shear force bending moment of mid udl

figure

sign convention

sign convention

STEPS FOR FINDING REACTIONS 

Here, In The Above Figure We Have Two Unknown Reaction Force RA  and  RD. 

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 +RD And Three Downward Forces Acting  On A Beam. They Are Point Load Of -5KN, -7KN, UDL of -3KN/m Respectively. 

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. 

Distance For The Moment Of Udl Load Will Be Half Of Its Span Of The Beam. 

E.G. ∑MC= +(RA*6)-(5*4)-(3*4*2) 

Here We Have Taken ∑M At C. 

As  you can see we had not included Nodal load of -7KN because it is acting on Node C and the distance will be zero i.e., (-7*0). 

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-5 – (3*4)-7+RD =0 

RA+RD= 24KN  –(1) 

LET US USE ∑M=0,  

∑MA=0 

(RA*0)+ (5*2) + (3*4*4) + (7*6)-9RD=0 

RD=(100/9)= 11.11KN 

FROM (Eqn-1), 

RA+11.11=24KN

RA= 12.89KN 

END SHEAR FORCE

FA=RA= 12.89KN 

FB=12.89-5= 7.89KN 

FC=12.89-5-(3*4)-7= -4.11KN 

FDL=18.423-5-(7*3)= -11.11KN 

FDR=12.89-5-(3*4)-7-11.11+11.11 = 0 

MID SHEAR FORCE

F(A-B) = 12.89KN 

F(B-C) = 12.89-5-(3*2) = 1.89KN 

F(C-D) = 12.89-5-(3*4)-7 = -11.11KN 

END BENDING MOMENTS

MA=(12.89*0)= 0        (AS, THE DISTANCE IS ZERO) 

MB=(12.89*2)= 25.78KNm 

MC=(12.89*6)-(5*4)-(3*4*2)= 33.34KNm 

MD=(12.89*9)-(5*7)-(3*4*5)-(7*3)= 0 

BENDING MOMENTS AT MID SPAN

M(A-B)=(12.89*1)= 12.89KNm 

M(B-C)=(12.89*4)-(5*2)-(3*2*1)= 35.56KNm 

M(C-D)=(12.89*7.5)-(5*5.5)-(3*4*3.5)-(7*1.5)= 16.675KNm 

SFD & BMD MANUAL

shear force bending moment

STAAD OUTPUT

reaction staad
end shear force staad
mid shear force staad
end bending moment staad
mid bending moment staad

Result Summary

FX = FORCE IN HORIZONTAL DIRECTION

FY = FORCE IN VERTICAL DIRECTION

MZ = BENDING MOMENT

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