SHEAR FORCE AND BENDING MOMENT

Definition:

Shear force: The force which tend to shear off the section and is obtained as the algebraic sum of all forces including the reactions acting normal to the axis of the beam either to the left or right of the beam.

Sign convention

Positive Shear Force
 Negative Shear Force

Sign convention for shear force in brief

Direction
Positive SF
Negative SF
Left side
Upward force
Downward force
Right side
Downward force
Upward force


Bending moment: The algebraic sum of all the moments of the forces and reactions in a beam towards either left side or right side of the support is known as bending moment.

Types of bending moment:

Sagging bending moment: The moment about the point which produces convexity below the centre line is called as sagging bending moment. It is also called as positive bending moment.


Consider the simply supported beam with central concentrated load “F”. The moment due to the force F produces a deflection in the downward direction which in turn results in the formation of sagging bending moment.

Hogging bending moment: The moment about the point which produces convexity above the centre line is called as hogging bending moment. It is also called as negative bending moment.
Consider the simply supported beam with central concentrated load “F”. The moment due to the force F produces a deflection in the upward direction which in turn results in the formation of hogging bending moment.

Differences between sagging and hogging bending moment


Sagging bending moment
Hogging bending moment
The moment about the point which produces convexity below the centre line is called as sagging bending moment.
The moment about the point which produces convexity above the centre line is called as hogging bending moment.
 
 
It is also termed as positive bending moment
It is also termed as negative bending moment
Top fibres of beam are subjected to compression.
Top fibers of beam are subjected to tension.
Bottom fibers of beam are subjected to tension.
Bottom fibers of beam are subjected to compression.

Sign conventions for bending moments

Direction
Positive BM
Negative BM
Left side
Clock wise
Anti Clock wise
Right side
Anti Clock wise
Clock wise



Problems on cantilever beams
1)

Step 1: Calculation of shear force
Shear force at point of 500N = 500N
Shear force at point of 800N = 500+800= 1300N
Shear force at point of 300N = 500+800+300=1600N
Shear force at point of 400N = 500+800+300+400=2000N
Shear force at point of fixed point A = 500+800+300+400=2000N

Step 2: Calculation of Bending Moment
Bending Moment at point of 500N = 0N-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point of 800N = 500X0.5= -250N-m
Bending Moment at point of 300N = -500X1-800X0.5= -900N-m
Bending Moment at point of  400N = -500X1.5-800X1-300X0.5= -1700N-m
Bending Moment at fixed point A= -500X2-800X1.5-300X1-400X0.5= -2700N-m


2)

Step 1: Calculation of shear force
Shear force at point of 800N = 800N
Shear force at point of 500N = 500+800= 1300N
Shear force at point of 300N = 500+800+300=1600N
Shear force at point of fixed point A = 500+800+300=1600N


Step 2: Calculation of Bending Moment
Bending Moment at point of 800N = 0N-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point of 500N = 800X0.8= -640N-m
Bending Moment at point of 300N = -800X1.5-500X0.7= -1550N-m
Bending Moment at fixed point A= -800X2-500X1.2-300X1-300X0.5= -2350N-m





3)


Step 1: Calculation of shear force
Shear force at point of 3KN = 3KN
Shear force at point of 2KN = 3+2= 5KN
Shear force at point of 1KN = 3+2+1=6KN
Shear force at point of fixed point A = 3+2+1=6KN


Step 2: Calculation of Bending Moment
Bending Moment point of 3KN = 0N-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point of 2KN = -3X2= -6KN-m
Bending Moment at point of 1KN = -3X3-2X1= -11KN-m
Bending Moment at fixed point A= -3X4-2X2-1X1=-17KN-m

4)
Step 1: Calculation of shear force
Shear force at point B= 0
Shear force at point of fixed point A = 10 X 2.5=25KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0 N-m(Note: this is end point of beam where beyond this point there is no force existed)
Bending Moment at fixed point A= 10 X 2.5 X 2.5/2 = -31.25 KN-m

5)
Step 1: Calculation of shear force
Shear force at point C= 0
Shear force at point B=3KN
Shear force at point of fixed point A = 3KN

Step 2: Calculation of Bending Moment
Bending Moment at point C= 0 N-m(Note: this is end point of beam where beyond this point there is no force existed)
Bending Moment at point B=2X1.5X0.75=-2.25 KN-m
Bending Moment at fixed point A= 2 X 1.5 X(0.75+0.5)= -3.75 KN-m


