Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 15)
15.
The reaction at support A of the beam shown in below figure, is
Discussion:
61 comments Page 3 of 7.
DIPAK said:
1 decade ago
Above explanation is wrong because R at su = wl/2.
So putting the value in formula so getting answer = 5T.
So putting the value in formula so getting answer = 5T.
Engr.Sharif Ullah khan said:
3 years ago
T.l = 10 * 1 = 10t,
Rb * 10 = 10*1(10/2),
Rb * 10 = 50,
Rb = 50/10,
Rb = 5,
Ra + Rb = 10.
Ra = 10-5 = 5.
Rb * 10 = 10*1(10/2),
Rb * 10 = 50,
Rb = 50/10,
Rb = 5,
Ra + Rb = 10.
Ra = 10-5 = 5.
AJAY said:
7 years ago
The correct answer is 5t, since SSB is symmetrical and total load is 1 * 10 = 10t, Ra=Rb=10/2= 5t.
Roshan said:
7 years ago
It is 2.5 t for A, 2.5t for B, and 5.0t will be support C. If you consider B as centre support.
Rohit Kumar Behera said:
6 years ago
It's 1.875 as calculated above , 3 pt theorom, calculate mb as in str and then proceed for ra
Satya said:
9 years ago
According to general assumption, Ra is 1st support from the left. And the answer is 15/8.
Kishore said:
1 decade ago
This is continuous beam use theorem of three moment to find the reactions at the support.
Srikanth B R said:
9 years ago
No need to calculate simple assumption at any support, the reaction will always be zero.
Hemant chauhan said:
7 years ago
This beam is simitricaly,
Hence ra +rb = total load ÷ 2
=> 1*10 = 10 ÷ 2 =5.
Hence ra +rb = total load ÷ 2
=> 1*10 = 10 ÷ 2 =5.
Ammu said:
7 years ago
They asking load only not moment. Moment at the support A will be zero, Not reaction.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers