Stress Analysis Homework

This homework assignment focuses on stress analysis problems. The solutions demonstrate calculations of principal stresses, shear stresses, and strain tensors using various methods, including the finite element method. The problems involve analyzing stress at different points in a cylinder under internal pressure and axial loads, determining if yielding occurs, and applying the finite element method to a simple beam problem. The solutions compare results obtained from different methods, highlighting the differences and reasons for discrepancies. The assignment covers fundamental concepts in mechanical engineering, particularly relevant to stress and strain analysis.

Answer (2)
Given data:
ri=0.5 m, di=1 m
r 0=0.52 m, d 0=1.04 m
P=100 bars
t=r 0−ri=0.52−0.50=0.02m
k = ro
ri = 0.52
0.50 =1.04
Stress at point (A)
Internal pressure,
Axial tension σx 1= p a
2 t =100 x 0.52
0.04 =1300 N
m 2
Hoop tension σy= p a
t =100 x 0.52
0.02 =2600 N
m2
Axial compression
Axial Normal stress σx 2= P
( k2−1 ) = 100
( 1.042−1 ) =1225.4901 N
m 2
Twisting moment:
Maximum out-plane shear stress, τ = Pr
2t = 100 x 0.52
0.04 =1300 N
m 2
Answer (a)
State of stress:
σx=σx 1+σx 2=1300+1225.5=2525.2 N
m2
σy=2600 N
m2
τ xy=1300 N
m2
σ Z=τ yz =τ xz=0
The maximum in-plane shear stress:
τ xy= √ ( σ x−σ y
2 )2
+ τxy
2= √1387.56+1690000=1300.50 N
m 2
1 | P a g e

Answer (b)
Principal stresses:
σ 1= σ x +σ y
2 + √ ( σ x−σ y
2 )2
+τ xy
2
¿ 2525.2+ 2600
2 +¿1300.50
¿ 3863.1 N /m2
σ 2= σx +σ y
2 − √ ( σ x−σ y
2 )2
+ τxy
2
¿ 2525.2+2600
2 −1300.50
σ 2=1261.7 N
m 2
σ 3=0
Answer (c)
τ max= ( σ 1−σ 3 )
2 =3863.1
2 =1931.55 N
m2
Yielding not occur because maximum shear stress less than yield stress.
Stress at point (B)
Internal pressure,
Axial tension σx 1= p a
2 t =100 x 0.50
0.04 =1200 N
m 2
Hoop tension σy= p a
t =100 x 0.50
0.02 =2500 N
m2
Twisting moment:
Maximum out-plane shear stress, τ = Pr
4 t = 100 x 0.50
0.04 =625 N
m2
Answer (A)
State of stress:
2 | P a g e

σx=1200 N
m2
σy=2500 N
m2
τ xy=625 N
m2
σ Z=τ yz =τ xz=0
The maximum in-plane shear stress:
τ xy= √ ( σ x−σ y
2 )2
+ τxy
2= √1387.56+1690000=901.73 N
m2
Answer (B)
Principal stresses:
σ 1= 1200+2500
2 + √ ( σ x−σ y
2 )
2
+τ xy
2
¿ 1850+901.73
¿ 2751.73 N / m2
σ 2= σx +σ y
2 − √ ( σ x−σ y
2 )2
+ τxy
2
¿ 1850−901.73
σ 2=948.27 N
m2
σ 3=0
Answer(c)
τ max= ( σ 1−σ 3 )
2 =2751.73
2 =1375.865 N
m2
Shear stress of given cylinder lower then yield stress, so yielding not occur.
3 | P a g e

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
1000 =0.001
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.
StrainTensor L= [ 0.001 0.0005 0
0.0005 1 0.0005
0 0.0005 1 ]
Answer (C)
As given m = 9500mm so the change in length = 500mm so that following strain
tensor obtained
ε = 500
1000 =0.5
Following matrix obtained
ε ij= [ 0.5 0.25 0.25
0.25 1 0
0.25 0 1 ]
σ =ε ij E
Converting into meter and obtained following equation
4 | P a g e

