Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 6 (Q.No. 12)
12.
A square beam and a circular beam have the same length, same allowable stress and the same bending moment. The ratio of weights of the square beam to the circular beam is
Discussion:
12 comments Page 1 of 2.
Chinna said:
5 years ago
For square beam Z1 =a^3/6
For circular beam Z2=πd^3/32.
(Since having the same length, same allowable stress and same bending moment section modulus of both beam equals z1=z2).
(a^3)/6=π(d^3)/32
(d=1.193a) (d in terms of a).
(Weights of two beams proportional to cross-sectional areas )
W1/W2 =A1/A2 (area of square beam=a^2).
( Area of circular bean=πd^(2)by 4)
=( a^2)/π(d^2)/4 ( substitute d value )
= 1/1.118
W1/W2=1/1.12 (answer).
For circular beam Z2=πd^3/32.
(Since having the same length, same allowable stress and same bending moment section modulus of both beam equals z1=z2).
(a^3)/6=π(d^3)/32
(d=1.193a) (d in terms of a).
(Weights of two beams proportional to cross-sectional areas )
W1/W2 =A1/A2 (area of square beam=a^2).
( Area of circular bean=πd^(2)by 4)
=( a^2)/π(d^2)/4 ( substitute d value )
= 1/1.118
W1/W2=1/1.12 (answer).
(1)
Moun said:
6 years ago
Thanks @Ankit.
Asit said:
6 years ago
Thanks for this explanation.
Karan Panjabi said:
7 years ago
Great explanation @Amal, Your approach is Very easy to understand.
SATEESH said:
8 years ago
Weight for R:S:C=0.8:1:1.2.
Shivani said:
8 years ago
Your explanation is simple and easy to understand Thanks @Amal.
Neeraj said:
9 years ago
@Ravi Raj Anand.
You take any value a = 1 or 2 o r3. Infinite answer come same.
Good solution @Ankit .
You take any value a = 1 or 2 o r3. Infinite answer come same.
Good solution @Ankit .
Ankit said:
9 years ago
B.M is same so,
Z1 = Z2.
a^3/6 = 3.14 d^3/32,
= .098 d^3.
Here a^3 = .588 d^3,
a = .8377 d.
Now sigma1 = sigma 2.
W1/w2 = A1/A2.
A1 = a^2 and A2 = 3.14 * d^2 /4.
And put the value f a = .8377 in A1.
Then A1/A2 = .8929 = 1/ 1.12.
Z1 = Z2.
a^3/6 = 3.14 d^3/32,
= .098 d^3.
Here a^3 = .588 d^3,
a = .8377 d.
Now sigma1 = sigma 2.
W1/w2 = A1/A2.
A1 = a^2 and A2 = 3.14 * d^2 /4.
And put the value f a = .8377 in A1.
Then A1/A2 = .8929 = 1/ 1.12.
RAVI RAJ ANAND said:
9 years ago
@Amal.
Can you explain that what is the reason behind to take a = 1.
I am little confused by taking the value.
Can you explain that what is the reason behind to take a = 1.
I am little confused by taking the value.
AMAL said:
10 years ago
M/I = sigma/y = E/R.
Here bm is same so I/Y = c.
z1 = z2.
Lets take z1 for square,
z1 = a^2/6.
Now let a = 1 (suppose).
So z1 = 0.166667.
z2 (circle) = pi/32*d^3.
= 0.098*d^3.
Equalize them z1 = z2.
0.16667 = 0.098*d^3.
d = 1.1933.
Therefore area = pi/4*(1.19)^2 = 1.118 = 1.12.
Now weight is directly proportional to area (sigma = wt/area).
So W1/w2 = A1/A2 where area of square = a^2 = 1.
= 1/1.12.
Here bm is same so I/Y = c.
z1 = z2.
Lets take z1 for square,
z1 = a^2/6.
Now let a = 1 (suppose).
So z1 = 0.166667.
z2 (circle) = pi/32*d^3.
= 0.098*d^3.
Equalize them z1 = z2.
0.16667 = 0.098*d^3.
d = 1.1933.
Therefore area = pi/4*(1.19)^2 = 1.118 = 1.12.
Now weight is directly proportional to area (sigma = wt/area).
So W1/w2 = A1/A2 where area of square = a^2 = 1.
= 1/1.12.
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