draw shear force and bending moment diagram of triangular and trapezoidal load manual vs staad pro

figure

draw shear force and bending moment diagram of given figure. 

SIGN CONVENTION 

sign convention
STEPS FOR FINDING REACTIONS

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

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 +RB 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. ∑MB= (RA*3) – (0.5*3*6*(3/3))

Here We Have Taken ∑M At B.

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 – ( *3*6) – (6*2) – ( *6*6) – ( *1.5*9) – ( *1.5*9) + RD=0

 

RA+RD= 52.5KN — (EQN -1)

 

∑ MA=0

 

(RA*0) + (0.5*3*6* ) + (6*2*(3+3)) + (0.5*6*6*( +3))

 

+ (0.5*1.5*9*( +9)) + (0.5*1.5*9*( +10.5)) – (RD*12)= 0

 

RD= 29.813KN

 

RA= 52.5-29.813= 22.688KN

 

Or, 

 

∑MD=0

 

(RD*0) -(0.5*1.5*9* ) – (0.5*1.5*9*( +1.5)) – (6*2*(3+3)) – (0.5*6*6*( +3)) – (0.5*3*6*( +9)) + (RA*12)= 0

 

RA= 22.688KN

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= 22.688KN

 

FB= 22.688 – (0.5*3*6)= 13.688KN

 

FC= 22.688 – (0.5*3*6)-(6*2) – (0.5*6*6)= -16.312KN

 

FDL= 22.688 – (0.5*3*6)-(6*2) – (0.5*6*6) – (0.5*1.5*9)

 

-(0.5*1.5*9)= -29.812KN

 

FDR= 22.688 – (0.5*3*6) – (6*2) – (0.5*6*6) – (0.5*1.5*9)

 

-(0.5*1.5*9) + 29.813= 0

To Calculate The Mid Shear Force

 Take Summation of forces acting on node or member.

CALCULATION OF MID SHEAR FORCE

F(A-B)= 22.688 – (0.5*1.5* ) = 20.438KN

 

F(B-C)= 22.688 – (0.5*3*6) – (3*2) – (0.5*3* ) = 3.188KN

 

F(C-D)= 22.688 – (0.5*3*6) – (6*2) – (0.5*6*6) – (0.5*1.5*9) =

 

-23.062KN

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= (22.688*0) = 0

 

MB= (22.688*3) – (0.5*3*6* ) = 59.064KNm

 

MC= (22.688*9) – (0.5*3*6*( +6)) – (6*2*3) – ( 0.5*6*6* ) = 

 

69.192KNm

 

MD= (22.688*12) – (0.5*3*6*( +9)) – (6*2*6)-(0.5*6*6*( +3))

 

-(0.5*1.5*9*( +1.5)) – (0.5*1.5*9* )= 0.006 = 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) = (22.688*1.5) – (0.5*1.5* * ) = 32.907KNm

 

M(B-C)= (22.688*6)-(0.5*3*6*( +3))-(3*2*1.5)-(0.5*3* * 

 

= 86.628KNm

 

M(C-D)= (22.688*10.5) – (0.5*3*6*( +7.5)) – (6*2*4.5)

 

-(0.5*6*6*( +1.5)) – (0.5*1.5*9*(1.5/3)) = 41.349KNm

shear force and bending moment diagram manual
shear force and bending moment diagram manual
STAAD OUTPUT
reaction staad
shear force staad
end bending moment staad
mid bending moment staad

STAAD RESULT SUMMARY

FX = FORCE IN HORIZONTAL DIRECTION

FY = FORCE IN VERTICAL DIRECTION

MZ = BENDING MOMENT

result summary staad

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