Shear Stress Analysis in Structural Components

The assignment content discusses various problems related to stress and strain in a rod under different conditions. It involves calculating the principal stresses, shear stresses, and yielding at points A and B. Additionally, it also deals with finite element analysis and obtaining accurate results by dividing elements at each mm variation of stress.

Answer (2):
Given data:
d=0.2m
L=1 m
T =20 π
Stress at point (A)
∑ Fx=0
V x−20 π =0
∑ Fy=0
V y=−20 π
∑ My=0
σ yb= N
A = 20
0.1 π = 200
π =63.66
σ y= M y
I
I = π R4
4 = π ( 0.1 )4
4 =0.25 π x 10−4 m4
y= 4 R
3 π
σ y= 20 π x 1 x 4 x 0.1
3 π x 0.25 π x 10−4
¿ 3.395305 x 104
σ y=63.66+33953.05
σ y=34016.71
Find, shear stress
τ y=V y Q
I t
Q=∑ A−1 y−1
¿ π ( 0.1 ) 2
2 ( 4 R
3 π )
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¿ π ( 0.1 )2
2 ( 4 x 0.1
3 π )
Q=6.66 x 10−4
I = π R4
4 = π ( 0.1 ) 4
4 =0.25 π x 10−4 m4
τ A = 20 π x 6.66 x 10−4
0.25 π x 10−4 x 0.2
τ A =2664 kN /m2
Answer (B)
As per given condition if maximum shear stress greater then yield stress then yield occur, but value
of maximum shear stress is lower than yield stress therefore yielding not occur at point “A”
Principal stress at point A,
σ 1,2=( σ x +σ y
2 )± √ ( σ x +σ y
2 )2
+ ( τxy )2
¿ 34016.71+ 0
2 ± √ ( 34016.71+0
2 )
2
+ ( 2664 )2
¿ 17008.355 ±34120.327
σ A=51128.682 kN
m2 (Tensile)
σ A=−17111.972 kN
m2 (Compressive)
Answer (c)
τ max=± √ ( σ x +σ y
2 )2
+τxy
2
¿ √ ( 34016.71+0
2 )2
+26642
¿ 34120.327
¿ 34. 120 Mpa
As per given condition if maximum shear stress greater then yield stress then yield occur, but value
of maximum shear stress is lower than yield stress therefore yielding not occur at point “A”
Calculation for point: B
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Answer (A)
Stress at the point (B)
∑ Fy=0
V y−(−20 π )=0
∑ My=0 ; Mx−20 π ( 1 ) =0
Mx=−20 π
∑ Fx=0
V x=0
∑ Fz=0
V Z =0
Shear stress at point B,
τ B= V y Q
I t
I = π R4
4 = π ( 0.1 ) 4
4 =0.25 π x 10−4 m4
Q=∑ A−1 y−1
¿ π ( 0.1 ) 2
2 ( 4 R
3 π )
¿ π ( 0.1 )2
2 ( 4 x 0.1
3 π )
Q=6.665 x 10−4
τ B= 20 π x 6.665 x 10−4
0.25 π x 10−4 x 0.2
¿−2666 kN /m2
σ B= M y
I = 20 π x ( 1 ) x ( 0.1 )
0.25 π x 10−4 =80,000 kN /m2
σ y=σ Z=τxz=τ yz=0
Answer (B)
Principal stress at point B,
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σ 1,2=( σ x +σ y
2 )± √ ( σ x +σ y
2 )2
+ ( τxy )2
¿ 80,000+ ¿
2 ± √ ( 80000+0
2 )2
+ (−2666 )2 ¿
σ 1,2=40000 ± 40088.74
σ 1,2=80088.74 (Tensile) kN /m2
σ 1,2=−88.74 ( Compressive )
Answer (c)
Yielding not occur at point “B”.
τ max=± √ ( σ x +σ y
2 )2
+τxy
2
¿ √ ( 80000+0
2 )2
+ 26662
¿ 40088.74 kN
m2
¿ 40.088 Mpa
As per given condition if maximum shear stress greater then yield stress then yield occur, but value
of maximum shear stress is lower than yield stress therefore yielding not occur at point “B”
Answer (3):
StrainTensor= [εxx εxy εxz
εyx εyy εyz
εzx εzy εzz ]
¿
[ ∂ u
∂ x
1
2 ( ∂u
∂ y + ∂ v
∂ x ) 1
2 ( ∂ u
∂ z + ∂ w
∂ x )
1
2 ( ∂ u
∂ y + ∂ v
∂ x ) ∂ v
∂ y
1
2 ( ∂ v
∂ z + ∂ w
∂ y )
1
2 ( ∂u
∂ z + ∂ w
∂ x ) 1
2 ( ∂ v
∂ z + ∂ w
∂ y ) ∂ w
∂ z
]Assume 1mm pressed after applied vertical force,
So the strain in vertical direction = change∈lenght
originallenght = 1
10 =0.1
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Whereas die fixed test specimen in horizontal direction and Z-direction.
Therefore specimen cannot expand in X and Z direction or X1 and X3 direction.
