For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum value of the bending moments, both positive and negative. Solution First of all we have to solve for the reaction at supports i.e RAand RB On applying the equilibrium condition we get RA = 3.15 kN and RB = 7.35 kN Now we will try to draw the SFD and BMD For SFD we will calculate the shear stress at point a , b and C Left of A = 0 , Right of A = 3.15 Left of B = - 4.35 RIght of B = 3 AT C = 0 hence the diagram will be something like this. Similarly we will calculate for the bending moment. BM at A = 0 BM at B = -1.8 kNm BM at C = 0 Now maximum bending moment we will write the equation of bending moment between point A and B M = RA X - (2.5 X2 / 2) Now for Maximum value we will differentiate it and equate it to zero therefore the value of X = 1.26 m Maximum negative bending moment will be at B as seen from the diagram.

1. For the beam shown, draw the shear and bending-moment diagrams, and determine the
magnitude and location of the maximum value of the bending moments, both positive and
negative.
Solution
First of all we have to solve for the reaction at supports i.e RAand RB
On applying the equilibrium condition we get RA = 3.15 kN and RB = 7.35 kN
Now we will try to draw the SFD and BMD
For SFD we will calculate the shear stress at point a , b and C
Left of A = 0 , Right of A = 3.15
Left of B = - 4.35 RIght of B = 3
AT C = 0 hence the diagram will be something like this.
Similarly we will calculate for the bending moment.
BM at A = 0
BM at B = -1.8 kNm
BM at C = 0
Now maximum bending moment
we will write the equation of bending moment between point A and B
M = RA X - (2.5 X2 / 2)
Now for Maximum value we will differentiate it and equate it to zero therefore the value of X =
1.26 m
Maximum negative bending moment will be at B as seen from the diagram