¿ 1000 x [ 0.5 0.25 0.25
0.25 1 0
0.25 0 1 ]
σ ij= [ 500 250 250
250 1000 0
0 25 1000 ] N /mm 2
F=σ x 1 x 1 x Area
Fx 1 x 1=500 x ( 10 x 10 )
Fx 1 x 1=50000 N
Answer (4):
Using two element each of 50 mm long in length, the finite element model in figure below. Nodes
and elements are numbered as shown. Noted that the area at the mid-point of the plate is as
discussed following:
A ( x )= A ( 0 ) (1− x
2 L )
A 1 ( 0 )=10 mm2
L=100 mm
Let us take area X = L,
A ( L )=10 (1− 1
2 )
A 2 ( L )=5 mm 2
So, the area at mid span A 3= A 1+ A 2
2 =10+5
2 = 7.5mm2
Area of top most ¿ position= 10+7.5
2 =8.75 mm 2
5 | P a g e

Area of bottom most position= 7.5+5
2 =6.25 mm 2
Boundary condition F 1=0 , F 3=1 N
Element stiffness matrix Ke= EA
L [ 1 −1
−1 1 ]
K1= 1000 x 8.75
50 [ 1 −1
−1 1 ]
K2= 1000 x 6.25
50 [ 1 −1
−1 1 ]
The global stiffness matrix K assembled
K= 1000
50 [ 1 −8.75 0
−8.75 15 −6.25
0 −6.25 6.25 ]
The element body force vector are
Fe= Alρ
2 [1
1 ]
F1= 8.75 x 50 x 1000
2 [1
1 ]
F2=6.25 x 50 x 1000
2 [ 1
1 ]
The force applied
F1= 8.75 x 50 x 1000
2 [1
1 ]
F1=[ 0.218750
0.218750 ]
F2=6.25 x 50 x 1000
2 [ 1
1 ]
F2=
[0.156250
0.156250 ]
6 | P a g e

The global load vector F is assembled from f 1 , f 2 and the point load F = 1N added
F= [ 0.218750
0.37500
1.15625 ]
Now using formula in order to obtained displacement F=kx
1000
50 [ 1 −8.75 0
−8.75 15 −6.25
0 −6.25 6.25 ][u 1
u 2
u 3 ]= [0.218750
0.37500
1.15625 ]
sinceu 1=0 due ¿ ¿ postion of vertical rod , eliminate first raw∧first coloum
1000 x 6.25
50 [ 15 −6.25
−6.25 6.25 ][u 2
u3 ]= [ 0.375
1.15625 ]
Solving above ¿
u 2=1.4 mm
u 3=3.35952 mm
To obtain stress
σ =EBq
σ 1=1000 x 1
50 [ −1 1 ] [ 0
1.4 ] =28 N
mm 2
σ 2=1000 x 1
50 [ −1 1 ] [ 1.4
3.35952 ]=39.1904 N
mm 2
Answer (b)
X = 50 mm, at mid span.
A ( 50 ) = A ( 0 ) ( 1− 50
200 )
¿ 10 ( 0.75 )
A ( 50 )=7.5 mm 2
A ( x )= A ( 0 ) (1− x
2 L )
A ( L )=10 (1− L
2 L )
¿ 5 mm 2
7 | P a g e

Lets take x = 99 mm
A ( 99 ) =A ( 0 ) (1− 99
2 x 100 )
¿ 10 ( 0.505 )
¿ 5.05 mm
σ = ρg ( L−x )
2 (1+ A ( L )
A ( x ) )
¿ ρg ( 100−50 )
2 ( 1+ A ( L )
A ( 0 ) )
¿ ρg ( 100−50 )
2 (1+ 5
10 )
σ =ρg ( 37.5 )
¿ 1000 x 9.8 x (37.5)
σ =36.75 N / mm 2
σ = ρg ( 100−99 )
2 ( 1+ A ( L )
A ( 99 ) )
¿ ρg ( 100−99 )
2 ( 1+ 5
5.05 )
ρg ( 100−99 )
2 ( 1.99 )
1000 x 10 x ( 1 )
2 ( 1.99 )
σ =99.50 N /mm 2
The finite element method and other calculative method have different answer because finite
element consider behaviour at the element level whereas number of elements increasing more
accurate results obtained. In calculative method considering entire body and so that results may
varies from finite element method.
8 | P a g e