Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 16)
16.
The beam shown in below figure is supported by a hinge at A and a roller at B. The reaction RA of the hinged support A of the beam, is
Discussion:
8 comments Page 1 of 1.
Shravani Patil said:
4 years ago
Rb * 5 + 5 * 2 = 2 * 8 * 4.
Rb*5 = 64 - 10.
Rb*5 = 54.
Rb = 10.8.
So, Ra + Rb = 21,
Hence, Ra = 10.2.
Rb*5 = 64 - 10.
Rb*5 = 54.
Rb = 10.8.
So, Ra + Rb = 21,
Hence, Ra = 10.2.
(3)
Abhishek said:
6 years ago
Compare both side moments at B,
(3*2*1.5) = (2*5*2.5) + (5*7) - (5Ra).
9 = 25+35-5Ra.
Ra=51/5.
Ra=10.2t.
(3*2*1.5) = (2*5*2.5) + (5*7) - (5Ra).
9 = 25+35-5Ra.
Ra=51/5.
Ra=10.2t.
Harsh C. Patel said:
8 years ago
Taking moment at A.
Anticlockwise moment = clockwise moment.
From the above an equation we obtain the value of Ra(support reaction at a).
Anticlockwise moment = clockwise moment.
From the above an equation we obtain the value of Ra(support reaction at a).
Abhay said:
10 years ago
Taking Moment at B,
ΣMB = -5x7 + Ra x 5- (2x5)x(5/2)+ (2x3)x3/2 = 0 (clockwise +ve).
-35 -25 + 9 = - Ra x 5.
Ra = 51/5 = 10.2 t.
ΣMB = -5x7 + Ra x 5- (2x5)x(5/2)+ (2x3)x3/2 = 0 (clockwise +ve).
-35 -25 + 9 = - Ra x 5.
Ra = 51/5 = 10.2 t.
(1)
Salunkhe Gunjan S. said:
1 decade ago
Rb*5+5*2 = 2*8*4.
Rb*5 = 64-10.
Rb*5 = 54.
Rb = 10.8.
So, Ra+Rb = 21,
Hence, Ra = 10.2.
Rb*5 = 64-10.
Rb*5 = 54.
Rb = 10.8.
So, Ra+Rb = 21,
Hence, Ra = 10.2.
ALFIFI said:
1 decade ago
The resultant of the distributed load between A and B is:
2 t/m * 5 m = 10 t ... act at midway between A and B.
The resultant of the distributed load between B and the right free end of the beam is:
2 t/m * 3 m = 6 t ... act at midway.
Applying the moments=0 rule at point B:
(Clockwise is positive).
0= -5(7) + 5Ra - 10(2.5) + 6(1.5).
Solving for Ra gives:
Ra = 10.2 t.
Answer is D.
2 t/m * 5 m = 10 t ... act at midway between A and B.
The resultant of the distributed load between B and the right free end of the beam is:
2 t/m * 3 m = 6 t ... act at midway.
Applying the moments=0 rule at point B:
(Clockwise is positive).
0= -5(7) + 5Ra - 10(2.5) + 6(1.5).
Solving for Ra gives:
Ra = 10.2 t.
Answer is D.
Sorabh said:
1 decade ago
Found out by moment and shear force method.
Jinal said:
1 decade ago
Ra*5+3*2*1.5 = 5*2*2.5+10+2*5*2.5.
Ra = 60-9/5.
Ra = 10.2.
Ra = 60-9/5.
Ra = 10.2.
(1)
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