6)
Step 1: Calculation of shear force
Shear force at point B= 3KN
Shear force at point of fixed point A = 4+3=7 KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0 N-m(Note: this is end point of beam where beyond this point there is no force existed)
Bending Moment at fixed point A= -(3 X 2)-(2 X 2 X 1) = -10 KN-m

7)


Step 1: Calculation of shear force
Shear force at point D = 8KN
Shear force at point C= 15+8= 23KN
Shear force at point B= 15+8= 23KN
Shear force at point of fixed point A =4X2+8+15=31KN

Step 2: Calculation of Bending Moment
Bending Moment at point D= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point C = -(8X2)= -16KN-m
Bending Moment at point B = -(8X3)-(15X1)= -39KN-m
Bending Moment at fixed point A= -(8X5)-(15X3)-(8X1)= - 93KN-m

8)

Step 1: Calculation of shear force
Shear force at point C =0KN
Shear force at point B= 2+1.5X0.5=2.75KN
Shear force at point of fixed point A =2+1.5X2=5KN

Step 2: Calculation of Bending Moment
Bending Moment at point C= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point B = -(1.5X0.5X0.25)= -0.1875KN-m
Bending Moment at fixed point A= -(1.5X2X1)-(2X1.5)= -6KN-m

9)


Step 1: Calculation of shear force
Shear force at point D = 0KN
Shear force at point C= 3+(2X0.25)=3.5KN
Shear force at point B= 2X1.25+3=5.5KN
Shear force at point of fixed point A =5.5KN

Step 2: Calculation of Bending Moment
Bending Moment at point D= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point C = -(2X0.25X0.25/2)= -0.0625KN-m
Bending Moment at point B = -(3X1)-(2X1.25X1.25/2)= -4.5625KN-m
Bending Moment at fixed point A= -(3X1.25)-(2X1.25X(1.25/2+0.25))= - 5.9375KN-m

10)

Step 1: Calculation of shear force
Shear force at point E = 2.5KN
Shear force at point D= 2.5KN
Shear force at point C= 2.5+1X2=4.5KN
Shear force at point B= 2.5+1X2+3=7.5KN
Shear force at point of fixed point A =7.5KN

Step 2: Calculation of Bending Moment
Bending Moment at point E= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point D = -(2.5X0.5)= -1.25KN-m
Bending Moment at point C= -(2.5X2.5)-(2X1)= -8.25KN-m
Bending Moment at point B= -(2.5X4)-(2X2.5)= -15KN-m
Bending Moment at fixed point A= -(2.5X5)-(2X3.5)-(3X1)= - 22.5KN-m

11)


Step 1: Calculation of shear force
Shear force at point D= 20KN
Shear force at point C=20+10=30KN
Shear force at point B=20+10+2(12) =69KN
Shear force at point of fixed point A =69KN

Step 2: Calculation of Bending Moment
Bending Moment at point D= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point C = -(20X2)= -40KN-m
Bending Moment at point B= -(2.0X4)-(10X2)-24= -124KN-m
Bending Moment at fixed point A= -(20X5)-(10X3)-(24X2)-15= - 193KN-m

12)
Step 1: Calculation of shear force
Shear force at point E = 2KN
Shear force at point D= 2KN
Shear force at point C= 20+2=22KN
Shear force at point B= 3+22=25KN
Shear force at point of fixed point A =25KN

Step 2: Calculation of Bending Moment
Bending Moment at point E= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point D = -(2X0.5)= -1KN-m
Bending Moment at point C= -(2X2.5)-(20X1)= -25KN-m
Bending Moment at point B= -(2X4)-(20X2.5)= -58KN-m
Bending Moment at fixed point A= -(2X5)-(20X3.5)-(3X1)= - 83KN-m

13)

Step 1: Calculation of shear force
Shear force at point B=0KN
Shear force at point of fixed point A =10KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at fixed point A= -(1/2X5x4)(1/3X4)= - 13.33KN-m


14)
Step 1: Calculation of shear force
Shear force at point F = 0KN
Shear force at point E= 15KN
Shear force at point D=15KN
Shear force at point C= 15KN
Shear force at point B= 15+10=25KN
Shear force at point of fixed point A =25KN

Step 2: Calculation of Bending Moment
Bending Moment at point F= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point E = 0KN-m
Bending Moment at point D= -(15X0.5)-20= -27.5KN-m
Bending Moment at point C= -(15X1)= -15KN-m
Bending Moment at point B= -(15X1.5)-(10X0.25)= -25KN-m
Bending Moment at fixed point A= -(15X2)-(10X0.75)= - 37.5KN-m


