find shear force and bending moment diagram of isoceles triangular load manual vs staad pro

figuure

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 One Downward Is Forces Acting On The Beam. i.e., UVL  Of  6KN/m.

Here, Base of Triangle = 6m

Height Of  Triangle= 6KN/m.

Divide The Isosceles Triangle Into Half To Form Two Right Angled Triangle.

Now,

Base of Triangle = 3m

Height Of  Triangle= 6KN/m.

Convert The UVL into Force by Calculating The Area Of Each  Triangle. (0.5*3*6)+(0.5*3*6)= 18KN.

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)

bending moment distance

e.g. ∑MB= +(RA*6) – (0.5*3*6*( +3)) – (0.5*3*6* )

Here We Have Taken ∑M At B.

Take The Summation Of All Moments Acting On The Beam (Both Clock Wise And Anticlock 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 – (0.5*3*6) – (0.5*3*6) + RB =0

RA+RB=18KN     –1

LET US USE ∑M=0,

∑MA=0

(RA*0) + (0.5*3*6*((2*3)/3)) + (0.5*3*6*((3/3)+3)) – 6RB= 0

RB= (54/6) =9KN.

FROM (Eqn-1),

RA+9=18KN

RA=9KN.

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=RA= 9KN

FBL=9 – (0.5*3*6) – (0.5*3*6)= -9KN

FBR=9 – (0.5*3*6) – (0.5*3*6) + 9= 0

To Calculate The Mid Shear Force

 Take Summation of forces acting on node or member.

CALCULATION OF MID SHEAR FORCE

F (A-B)=9 – (0.5*3*6)= 0

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=(9*0) = 0        (AS, THE DISTANCE IS ZERO)

MB=(9*6) – (0.5*3*6*((3/3)+3)) – (0.5*3*6*((2*3)/3))=  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 – MOMENTS

M (A-B)= (9*3)-(0.5*3*6*(3*3))= 18KNm

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