Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 16)
16.
In the cantilever truss as shown in below figure, the horizontal component of the reaction at A, is
Discussion:
15 comments Page 1 of 2.
Mahesh said:
3 years ago
Answer:A.
(5+5+5)*10 = 5h,
15 * 10 = 5h,
150 = 5h,
h = 150÷5,
h = 30.
(5+5+5)*10 = 5h,
15 * 10 = 5h,
150 = 5h,
h = 150÷5,
h = 30.
(3)
Jahandad khan said:
4 years ago
@Suresh.
Where is point B? Explain it clearly.
Where is point B? Explain it clearly.
(1)
Nikhil Dongre said:
5 years ago
10 is the distance or load?
(1)
Nikhil Dongre said:
5 years ago
@Suresh.
Where is the point B? Explain about it.
Where is the point B? Explain about it.
Chivi said:
6 years ago
(5+5+5)*10 = h*5.
h = 150÷5,
h = 30ton.
h = 150÷5,
h = 30ton.
Niraj said:
7 years ago
10x15 = Hx5.
H = 30 tonnes.
H = 30 tonnes.
(1)
Tayu said:
8 years ago
You are correct @Aung.
(1)
Aung Myin Bwar said:
8 years ago
Considering Zero moment at point A, B is point below point A.
Rbx=horizontal reaction at B, Rax=horizontal reaction at A
Ma=moment at A.
E Ma=0
(-Rbx5)+(15x10)=0
-5Rbx= -10x15=150
Rbx=30 ton.
E R=0 ->+.
-Rax+Rbx=0.
Rax=Rbx=30 ton.
Rbx=horizontal reaction at B, Rax=horizontal reaction at A
Ma=moment at A.
E Ma=0
(-Rbx5)+(15x10)=0
-5Rbx= -10x15=150
Rbx=30 ton.
E R=0 ->+.
-Rax+Rbx=0.
Rax=Rbx=30 ton.
(4)
Suresh said:
9 years ago
Because it has only axial load and there is 3 axial load to be applicable in a space of 5m from A point so total is 10 each tonne for 3 (10 * 3 = 30 tonnes).
Krishna said:
9 years ago
Take sum of moments at other rigid joint = 0.
10 * 15 = h * 5.
Then, H = 30.
10 * 15 = h * 5.
Then, H = 30.
(2)
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