15)


Step 1: Calculation of shear force
Shear force at point B=0KN
Shear force at point of fixed point A =12KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at fixed point A= -(1/2X6x4)(1/3X4)= - 24KN-m

16)

Step 1: Calculation of shear force
Shear force at point B=0KN
Shear force at point of fixed point A=2+1(4) =6KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at fixed point A= -(2X4)-(1X4X2)= - 16KN-m

17)



Step 1: Calculation of shear force
Shear force at point B=3 KN
Shear force at point C = 3 KN
Shear force at point of fixed point A=3+1(2)+2 =7 KN

Step 2: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C= 3X4=12KN-m
Bending Moment at fixed point A= -(3X6)-(1X2X1)-(2X1)= - 22KN-m

18)


Step 1: Calculation of shear force
Shear force at point D= 0KN
Shear force at point C=0.5X2X20=20KN
Shear force at point B=20+30 =50KN
Shear force at point of fixed point A =50KN

Step 2: Calculation of Bending Moment
Bending Moment at point D= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point C = -(0.5X20X2X1/3(2))= -13.33KN-m
Bending Moment at point B= -(0.5X20X2X(1/3(2)+1)))+(100)= -66.66N-m
Bending Moment at fixed point A= -(0.5X20X2X(1/3(2)+3)-(30X2)= - 133.33KN-m

19)


Step 1: Calculation of shear force
Shear force at point E= 0KN
Shear force at point D=0KN
Shear force at point C= 20KN
Shear force at point B= 20+10=30KN
Shear force at point of fixed point A =30KN

Step 2: Calculation of Bending Moment
Bending Moment at point E= 0KN-m(Note: this is end pont of beam where beyond this point there is no force existed)
Bending Moment at point D= 100= 100KN-m
Bending Moment at point C=  0KN-m
Bending Moment at point B= -(20X1)= -20KN-m
Bending Moment at fixed point A= -(20X2)-(10X1)= - 50KN-m

Problems on simply supported beams

1)

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 4+10+7
Ra+ Rb= 21KN
ΣMa =0
Rb(8)-7(6)-10(4)-4(1.5)=0
Rb = 11 KN ; Ra = 10 KN

Step 2: Calculation of shear force
Shear force at point B = -11KN
Shear force at point C= -11+ 7= -4KN
Shear force at point D=-11+ 7+10 = 6KN
Shear force at point E= -11+ 7+10+4 = 10KN
Shear force at point A=10KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C= (11X2)= 22KN-m
Bending Moment at point D=  (11X4)-(7X2)=30KN-m
Bending Moment at point E= (11X6.5)-(7X4.5)-(10X2.5)= 15KN-m
Bending Moment at fixed point A=(11X8)-(7X6)-(10X4)-(4X1.5)  = 0KN-m

2)


Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 2+5+4
Ra+ Rb= 11KN
ΣMa =0
Rb(6)-4(4.5)-5(3)-2(1.5)=0
Rb = 6 KN ; Ra = 5 KN

Step 2: Calculation of shear force
Shear force at point B = -6KN
Shear force at point E= -6+4= -2KN
Shear force at point D=-6+ 4+5 = 3KN
Shear force at point C= -6+ 4+5+2 = 5KN
Shear force at point A=5KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E= (6X1.5)= 9KN-m
Bending Moment at point D=  (6X3)-(4X1.5)=12KN-m
Bending Moment at point C= (6X4.5)-(4X3)-(5X1.5)= 7.5KN-m
Bending Moment at fixed point A=(6X6)-(4X4.5)-(5X3)-(2X1.5)  = 0KN-m


3)

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 3+6
Ra+ Rb= 9 KN
ΣMa =0
Rb(6)-6(4)-3(2)=0
Rb = 5KN ; Ra = 4 KN


Step 2: Calculation of shear force
Shear force at point B = -5KN
Shear force at point D= -5+6= 1KN
Shear force at point C= -5+6+3 = 4KN
Shear force at point A= 4KN


Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (5X2)= 10KN-m
Bending Moment at point C=  (5X4)-(6X2)=8KN-m
Bending Moment at fixed point A=(5X6)-(6X4)-(3X2) = 0KN-m

