shear force and bending moment diagram of fixed beam easily

draw shear force and bending moment diagram by slope and deflection method

SIGN CONVENTION

steps of slope and deflection method

fixed end moment

slope and deflection equation

applying condition of equilibrium

final moments

FINDING SUPPORT REACTIONS BY SEPARATING EACH SPAN

SPAN AB

∑MA=0

-23.55+(RA*0) +(15*4*2) +12.9-(RB*4) =0

RB1= 27.34KN

∑FY=0

RA-(15*4) +27.34=0

RA= 32.66KN

SPAN BC

∑MB=0

-12.9+(RB2*0) +(10*2) +1.96-(RC*5) =0

RC=1.81KN

∑FY=0

RB2-10+RC=0

RB2-10+1.81=0

RB2= 8.19KN

FINAL SUPPORT REACTIONS

RA= 32.66KN

RB= RB1+RB2= 27.34+8.19= 35.53KN

RC= 1.81KN

END SHEAR FORCE

FA= 32.66KN

FBL= 32.66- (15*4) = -27.34KN

FBR= 32.66- (15*4) +35.53= 8.19KN

FC= 32.66- (15*4) +35.53-10+1.81= 0

To Calculate The Mid Shear Force

 Take Summation of forces acting on node or member.

MID SHEAR FORCE

F(A-B) = 32.66- (15*2) = 2.66KN

F(B-C) = 32.66- (15*4) +35.53-10= -1.81KN

END BENDING MOMENTS

MAB= -23.55KNm

MBA= 12.9KNm

MBC= -12.9KNm

MCB= 1.96KNm

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.

MID BENDING MOMENTS

M(A-B) = -23.55+(32.66*2) -(15*2*1) = 11.77KNm

M(B-C) = -23.55+(32.66*6.5) -(15*4*4.5) +(35.53*2.5) -(10*0.5) = 2.565KNm

shear force and bending moment diagram manual

shear force and bending moment diagram manual

shear force at different sections of the beam

SPAN AB

F(0m)= 32.66KN

F(1m)= 32.66-(15*1)= 17.66KN

F(2m)= 32.66-(15*2)= 2.66KN

F(3m)= 32.66-(15*3)= -12.34KN

F(4m)= 32.66-(15*4)= -27.34KN

SPAN BC

F(0m)= 32.66-(15*4)+35.53= 8.178KN

F(1.25m)= 32.66-(15*4)+35.53= 8.178KN

F(2.5m)= 32.66-(15*4)+35.53-10= -1.82KN

F(3.75m)= 32.66-(15*4)+35.53-10= -1.82KN

F(5m)L= 32.66-(15*4)+35.53-10= -1.82KN

bending moment at different sections of the beam

SPAN AB

M(0m)= -23.55KNm

M(1m)= -23.55+(32.66*1)-(15*1*0.5)= 1.61KNm

M(2m)= -23.55+(32.66*2)-(15*2*1)= 11.77KNm

M(3m)= -23.55+(32.66*3)-(15*3*1.5)= 6.93KNm

M(4m)= -23.55+(32.66*4)-(15*4*2)= -12.91KNm

SPAN BC

M(0m)=- 23.55+(32.66*4)-(15*4*2)= -12.91KNm

M(1.25m)= -23.55+(32.66*5.25)-(15*4*3.25)+(35.53*1.25)= -2.68KNm

M(2.5m)= -23.55+(32.66*6.5)-(15*4*4.5)+(35.53*2.5)-(10*0.5)= 2.54KNm

M(3.75m)= -23.55+(32.66*7.75)-(15*4*5.75)+(35.53*3.75)-(10*1.75)= 0.26KNm

M(5m)=-23.55+(32.66*9)-(15*4*7)+(35.53*5)-(10*3)= -2.02KNm

staad output

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