Comparison of Different Methods for Stress Analysis

The assignment content discusses stress and displacement calculations for a cylindrical rod using different methods, including conventional and Finite Element Methods (FEM). The problem involves calculating stresses at points A and B and displacements between three nodal points along the bar. The solutions obtained through both methods are compared, highlighting the advantages of FEM in achieving more accurate results.

Year: 2013/14
Answer (2):
Given data:
d=0.2m
L=1 m
T =10 π
Stress at point A,
σ x=σ y=σ Z=τxy =τ xz=τ yz =0
Stress at the point (B)
∑ Fx=0
V x−20 π =0
∑ Mx=0 ; Mx−20 π ( 1 ) =0
Mx=20 π
∑ Fy=0
V y=0
∑ Fz=0
V Z =0
Stress at point A,
σ x=σ y=σ Z=τxy =τ xz=τ yz =0
Principal stress at point A,
Since the point A is the origin of the cylinder rod and there are no external applied load, so
the principal stress as following,
σ 1=0 ( Tensile , compressive )
τ =0
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Shear stress at point B,
τ B= V x 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
( σ x+ σ y
2 )
2
+ ( τxy ) 2
Principal stress at point B,
σ 1,2=( σ x +σ y
2 )± √ ( σ x +σ y
2 )2
+ ( τxy )2
¿ 80,000+0
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 )
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Answer (c)
Yielding not occur at point “A”.
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 (4):
(A) The bar element separated by three nodal point. At the fixed point of bar, middle
point of bar and at the end point of bar.
A=1mm2
The following equation obtain in order to find the displacement and stress.
u ( x )=(1− x
l )u1 + ( x
l )u2
u= [ N ] { d }
Where N=[ ( 1− x
l ) x
l ]
{ d }= [u1
u2 ]
ε x= du
dx
¿ ( −1
l ) u1 + ( 1
l ) u2
[ −1
l
1
l ] {u1
u2 }
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[ B ] =[ −1
l
1
l ]
σ x=E εx
¿ E [ B ] { d }
σ x= [ D ] [ B ] { d }
Ke= [ 1 −1
−1 1 ]
The force element equation f =∫
0
l
[ N ] T
q dt
¿ [ K ]e d [u1
u2 ]
Now , [ K ]e= AE
l [ 1 −1
−1 1 ]
[ K ] ( d ) = ( f )
AE
l [ 1 −1
−1 1 ] [u1
u2 ]= { f }
f e=∫
0
l
[ N ]
T
q dt
f 1
e=∫
0
l
[ ¿
1− x
l
x
l
] ρAg dx ¿
¿∫
0
l /2
[¿
1− x
l
x
l
]ρAg { l
2 }dx +∫
l /2
l
[¿
1− x
l
x
l
] { l
2 } ρAgdx ¿ ¿
¿
[1− x
l
x
l ] ρAgl
¿ (1− x
l ) ρAgl+ x
l ( ρAgl )
f 1
e=ρAgl
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f 2
e=∫
l /2
l
[¿ 1− x
l ] ρAg dx +∫
l
2
l
[¿ x /l] ρAg dx ¿¿
[ 1− x
l ] l
2
l
ρAg dx + [ x
l ] l
2
l
ρAg dx
¿ lρAg
Ke= AE
l
2
[ 1 −1 0
−1 2 −1
0 −1 1 }{u1
u2
u3
}=
[ lρg
4
lρg
2
lρg
4 ]Ke= 2 AE
l ¿
u2= 3
4
l2 ρg
E = 3
4 x 1002 x 10−6 x 9806.65
4 x 1000 =0.024516 mm
u3= l2 ρg
2 E =1002 x 10−6 x 9806.65
2 x 1000 =0.04903325 mm
u1=0 ,u2=0.024516
put the value∈following equation
ε x= du
dx =+ ( 1
100 )0.04903325
σ x=E εx
¿ 1000 x 0.0004903325
σ x=4.903325 kN /mm 2
Use value of u2=0.024516 mm ,u3=0.04903325 mm
ε x= du
dx = ( −1
100 )0.024516+( 1
100 )0.04903325
¿−0.00024516+0.0049033
ε x=0.00514846
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σ x=E εx
¿ 1000 x 0.00514846
σ x=5.14846 kN
mm 2
Answer (B)
σ n=ρcv
v= √2 gh= √2 x 9.8 x 0.05=0.9899 m
s 2
c= √ E
ρ
¿ √ 1000
1000
¿ 1
σ n=(1000 x 1 x 0.9899)100
σ n=9.899 kN
mm 2
As shown in above two solution there are noticeable difference by using different method
such as Finite element method and conventional method. If require more accurate result then
it is recommend to use Finite element method
As the above obtained value the result can be improve by divide the entire segment into
discretise parts. As per problem requirement the segment have divided in to two segment
only, so displacement and stress obtain two points only. Except fixed position. If the
cylindrical segment divided into more than two segment, more accurate results obtain. As
segment divided into small parts more accurate results obtain.
*that is also define in finite element analysis definition. The FEA deals with the small
segment in order to get accurate answer in analytical method.
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