Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 22)
22.
A triangular section having base b, height h, is placed with its base horizontal. If the shear stress at a depth y from top is q, the maximum shear stress is
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 1 of 3.

Prem prakash said:   5 years ago
What is the use of q? Please explain.

Harsh Sharma said:   6 years ago
Qmax= (3/2)* Qavg -----------> (i)
Qavg = S/A.
Where S= Shear Force.
And A= (b*h)/2.

Hence, Qavg= 2S/(b*h) ----------> (ii)

Now from (i) & (ii).
Qmax = (3/2)* {2S/(b*h)}.

i.e. Qmax = 3S/(b*h).
(8)

Sachin s naik said:   6 years ago
Thank you @Krishna.

PRASENJIT said:   6 years ago
For triangular section:

Qmax = Qav*h^2/(12*I) ---> Eqn 1.

Again I for triangular section= 1/36 bh^3.
Now you put the value of "I" in equation no.1.
You will get the answer as option A.
Here Qav is denoted by S.

Amu bokato said:   6 years ago
Thank you @Krishna.

Sakib said:   7 years ago
Why 1.5? It should be multiflied by 4/3.
(1)

Khemraj bind said:   7 years ago
S = shear stress.
Area of triangle=0.5 * bh.

Maximum shear stress of triangle = Average shear stress of triangle 3/2.
= (3/2) * S/A,
= (1.5 S) /0.5 bh,
= 3S/bh.
The correct answer is option A.

Krutarth said:   7 years ago
Why 1.5* S?

Ziya said:   8 years ago
S = Shear force.

Maximum shear stress = 3S/bh.

Sheela said:   8 years ago
Thanks @Krishna.


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