by the original dimension is called strain.
Answer and Explanation: 1
Given Data:
The uniform distributed load is: w
= 300 N/m
The area of cross section of rod is: A
2
= 20 mm
The figure below represents the free body diagram of ABD bar.
Free Body Diagram
The angle θ is,
AC
tan θ =
AB
1.5 m
tan θ =
2 m
θ = 36.869
∘
The concentrated point load of uniform load is,
W = w (AD)
W = (300 N/m) (4 m)
W = 1200 N
The concentrated point load acts at centroid.
Take the moment about A,
∑ MA = 0
W (AB) − P sin θ (AB) = 0
Here, P is the force in wire.
Substitute the values in the above equation.
(1200 N) (2 m) − P sin 36.869
∘
(2 m) = 0
P = 2000.041 N
The modulus of elasticity of the 6061-T6 aluminium is: E
The length of BC is,
= 69 GPa
BC = √AB
2
+ AC
BC = √(2 m)
2
2
+ (1.5 m)
2
BC = 2.5 m
The expression elongation in the wire is,
P (BC)
δBC =
AE
Substitute the values in the above equation.
1000 mm
(2000.041 N) (2.5 m ×
)
1 m
δBC =
2
1000 N/mm
2
(20 mm ) (69 GPa ×
)
1 GPa
δBC = 3.623 mm
The figure below represents the deformation of end D.
Deflection at D
The elongation in CB is,
δBC = BC
1 m
′
− BC
′
3.623 mm (
) = CB − 2.5 m
1000 mm
CB
′
= 2.503623 m
From the cosine rule, the displaced length of AB is,
′
(CB )
2
= (AC)
2
′
+ (AB )
2
∘
′
− 2 (AC) (AB ) cos(90
+ θ)
Substitute the values in the above equation.
l(2.503623)
∘
2
= (1.5 m)
2
+ (2 m)
2
∘
− 2 (1.5 m) (2 m) cos(90
−3
cos(90
+ ϕ) = −3.0213 × 10
∘
90
+ ϕ = 90.3464
ϕ = 0.1731
∘
∘
From the figure,
δ
tan ϕ =
AD
Substitute the values in the above equation.
tan 0.1731
∘
δ
=
4 m
δ = 0.0120846 m
Thus, the vertical deflection at D is 0.0120846
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Learn more about this topic:
.
m
+ ϕ)
What is Deformation? - Definition, Types & Process
from
Chapter 10 / Lesson 10
70K
Learn about the deformation definition, brittle deformation, and types of deformation. See different stresses
responsible for the deformation of the crust.
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