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)
(compressive)
(tensile) and (compressive)none of the above
Discussion:
5 comments Page 1 of 1.
Mbkavad said:
1 decade ago
Sin rule ab w*(sin60/sin30) = root 3. Similar to Wac.
GVR Murty said:
10 years ago
Answer: Fab=W, Fbc=-2W;
Equation: Fab/Sin240=Fbc/Sin90=-W/Sin30.
Equation: Fab/Sin240=Fbc/Sin90=-W/Sin30.
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.
--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.
GAURAV said:
6 years ago
I got another answer, can anyone explain?
(2)
Rahul Thakur said:
5 years ago
Tan30° =w/CD = 1/√3 = CD = √3W.
Cos30 = CD/BC = √3/2= √3W/BC .
So, BC=2W.
Cos30 = CD/BC = √3/2= √3W/BC .
So, BC=2W.
(1)
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