Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 7 (Q.No. 36)
36.
For a given stress, the ratio of moment of resistance of a beam of square cross-section when placed with its two sides horizontal to the moment of resistance with its diagonal horizontal, is
Discussion:
11 comments Page 1 of 2.
Sigh said:
9 years ago
Very nice explanation of the answer @Shobhit.
(2)
Sushil said:
1 decade ago
As per my calculation option (a) is correct, if anybody know the proof as per option (d) please comment.
Hemakumar said:
1 decade ago
Ya answer is right. Compare the MI of the square and diamond.
Venkat said:
9 years ago
Can anyone provide the proof for the solution?
Shobhit said:
9 years ago
Let a = side of sq beam.
Section modulus of sq beam with its two sides horizontal(Z1) = a^4/{12 * a/2} = a^3/6.
Section modulas of sq beam with its diagonal horizontal(Z2) = a^4/[12*a/sqrt2]
= a^3/{6sqrt2.
Now the moment of resistanse of a section is directly proportional to their section modulas so
M1/M2 = Z1/Z2 = (a^3 * 6sqrt2) / (6 * a^3) = sqrt2.
Section modulus of sq beam with its two sides horizontal(Z1) = a^4/{12 * a/2} = a^3/6.
Section modulas of sq beam with its diagonal horizontal(Z2) = a^4/[12*a/sqrt2]
= a^3/{6sqrt2.
Now the moment of resistanse of a section is directly proportional to their section modulas so
M1/M2 = Z1/Z2 = (a^3 * 6sqrt2) / (6 * a^3) = sqrt2.
Afaq said:
9 years ago
Can any one elaborate this?
Bivash Mondal said:
7 years ago
M1/M2=fz1/fz2=z1/z2=y2/y1=a√2/a=√2,
Where a is the side of the square,
Z=I/Y.
Where a is the side of the square,
Z=I/Y.
Rupesh said:
7 years ago
Nice explanation, Thanks @Bivash.
Soudamini.B.S. said:
7 years ago
Let us consider the side of a square to be 'h' m
For square section,
I1=(h^4)/12.
Y1(max)=h/2.
M=(f/y)*l.
M1=(f*h^3)/6.
For rhombus i.e along the diagonal of a square,
I2=h^4/12.
But y(max)=(h*√2)/2=h/√2.
M=(f/y)* I.
M2=(√2fh^3)/12.
M1/M2= √2.
For square section,
I1=(h^4)/12.
Y1(max)=h/2.
M=(f/y)*l.
M1=(f*h^3)/6.
For rhombus i.e along the diagonal of a square,
I2=h^4/12.
But y(max)=(h*√2)/2=h/√2.
M=(f/y)* I.
M2=(√2fh^3)/12.
M1/M2= √2.
Karan said:
6 years ago
@Soudamini. B.S.
Great explanation.
Great explanation.
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