shear force and bending moment diagram of downward load

draw shear force and bending moment diagram of given figure. 

SIGN CONVENTION

STEPS FOR FINDING REACTIONS

Here, In The Above Figure We Have Two Unknown Reaction Force RA And RH.

To Solve Two Unknowns We Need Two Equations.

1.   Apply First Equilibrium Equation

∑FY( Summation Of Vertical Forces)= 0

Here, TWO Vertical Reaction Forces Are Acting Upward. They Are +RA And +RH And Three Downward  Forces Are Acting On The Beam AB, BC, CD.

Here, Down ward load on beam AB= (0.5*3*6)= -9KN

Down ward load on beam AB= (6*2) + (0.5*6*6)= -30KN

Down ward load on beam AB= (0.5*1.5*9) + (0.5*1.5*9) = -13.5KN

Take The Summation Of All Vertical Forces Acting On The Beam (Both Upward And Downward)

2.   Apply Second Equilibrium Equation

∑M(Summation Of Moments)= 0

Formula Of Moment= Force*Distance.

Force= Area of Triangle= 0.5*3*6= 9KN

if we are taking moment at the perpendicular side of triangle

= F*(b/3)

if we are taking moment at the vertex side of triangle= F*(2b/3)

e.g. ∑ME= (RA*7) – (5*5) – (3*2*(1+3)) – (0.5*3*6* )

Here We Have Taken ∑M At E.

Take The Summation Of All Moments Acting On The Beam (Both Clock Wise And Anti clock Wise)

(Better See The Sign Convention For Positive And Negative Moment Or Clockwise And Anticlockwise Moment).

By Solving The Two Equations We Can Calculate Both The Support Reaction.

Once We Got The Values Of Reaction We Can Calculate Shear Force And Bending Moment.

FINDING THE REACTIONS

∑ FY=0

RA-5-(2*3)-(0.5*3*6)-(3*2)-(0.5*3*6)-(0.5*1.5*9)-(0.5*1.5*9)+RH=0

RA+RH= 48.5KN — (EQN-1)

∑MA= 0

(RA*0) + (5*1) + (2*3*3) + (0.5*3*6*( +4)) + (3*2*(1.5+7))

+ (0.5*3*6*( +7)) + (0.5*1.5*9*( +10)) + (0.5*1.5*9*( +11.5))

+ 4 – (RH*15)=0

RH= 24.55KN

RA+24.55=48.5KN

RA= 23.95KN

Or,

∑MH=0

(RH*0) + 4 – (0.5*1.5*9*( +2)) – (0.5*1.5*9*( +3.5)) – (2*3*6.5)

-(0.5*3*6*(  +5)) – (0.5*3*6*(  +8)) – (3*2*12) – (5*14) + (RA*15)= 0

RA= 23.95KN

To Calculate The End Shear Force

 Take summation of forces acting on node or member.

In the below calculation L indicates left and R indicates right.

CALCULATION OF END SHEAR FORCE

FA= 23.95KN

FB=23.95 – 5= 18.95KN

FC=23.95 – 5= 18.95KN

FD=23.95 – 5 – (2*3)= 12.95KN

FE=23.95 – 5 – (2*3) – (0.5*3*6)= 3.95KN

FF=23.95 – 5 – (2*3) – (0.5*3*6) – (3*2) – (0.5*3*6)= -11.05KN

FG=23.95 – 5 – (2*3) – (0.5*3*6) – (3*2) – (0.5*3*6) – (0.5*1.5*9) -(0.5*1.5*9)= -24.55KN

FG=23.95 – 5 – (2*3) – (0.5*3*6) – (3*2) – (0.5*3*6) – (0.5*1.5*9) -(0.5*1.5*9) + 24.55= 0

CALCULATION OF MID SHEAR FORCE

F(A-B)= 23.95KN

F(B-C)=23.95 – 5= 18.95KN

F(C-D)=23.95 – 5 – (1*3)= 15.95KN

F(D-E)=23.95 – 5 – (2*3) – (0.5*1.5* )= 10.7KN

F(E-F)=23.95 – 5 – (2*3) – (0.5*3*6) – (1.5*2) – (0.5*1.5* )= -1.3KN

F(F-G)=23.95 – 5 – (2*3) – (0.5*3*6) – (3*2) – (0.5*3*6) – (0.5*1.5*9)=

-17.8KN

F(G-H)=23.95 – 5 – (2*3) – (0.5*3*6) – (3*2) – (0.5*3*6) – (0.5*1.5*9)

-(0.5*1.5*9)= -24.55KN

To Calculate The End Bending Moment

 Calculate moments acting on each node or member and add the remaining distance up to which node you want to calculate the moment.

CALCULATION OF END BENDING MOMENTS

MA=(23.95*0)= 0

MB=(23.95*1)= 23.95KNm

MC=(23.95*2) – (5*1)= 42.9KNm

MD=(23.95*4) – (5*3) – (2*3*1.5)= 74.8KNm

ME=(23.95*7) – (5*6) – (2*3*4) – (0.5*3*6* )= 104.65KNm

MF=(23.95*10) – (5*9) – (2*3*7) – (0.5*3*6*( +3)) – (2*3*1.5)

-(0.5*3*6* )= 98.5KNm

MG=(23.95*13) – (5*12) – (2*3*10) – (0.5*3*6*( +6)) – (2*3*4.5)

-(0.5*3*6*( +3)) – (0.5*1.5*9*( +1.5)) – (0.5*1.5*9*( ))= 45.1KNm

MH=(23.95*15) – (5*14) – (2*3*12) – (0.5*3*6*( +8)) – (2*3*6.5)

-(0.5*3*6*( +5)) – (0.5*1.5*9*( +3.5)) – (0.5*3*6*( +2)) + 4= 0

To Calculate The Mid Bending Moment

Calculate moments acting on each node or member and add the remaining distance upto which node you want to calculate the moment.

CALCULATION OF MID BENDING MOMENTS

M(A-B)=(23.95*0.5)= 11.975KNm

M(B-C)=(23.95*1.5) – (5*0.5)= 33.425KNm

M(C-D)=(23.95*3) – (5*2) – (1*3*0.5)= 60.35KNm

M(D-E)=(23.95*5.5) – (5*4.5) – (2*3*2.5)

-(0.5*1.5* * )= 93.1KNm

M(E-F)=(23.95*8.5) – (5*7.5) – (2*3*5.5) – (0.5*3*6*( +1.5))

-(1.5*2*0.75) – (0.5*1.5* * )= 107.2KNm

M(F-G)=(23.95*11.5) – (5*10.5) – (2*3*8.5) – (0.5*3*6*( +4.5))

-(3*2*3) – (0.5*3*6*( +1.5)) – (0.5*1.5*9* )= 78.55KNm

M(G-H)= (23.95*14) – (5*13) – (2*3*11) – (3*2*5.5)

-(0.5*3*6*( +4)) – (0.5*3*6*( +7)) – (0.5*1.5*9*( +2.5))

-(0.5*1.5*9*( )+1))= 20.55KNm

shear force and bending moment diagram manual
shear force and bending moment diagram manual
staad output

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