sfd and bmd of fixed beam with uvl loading

To draw sfd and bmd diagram of fixed beam with uvl loadings such as right triangle, trapezoidal and isosceles triangle.

first we need to calculate fixed end moments, slope deflection equation then applying equilibrium equation to find the value of slope (ϴ) and at last finding final moments or end bending moments.

draw sfd and bmd diagram by slope and deflection method

SIGN CONVENTION

steps of slope and deflection method

fixed end moment

slope and deflection equation

equilibrium equation

final moments

REACTIONS BY SEPERATING EACH SPAN

SPAN AB

∑MB=0

-13.22+(RA*6)-(0.5*6*9*(6/3))+11.362=0

RA= 9.308KN

∑FY =0

RA-(0.5*6*9) +RB1=0

9.308-27+RB1=0

RB1= 17.692KN

SPAN BC

∑MB=0

(0.5*1.5*12((2*1.5)/3))+(0.5*1.5*12*((1.5/3)+1.5))-(RC1*3)

-11.362+8.662=0

RC1=8.1KN

∑FY=0

RB2-(0.5*1.5*12)-(0.5*1.5*12)+RC1=0

RB2-18+8.1=0

RB2=9.9KN

SPAN CD

∑MD=0

-8.662+(RC2*6)-(6*3*3)-(0.5*6*3*(6/3))+16.369=0

RC2= 10.715KN

∑FY=0

RC2-(6*3)-(0.5*6*3)+RD=0

10.715-18-5+RD=0

RD=16.285KN

FINAL REACTIONS

RA= 9.308KN

RB=RB1+RB2=17.692+9.9= 27.592KN

RC=RC1+RC2F8.1+10.715= 18.815KN

RD= 16.285KN

END SHEAR FORCE

FA= 9.308KN

FBL=9.308-(0.5*6*9) = -17.692KN

FBR=9.308-(0.5*6*9) +27.592= 9.9KN

FCL=9.308-(0.5*6*9) +27.592-(0.5*1.5*12) -(0.5*1.5*12) = -8.1KN

FCR=9.308-(0.5*6*9) +27.592-(0.5*1.5*12) -(0.5*1.5*12) = -8.1+18.815= 10.715KN

FDL=9.308-(0.5*6*9) +27.592-(0.5*1.5*12) -(0.5*1.5*12) = -8.1+18.815-(6*3) -(0.5*6*3) = -16.285KN

FDR=9.308-(0.5*6*9) +27.592-(0.5*1.5*12) -(0.5*1.5*12) = -8.1+18.815-(6*3) -(0.5*6*3)+16.285= 0

To Calculate The Mid Shear Force

Take Summation of forces acting on node or member.

MID SHEAR FORCE

F(A-B)= 9.308-(0.5*(6/2)*(9/2))= 2.558KN

F(B-C)= 9.308-(0.5*6*9)+27.592-(0.5*1.5*12) = 0.9KN

F(C-D)= 9.308-(0.5*6*9)+27.592-(0.5*1.5*12)-(0.5*1.5*12)+18.815-(3*3)-(0.5*(3/2)*(3/2))= -0.535KN

FINAL MOMENT VALUE WILL BE THE END BENDING MOMENTS

END BENDING MOMENTS

MAB= -13.22KNm

MBA= 11.362KNm

MBC= -11.362KNm

MCB= 8.662KNm

MCD= -8.662KNm MDC= 16.369KNm

To Calculate The Mid 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

MID BENDING MOMENTS

note

To Draw the bending moment diagram first we have to draw free moment diagram, and then we have to mark all the values of final moments which we got by solving the slope deflection equation. join all the marked points to get the net bending moment diagram.

sfd and bmd diagram manual

sfd and bmd diagram manual

staad output

Result Summary

FX = FORCE IN HORIZONTAL DIRECTION

FY = FORCE IN VERTICAL DIRECTION

MZ = BENDING MOMENT

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