AE - May XXXXXXXXXXMec-A6-2

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AE - May XXXXXXXXXXMec-A6-2

Answered Same Day Dec 23, 2021

Solution

Robert answered on Dec 23 2021

124 Votes

4) Given data:
Yield Stress
MPa
Y
300
=
s
Compressive Force P = 180 KN
Applies Torque
m
KN
T
-
=
20
Factor Of Safety
2
=
n
a) Now as we know for the square cross-section
÷
÷
ø
ö
ç
ç
è
æ
=
9
2
3
T
t
So after Putting the value of T we get
2
3
90
m
KN
÷
ø
ö
ç
è
æ
=
t
Now Using P Longitudinal Stress
2
2
180
m
KN
÷
ø
ö
ç
è
æ
=
s
Now Using the Principal Stress formula We get,
))
)
2
((
2
2
2
1
t
s
s
s
+
+
=
……………….(1)
))
)
2
((
2
2
2
2
t
s
s
s
+
-
=
………………..(2)
Now according to Max Shear Stress theory

n
Y
s
t
£
max
Now
÷
ø
ö
ç
è
æ
-
=
2
2
1
max
s
s
t
So from eq. (1) and (2) we get,
))
)
2
((
2
2
max
t
s
t
+
=
Now Putting the values of above we get,
2
1000
*
300
))
)
2
((
2
2
=
+
t
s
Or,
150000
))
)
90
(
)
2
180
((
2
3
2
2
=
+
Solving the above equation we will get the value of ‘b’.
) If the load P acts parallel to the cross-section at the same point then only thing will be changed is that now it will produce bending-moment ‘M’ which will be equal to P*L where ‘L’ will be the length of the beam.
Now as we know
y
I
M
s
=
Where I is the moment of inertia about the C.G and y will be
2 in this...

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AE - May XXXXXXXXXXMec-A6-2

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