Stress Analysis Assignment: Point A and B, Spring Semester

This assignment provides a detailed analysis of stress at points A and B. It includes calculations of shear stress, principal stress, and an assessment of yielding conditions. The solution explains the determination of stress values and whether yielding occurs based on the given parameters. The assignment's content is specifically designed to help students understand the application of stress analysis principles. Desklib provides this assignment to aid students in their studies.

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”