Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 17)
17.
A member is balanced at its end by two inclined members carrying equal forces. For equilibrium the angle between the inclined bars must be
Discussion:
8 comments Page 1 of 1.
Ranjit ray said:
9 years ago
Lamis's theorem can be used for this concept.
P/sinP = Q/sinQ = R/sinR. if the angle are 120degree i.e.angleP = Q = R = 120 the magnitude of all forces in three members are same which gives equilibrium of the joint.
P/sinP = Q/sinQ = R/sinR. if the angle are 120degree i.e.angleP = Q = R = 120 the magnitude of all forces in three members are same which gives equilibrium of the joint.
(3)
Arachne said:
9 years ago
Explain it please.
(1)
Vipin sainath said:
5 years ago
@ Ranjit Ray,
Sin 120 is 0.89 and Sin 90 is 1.Then How do you tell 120 is right. I think Sin 90 is correct.
Sin 120 is 0.89 and Sin 90 is 1.Then How do you tell 120 is right. I think Sin 90 is correct.
(1)
Rajjj said:
1 decade ago
Any body explain?
Shakeel said:
9 years ago
Thanks @Ranjit Ray.
Praveen said:
7 years ago
Thanks @Ranjit.
Abubakar said:
5 years ago
@Ranjit Ray.
They have mentioned two inclined members not three. If it's 3 inclined members then it would be 120.
They have mentioned two inclined members not three. If it's 3 inclined members then it would be 120.
RAHUL said:
5 years ago
@All.
In question, there is 3 member, 1 is balanced by two other, all three is in equilibrium if angle b/w them is 120 degree.
In question, there is 3 member, 1 is balanced by two other, all three is in equilibrium if angle b/w them is 120 degree.
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