Question

10.A beam 5.9 m long is fixed at right end (point C) and free at the left end (point A).
A uniformly distributed load with a magnitude of 2.3 KN/m is carried by the beam
from the left to a 3.3 m span at point B. The distributed load is increased to 4.6 KN/m from point B to point C.
a. What is the magnitude of the vertical reaction (KN) at point C?
b. What is the magnitude of the moment (KN-m) at point C?
c. What is the magnitude of the shear force (KN) at point B? (indicate the sign if
negative)
d. What is the magnitude of the shear force (KN) at point C? (indicate the sign if
negative)
e. What is the magnitude of the moment (KN-m) at point B? (indicate the sign if
negative)
f. What is the magnitude of the moment (KN-m) at point C? (indicate the sign if
negative)

Answer 1

10: Given :- 2.3 kN/m 4.6 kN/m mind A K ch -3.3m. * 2.6m 5.9m Vc k 9) Magnitude of vertical reaction(kn) @ pointc; By using equilibrium equation, Efy=0 le, summation d all vertical reactions - 0. ve = (2:3* 3.3) + (4.6*2.6) Y =(t). 19.55kN. In Cantilever, the vertical reaction @ fixed end is considered as shear force also. . d) Magnitude of Shear force are = 19.55KN. b, f) Magnitude of moment (kum) @ pointc :- By using equilibrium equation, EM = 0.

2 Moment @c = {(23* 23*26423)}+{ucxaf'} , (2.3*3.3) KN (4 6x2.2 kN. & 2 me # K 33 K-265 Me = {(32.25) + (31.1)} Mc. = -63.35 KN-m - : The moment is c) Magnitude is hogging negative sign - shear force @ Bi (2.3*33) KN (4.6K2.6) KN. of 서 A B Shear force @B: Sum of reactions at left side (or) right side of point Isf @B 7.59 KN Cleft) =)

3 Magnitude of Manent ha point Bo- MB = -2.3*3.3*3.3 MB = - 12.52 KN-m 2 The moment is hogging , it is negative