4)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 10(6)
Ra+ Rb= 60 KN
ΣMa =0
Rb(12)-10(6)(3)=0
Rb = 15KN ; Ra = 45 KN

Step 2: Calculation of shear force
Shear force at point B = -15KN
Shear force at point C= -15KN
Shear force at point A= -15+10(6)= 45 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C=  (15X6)=90KN-m
Bending Moment at fixed point A=(15X12)-(10X6X3)= 0KN-m


5)

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 5(2)+ 10(2)
Ra+ Rb= 30 KN
ΣMa =0
Rb(6)-5(2)(4+1)-10(2)1=0
Rb = 11.67KN ; Ra = 18.33 KN

Step 2: Calculation of shear force
Shear force at point B = -11.67KN
Shear force at point D= -11.67+5X2= -1.67KN
Shear force at point C= -1.67 KN
Shear force at point A= -1.67+ 10X2 = 18.33KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (11.67X2)-(5X2X1)= 13.34KN-m
Bending Moment at point C=  (11.67X4)-(5X2)(1+2)=16.68KN-m
Bending Moment at fixed point A=(11.67X6)-(5X2)(1+4)-(10X2X1) = 0 KN-m

Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-10(x)=0
18.33-10(x)=0
x=1.833m
Maximum BM corresponding to x
Ra(x)-10(x2/2)
18.33(1.833)-10(1.833x1.833/2)
16.79KN-m = Mmax


6)
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 50+ 10(4)+40
Ra+ Rb= 130 KN
ΣMa =0
Rb(10)-40(6)-10(4)(2+2)-50x2=0
Rb = 50 KN ; Ra = 80 KN

Step 2: Calculation of shear force
Shear force at point B = -50KN
Shear force at point D= -50+40= -10KN
Shear force at point C=  -50+40+10x4+50= 80 KN
Shear force at point A = 80 KN


Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D= (50X4)= 200 KN-m
Bending Moment at point C=  (50X8)-(40X4)-(10x4x2)=160 KN-m
Bending Moment at fixed point A = 0 KN-m

Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-50-10(x-2)=0
80-50-10(x-2)=0
x=5 m
Maximum BM corresponding to x
Ra(x)-50x3-10x3x1.5
205 KN-m = Mmax

7)


Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 4+ 8+8
Ra+ Rb= 20 KN
ΣMa =0
Rb(10)-8(9)-8(5)-4(2)=0
Rb = 12 KN ; Ra = 8 KN

Step 2: Calculation of shear force
Shear force at point B = -12 KN
Shear force at point F= -12+8= -4 KN
Shear force at point E=  -4 KN
Shear force at point D=  -4+8= 4 KN
Shear force at point C= 4+4= 8 KN
Shear force at point A = 8 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point F= (12X1)= 12 KN-m
Bending Moment at point E=  (12X3)-(8x2)=20 KN-m
Bending Moment at point D=  (12X7)-(8x6)-(8x2)=20 KN-m
Bending Moment at point C=  (12X8)-(8x7)-(8x3)-(4x2)=8 KN-m
Bending Moment at fixed point A = 0 KN-m


Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-4-2(x-3)=0
8-4-2(x-3)=0
x=5 m
Maximum BM corresponding to x
Ra(x)-4x3-4x1
24 KN-m = Mmax


8)

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 25+ 10(6)
Ra+ Rb= 85 KN
ΣMa =0
Rb(6)-25(2)-10(6)(6/2)=0
Rb = 38.33 KN ; Ra = 46.67 KN

Step 2: Calculation of shear force
Shear force at point B = -38.33 KN
Shear force at point C=-38.33+10(4)+25=  26.67 KN
Shear force at point A =-38.33+10(6)+25 =46.67 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C=  (38.33X4)-(10x4x2)-(25x2)=73.33 KN-m
Bending Moment at fixed point A = 0 KN-m


Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-25-10(x)=0
46.67-25-10(x)=0
x=2.167 m
Maximum BM corresponding to x
Ra(x)-25x0.167-10x2.167x2.167/2
73.48 KN-m = Mmax

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 0
Ra = -Rb
ΣMa =0
Rb(6)-24=0
Rb = 4 KN ; Ra = -4KN

Step 2: Calculation of shear force
Shear force at point B = -4 KN
Shear force at point C=  -4 KN
Shear force at point A =-4 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C =  Right side of C=4(4)=  16 KN-m
                                             =   Left side of C = 4(2) = -8 KN-m
Bending Moment at fixed point A = 0 KN-m



Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 45+ 10(3)
Ra+ Rb= 75 KN
ΣMa =0
Rb(6)-45(3)-10(3)(3/2)-120=0
Rb = 50 KN ; Ra = 25 KN

Step 2: Calculation of shear force
Shear force at point B = -50 KN
Shear force at point D=-50 KN
Shear force at point C=-50+45 = -5 KN
Shear force at point A =-5+10(3) =25KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point D=  from B = 50x1.5-120= -45 KN-m
Bending Moment at point C = 50x3-120 =30 KN-m
Bending Moment at fixed point A = 0 KN-m

Step 4: Calculation of point of zero shear force x(point of max BM)
Ra-10(x)=0
25-10(x)=0
x=2.5 m
Maximum BM corresponding to x
Ra(x)-10x2.5x2.5/2
31.25 KN-m = Mmax



Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 40+ 5(2)
Ra+ Rb= 50 KN
ΣMa =0
Rb(7)-40(2)-5(2)((2/2)+2))-80=0
Rb = 27.14 KN ; Ra = 22.86 KN

Step 2: Calculation of shear force
Shear force at point B = -27.14 KN
Shear force at point C = -27.14 KN
Shear force at point D= -27.14 KN
Shear force at point E= -27.14+5(2)+40 = 22.86 KN
Shear force at point A = 22.86 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point C =27.14x1-80=-52.86 KN-m
Bending Moment at point D =27.14x3-80=1.42 KN-m
Bending Moment at point E = 27.14x5-80-5(2)(2/2)=50.7 KN-m
Bending Moment at fixed point A = 0 KN-m



Step 1: Calculation of the reactions
ΣH = 0
Ha = 100cos60+200cos45+300cos30
Ha = 451.23KN
ΣV = 0
Ra+ Rb= 100sin60+200sin45+300sin30
Ra+ Rb= 378KN

ΣMa =0
Rb(4)-300(3sin30)-200(2)(sin45)-100sin60=0
Rb = 204.86 KN ; Ra = 173.14KN

Step 2: Calculation of shear force
Shear force at point B = -204.86 KN
Shear force at point E= -204.86+300sin30= -54.86 KN
Shear force at point D= -204.86+300sin30+200sin45= 86.56KN
Shear force at point C= 204.86+300sin30+200sin45+100sin60 = 173.14 KN
Shear force at point A =173.14KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E=204.86x1= 204.86 KN-m
Bending Moment at point D =204.86x2-300sin30 =259.72 KN-m
Bending Moment at point C =204.86x3-300(2)sin30-200(2)sin45=-126.84 KN-m
Bending Moment at fixed point A = 0 KN-m
Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 0.5(5)(6)
Ra+ Rb= 15KN

ΣMa =0
Rb(5)-0.5(5)(6)(1/3)(5)=0
Rb = 5 KN ; Ra = 10KN

Step 2: Calculation of shear force
Shear force at point B = -5 KN
Shear force at point B = 10 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point A= 0KN-m

Step 4: Calculation of point of zero bending moment
-Rb + (0.5)(x)(6x/5) = 0 
x = 2.88m
Maximum bending moment = Rb(x) - (0.5)(x)(6x/5)(x/3)
Mmax = 9.62KN-m

Step 1: Calculation of the reactions
ΣH = 0
ΣV = 0
Ra+ Rb= 30+ 0.5(3)(20)
Ra+ Rb= 60KN

ΣMa =0
Rb(6)-30(5)-0.5(3)(20)(1+(2/3(3))=0
Rb = 40KN ; Ra = 20 KN

Step 2: Calculation of shear force
Shear force at point B = -40 KN
Shear force at point E = -40+30 = -10 KN
Shear force at point D= -10 KN
Shear force at point C= -10+(0.5)(3)(20)= 20 KN
Shear force at point A = 20 KN

Step 3: Calculation of Bending Moment
Bending Moment at point B= 0KN-m
Bending Moment at point E= 40X1=40 KN-m
Bending Moment at point D =40x2-30x1=50KN-m
Bending Moment at point C = 40x5-30x4-0.5(20)(3)(2/3(3))=20KN-m
Bending Moment at fixed point A = 0 KN-m

Step 4: Calculation of point of zero bending moment
Ra - (0.5)(x-1)(20/3)(x-1) = 0 
x = 1.45m
Maximum bending moment = Ra(x) - (0.5)(x-1)(20/3)(x-1)(2/3)(x-1)
Mmax = 28.79KN-m














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