Shear Stress Yielding Analysis

The assignment discusses the concept of stress and yielding in mechanics. It highlights that yielding does not occur at point 'A' because the maximum shear stress is lower than the yield stress. Similarly, it concludes that yielding also does not occur at point 'B' due to the same reason.

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 yb= 20 π x 1 x 4 x 0.1
3 π x 0.25 π x 10−4
¿ 3.3953 05 x 104
σ y=63.66+33953.05
σ y=34016.71
Find, shear stress
τ By=V y Q
I t
Q=∑ A−1 y−1
¿ π ( 0.1 ) 2
( 4 R
3 π )
¿ π ( 0.1 ) 2
( 4 x 0.1
3 π )

Q=13.33 x 10−4
I = π R4
4 = π ( 0.1 ) 4
4 =0.25 π x 10−4 m4
τ A = 20 π x 13.33 x 10−4
0.25 π x 10−4 x 0.2
τ A =5332 kN /m2
Answer (B)
Yielding not occur at point “A”.
τ max=± √ ( σ x +σ y
2 )2
+ τxy
2
¿ √ ( 80000+0
2 )2
+ 53322
¿ 40353.81 kN
m2
¿ 40.353 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”
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
+ ( 5332 ) 2
¿ 17008.355 ±34432.060
σ A= 51440.4153 kN
m2 (Tensile)
σ A=−17423.705 kN
m2 (Compressive)
Answer (c)
Yielding not occur at point “A”.
τ max=± √ ( σ x +σ y
2 )2
+ τxy
2

¿ √ ( 34016.71+0
2 )2
+53322
¿ 34432.060
¿ 34.432 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”
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
( 4 R
3 π )
¿ π ( 0.1 ) 2
( 4 x 0.1
3 π )
Q=13.33 x 10−4
τ B= 20 π x 13.33 x 10−4
0.25 π x 10−4 x 0.2
¿−5332 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,
σ 1,2=( σ x +σ y
2 )± √ ( σ x +σ y
2 )2
+ ( τxy )2
¿ 80,000+ ¿
2 ± √ ( 80000+0
2 )2
+ ( 5332 )2 ¿
σ 1,2=40000 ± 40353.81
σ 1,2=80353.81(Tensile)kN /m2
σ 1,2=−353.81 ( Compressive )
Answer (c)
Yielding not occur at point “A”.
τ max=± √ ( σ x +σ y
2 )2
+ τxy
2
¿ √ ( 80000+0
2 )2
+ 53322
¿ 40353.81 kN
m2
¿ 40.353 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”