Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 2 (Q.No. 7)
7.
A bar of square section of area a2 is held such that its one of its diagonals is vertical. The maximum shear stress will develop at a depth h where h is
Discussion:
7 comments Page 1 of 1.
Dev said:
2 years ago
For rhombus, max shear occurs at 3d/8 from apex where d is diagonal now d = √ a (side)
substitute d=a^(1÷2) .
We get 3a/(4×2^(1÷2)).
substitute d=a^(1÷2) .
We get 3a/(4×2^(1÷2)).
Saeed said:
3 years ago
it's simple.
The shape of the bar is rhombus so for this max shear =9/8 avg.
3 &radic2/4 =9/8.
The shape of the bar is rhombus so for this max shear =9/8 avg.
3 &radic2/4 =9/8.
(1)
Pramit Samanta said:
5 years ago
The correct Answer is 3a/4√2.
(1)
Priya Kumari said:
5 years ago
If we assume 'a' side of a square and join opposite vertex of square then we will obtain 4 right angle triangle. Using Pythagoras theorem base and height of a single right triangle will be a/√2 = avg. Shear stress.
Now we know that centroid of a traingle is 2/3 (a/√2) from vertex = avg. Shear stress. Formula of maximum shear stress is 9/8 * avg shear stress = 9/8* 2/3* a/√2 = 3a/4√2.
Now we know that centroid of a traingle is 2/3 (a/√2) from vertex = avg. Shear stress. Formula of maximum shear stress is 9/8 * avg shear stress = 9/8* 2/3* a/√2 = 3a/4√2.
(1)
Abhishek said:
5 years ago
It's a rhombus when diagonal vertical. So Maximum shear stress is 9/8 avg shear stress.
That's why its answer is b.
That's why its answer is b.
(1)
Rosni said:
8 years ago
Please clarify this with an explanation.
(1)
Firas jandali said:
1 decade ago
The question makes no sense lacking clarity, let alone the answer. Clarify the question then the answer becomes easy.
(1)
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