simple structural analysis of frame structure

The structural analysis of a structure includes calculating the internal and external forces acting on the body such as sheraf force and bending moment and plotting the diagram for it.

draw sfd and bmd diagram of frame structure having udl load of 8KN/m and uvl load of 12KN/m

SIGN CONVENTION

steps of slope and deflection method

fixed end moment

slope and deflection equation

equilibrium equation

final moments

Reaction at each span

SPAN AB

∑MB=0

3.603+7.206-3FA=0

FA=3.603KN

∑Fx=0

FA+FD=0

3.603+FD=0

FD=-3.603KN

SPAN BC

∑MC=0

-7.208+9.485-(8*4*2)+4RB=0

RB= 15.431KN

∑FY=0

RB-(8*4)+RC1=0 RC1= 16.569KN

SPAN CD

∑MC=0

-9.485+1.404-3RD+(0.5*3*12*((2*3)/3))=0

RD= 9.306KN

∑FY=0

RC2+RD-(0.5*3*12)=0

RC2= 8.694KN

or

∑MD=0

-9.485+1.404+3RC2-(0.5*3*12*(3/3))=0

RC2= 8.694KN

SPAN DE

∑MD=0

-1.404-0.702-3FE=0

FE= -0.702KN

∑Fx=0

FE+FD=0

FD= 0.702KN

∑Fx=0

FA+FE+FC=0

3.603-0.702+FC=0

FC= -2.901KN

FINAL REACTIONS

HORIZONTAL FORCE

FA= 3.603KN

FE= -0.702KN

FC= -2.901KN

VERTICAL FORCE

RA=RB= 15.431KN

RC=RC1+RC2=16.569+8.694= 25.263KN

RE= 9.306KN

END SHEAR FORCE

SPAN AB

FA= 3.603KN

FBL= 3.603KN

FBR=3.603-3.603= 0

SPAN DE

FE= -0.702KN

FDL= -0.702KN

FDR=-0.702KN+0.702= 0

SPAN BC & CD

FB=RB= 15.431KN

FCL=15.431-(8*4)= -16.569KN

FCR=15.431-(8*4)+25.263= 8.694KN

FDL=15.431-(8*4)+25.263-(0.5*3*12)= -9.306KN

FDR=15.431-(8*4)+25.263-(0.5*3*12)+9.306= 0

To Calculate The Mid Shear Force

Take Summation of forces acting on node or member.

MID SHEAR FORCE

F(A-B)= 3.603KN 

F(B-C)= 15.431-(8*2)= -0.569KN 

F(C-D)= 15.431-(8*4)+25.623-(0.5*1.5*(12/2))= 4.554KN 

F(D-E)= -0.702KN 

FINAL MOMENT VALUE WILL BE THE END BENDING MOMENTS

END BENDING MOMENTS

MAB= 3.603KNm 

MBA= 7.208KNm 

MBC= -7.208KNm 

MCB= 9.485KNm 

MCD= -9.485KNm 

MDC= 1.404KNm 

MDE= -1.404KNm 

MED= -0.702KNm 

MID BENDING MOMENTS

M(A-B)= 3.603-(3.603*15)= -1.802KNm 

M(B-C)= -7.028+(15.431*2)-(8*2*1)= 7.834KNm 

M(C-D)= -7.028+(15.431*5.5)-(8*4*3.5)+9.485-9.485+(25.263*1.5)-(0.5*1.5*(12/2)*(1.5/3))= 1.487KNm 

M(D-E)= -0.702-(-0.702*1.5)= 0.351KNm 

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.

structural analysis diagram manual

structural analysis diagram

staad output

Result Summary

FX = FORCE IN HORIZONTAL DIRECTION

FY = FORCE IN VERTICAL DIRECTION

MZ = BENDING MOMENT

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