Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 1 (Q.No. 21)
21.
In case of a simply supported rectangular beam of span L and loaded with a central load W, the length of elasto-plastic zone of the plastic hinge, is
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Anshul Gupta said:   1 decade ago
Shape factor = 1.5.

Plastic moment/elastic moment = 1.5.

1.5 = L/(2x).

x = L/3.

Arghya said:   1 decade ago
I have a problem in that why we consider the elastic moment as 2x?

Sagnik said:   1 decade ago
It is 1.5 = Mp/Me.
1.5 = (L/2)/X.

As plastic hinge will form at L/2.

Nitu said:   10 years ago
Length of plastic hinge = Distance between Mp and My.

You can use a formula too = L (1-(1/shape factor)).

Shape factor for rectangular = 1.5.

Plastic length = L/3.
(1)

Zubair Ahmad said:   9 years ago
S.F = Plastic Moment/Elastic moment.
1.5(for rect. section) = L/2/x (plastic hinge at (L/2).
1.5 = L/2x.
x = L/1.5X2.
x = L/3.

Rana Muhammad Ashraf said:   9 years ago
X is what distance? can anyone explain.

Chandan said:   9 years ago
X is the distance from one end to plastic hinge.

Steve said:   7 years ago
I think x is any distance along the beam where strain can occur under a considerable amount of stress. hence, the modulus of elasticity is applicable here. Also, L/2 is where ultimate failure is likely to occur and hence it is where the plastic hinge is located, hence, the plastic modulus is applicable here and that is the idea of having the ratio as 0.5L/X=1.5.

Therefore X=0.33333333L=L/3.

Nungshi said:   7 years ago
Lp/L =(1-1/Ks) where Ks is shape factor. (For centrally concentrated load).

Lp/L = √(1-1/Ks) (for udl).
(1)

Dhananjay said:   7 years ago
Thanks for explaining @Nitu.


Post your comments here:

Your comments will be displayed after verification.