Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 2 (Q.No. 29)
29.
A square column carries a load P at the centroid of one of the quarters of the square. If a is the side of the main square, the combined bending stress will be
Discussion:
9 comments Page 1 of 1.
62908 said:
6 years ago
Total stress=direct stress+bending stress due to eccentricity at x-x axis And y-y axis.
Here, find the maximum bending stress.
So bending stress (maximum)
= {(p.ex)/z + (p.ey)/z}.......(1no.)
So here,( ex = b/4) ,and (ey=b/4). ( Z=I/y ,)
So,
I= b^4/12 and y= b/2.
So, Z ={(b^4/12)x(2/b)}=(b^3)/6.
Then put the value ...(1no.) eq.
={Px(b/4)}/{(b^3)/6}+{Px(b/4)}/{(b^3)/6}
={pb/4x6/(b^3)}+{pb/4x6/(b^3)}
={3p/(2b^2)}+{3p/(2b^2)}
=2x{3p/(2b^2)}
={3p/(b^2)} answer.
Here, find the maximum bending stress.
So bending stress (maximum)
= {(p.ex)/z + (p.ey)/z}.......(1no.)
So here,( ex = b/4) ,and (ey=b/4). ( Z=I/y ,)
So,
I= b^4/12 and y= b/2.
So, Z ={(b^4/12)x(2/b)}=(b^3)/6.
Then put the value ...(1no.) eq.
={Px(b/4)}/{(b^3)/6}+{Px(b/4)}/{(b^3)/6}
={pb/4x6/(b^3)}+{pb/4x6/(b^3)}
={3p/(2b^2)}+{3p/(2b^2)}
=2x{3p/(2b^2)}
={3p/(b^2)} answer.
(2)
Steve said:
7 years ago
I think it is 4p/a^2.
(1)
Abhik said:
6 years ago
Here, it is asked combined stress i.e, P/A2 + 2p/a2.
(1)
Naren limbu said:
6 years ago
Stress=p/A+-(mx/z+my/z).
= P/a2+3p/a2
=4p/a2.
= P/a2+3p/a2
=4p/a2.
(1)
Alex khanal said:
12 months ago
Combined bending stress = Mx/Z+My/Z.
= (P*a/4)/(a3/6) + (P*a/4)/(a3/6),
= 3P/a2.
= (P*a/4)/(a3/6) + (P*a/4)/(a3/6),
= 3P/a2.
(1)
Raja said:
8 years ago
Please explain it.
Chhaya said:
8 years ago
It is (p/a^2)+Mxx/Z+Myy/Z.
Bhushan said:
8 years ago
But it is 4p/a^2.
Smt u .s. lakshmi said:
6 years ago
Please, I don't understand need some more explanation.
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