Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 6 (Q.No. 23)
23.
In a framed structure, as shown in the below figure, the forces in the members AB and BC are respectively
(tensile) and 2W (compressive)
2W(tensile) and (compressive)
(tensile) and (compressive)
none of the above
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
5 comments Page 1 of 1.

Rahul Thakur said:   5 years ago
Tan30° =w/CD = 1/√3 = CD = √3W.
Cos30 = CD/BC = √3/2= √3W/BC .
So, BC=2W.
(1)

GAURAV said:   6 years ago
I got another answer, can anyone explain?
(2)

KRUNAL PRAJAPATI said:   8 years ago
Full Solution:

--AB/sin(360-(90+30)) = BC/sin(90) = W/sin(30).
--AB/sin(240) = BC/sin(90)= W/sin(30).
--AB/sin(180+(90-30)) = BC/sin(90)= W/sin(30).
--AB/[-sin(90-30)] = BC/sin(90)= W/sin(30).
--AB/[-cos(30)] = BC/sin(90)= W/sin(30).
--2AB/(root3) = BC = 2W.

So, the ans is AB=(root3) * W & BC=2W.

GVR Murty said:   10 years ago
Answer: Fab=W, Fbc=-2W;

Equation: Fab/Sin240=Fbc/Sin90=-W/Sin30.

Mbkavad said:   1 decade ago
Sin rule ab w*(sin60/sin30) = root 3. Similar to Wac.

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