The remaining component values are 1.
¿ [ 1 0.5 0
0.5 0.1 0.5
0 0.5 1 ]
Answer (B)
f ( m ) =l
f ( x 2 ) =x 1
[ ∂ u
∂ x 1
1
2 ( ∂u
∂ x 2 + ∂ v
∂ x 1 ) 1
2 ( ∂ u
∂ x 3 + ∂ w
∂ x 1 )
1
2 ( ∂ u
∂ x 2 + ∂ v
∂ x 1 ) ∂ v
∂ x 2
1
2 ( ∂ v
∂ x 3 + ∂ w
∂ x 2 )
1
2 ( ∂ u
∂ x 3 + ∂ w
∂ x 1 ) 1
2 ( ∂ v
∂ x 3 + ∂ w
∂ x 2 ) ∂ w
∂ x 3 ]let us consider u=l , v=m∧w=10 x 3
¿
[ ∂l
∂ x 1
1
2 ( ∂ l
∂ x 2 + ∂ m
∂ x 1 ) 1
2 ( ∂ l
∂ x 3 + ∂(10 x 3)
∂ x 1 )
1
2 ( ∂ l
∂ x 2 + ∂ m
∂ x 1 ) ∂ m
∂ x 2
1
2 ( ∂ m
∂ x 3 + ∂(10 x 3)
∂ x 2 )
1
2 ( ∂ l
∂ x 3 + ∂(10 x 3)
∂ x 1 ) 1
2 ( ∂ m
∂ x 3 + ∂(10 x 3)
∂ x 2 ) ∂ 10 x 3
∂ x 3
]
¿
[ 1 1
2 ( 1+1 ) 1
2 ( 0+0 )
1
2 ( 1+1 ) 1 1
2 ( 0+0 )
1
2 ( 0+0 ) 1
2 ( 0+0 ) 0 ]
¿ [ 1 1 0
1 1 0
0 0 0 ]
For compressive stress σ =Eε
σ =E [1 1 0
1 1 0
0 0 0 ] N /mm2
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Answer (C)
As given m = 9.5 so the change in length = 0.5mm so that following strain
tensor obtained
ε ij= [ 1 0.025 0
0.025 0.05 0.025
0 0.025 1 ]
σ =ε ij E
¿ 1000 x [ 1 0.025 0
0.025 0.05 0.025
0 0.025 1 ]
σ ij= [ 1000 25 0
25 50 25
0 25 1000 ] N /mm 2
F=σ x 1 x 1 x Area
Fx 1 x 1=1000 x ( 10 x 10 )
Fx 1 x 1=105 N
Answer (4)
Given data:
E=3 Gpa=3000 N
mm 2
ρ=1000 kg
m3
α =100 x 10−6
℃
∆ T =1 ℃
Note: the rod is dividing into two segment with three nodes point.
The element stiffness matrix are,
K1= AE
50 [ 1 −1
−1 1 ]
K2= AE
50 [ 1 −1
−1 1 ]
Thus,
The global stiffness matrix K assembled
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K= AE
50 [ 1 −1 0
−1 2 −1
0 −1 1 ]
Now, in assembling F, both temperature and point load effects have to be consider. The element
temperature force due to ∆ T =1 ℃ are obtained
θ1= AEαT [ −1
1 ]
And
θ2= AEαT [ −1
1 ]
Upon assembling θ1∧θ2 and the point load, we get
F= AEαT [−1
0
1 ]
F=Kx using equation∈order ¿ get displacement
AE
50 [ 1 −1 0
−1 2 −1
0 −1 1 ][ u 1
u 2
u 3 ] =AEαT [ −1
0
1 ]
elemenating first raw∧first coloum
AE
50 ( 2u 2−u 3 ) =0
2 u 2=u 3
−u 2+u 3=αT
put value of u 3=2u 2∈above equation
u 2=αT
u 2=10−4 mm
u 3=2 x 10−4 mm
In evaluate elements stresses, we have using following equation,
σ 1= E
l 1 [−1 1 ] [ 0
10− 4 ]−EαT
¿ 3000
50 ( 10−4 )−3000 x 10−4 x 1
¿ 60 x 10−4−3000 x 10−4
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σ 1=−0.294 N
mm 2 … …(1)
σ 2= E
l 1 [ −1 1 ] [ 10− 4
2 x 10−4 ]−EαT
¿ 3000
50 (−10−4 +2 x 10−4 )−3000 x 10−4 x 1
σ 2=−0.294 N
mm 2 … … (2)
Answer (B)
σ =−ρg ( L−x )
Let x=50 mm ,
σ x=50=−1000 x 10 x ( 0.1−0.05 )
σ =−0.5 N
mm 2 … . ( 3 )
x=99 mm
σ x=99=−1000 x 10 x ( 0.1−0.099 )
¿−10 N
mm 2 … … … …(4 )
Compare results of (1),(2) with (3) and (4) . The results are major difference. In order to get
accurate result more elements need to divide at each mm variation of stress. The finite element
analysis use for obtaining accurate solution with least error. Whereas given calculative method
considering entire body so that probable error occur.
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