shear force bending moment of udl and point

DRAW SHEAR FORCE BENDING MOMENT DIAGRAM OF GIVEN FIGURE. 

  • WE KNOW, THERE ARE THREE EQUILIBRIUM EQUATION

∑FX=0, (SUMMATION OF HORIZONTAL FORCE) 

∑FY=0, (SUMMATION OF VERTICAL FORCE) 

∑M=0, (SUMMATION OF MOMENTS) 

SIGN CONVENTION

STEPS FOR FINDING REACTIONS 

Here, In The Above Figure We Have Two Unknown Reaction Force  RA  And RH. 

1.   Apply First Equilibrium Equation 

∑FY( Summation Of Vertical Forces)= 0 

Here, TWO Vertical Reaction Forces Are Acting Upward. They Are +RA  And +RH and Three Vertical Loads are acting downward 

2.   Apply Second Equilibrium Equation 

∑M(Summation Of Moments)= 0 

Formula Of Moment= Force*Distance. 

(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. 

FINDING THE REACTIONS

∑FY=0, 

RA-5 – (7*3)-(4*2)-3+RH =0 

RA+RH=37Kn     (Eqn-1) 

LET US USE ∑M=0,  

∑MA=0 

(RA*0)+ (5*2) + (7*3*5.5) + (4*2*10) + (3*12)-13RH=0 

RH= (241.5/13) =18.577KN. 

FROM (Eqn-1), 

RA+18.577=37Kn 

RA=18.423KN

Or, 

∑MH=0 

(RH*0)- (3*1)-(4*2*3)-(7*3*7.5)-(5*11) +13RH=0 

RA= (239.5/13) =18.423KN

END SHEAR FORCE 

FA= RA= 18.423KN 

FB= 4.5-6=18.423-5= 13.423KN 

FC= 18.423-5= 13.423KN 

FD= 18.423-5-(7*3) = -7.577KN 

FE= 18.423-5-(7*3) = -7.577KN 

FF= 18.423-5-(7*3)-(4*2) = -15.577KN 

FG= 18.423-5-(7*3)-(4*2)-3 = -18.577KN 

FH= 18.423-5-(7*3)-(4*2)-3+18.577 = 0 

MID SHEAR FORCE

F(A-B)= 18.423KN 

F(B-C)= 18.423-5 = 13.423KN 

F(C-D)= 18.423-5-(7*1.5) = 2.932KN 

F(D-E)= 18.423-5(7*3) = -7.577KN 

F(E-F)= 18.423-5-(7*3)-(4*1) = -11.577KN 

F(F-G)= 18.423-5-(7*3)-(4*2) = -15.577KN 

F(G-H)= 18.423-5-(7*3)-(4*2)-3 = -18.577KN 

END BENDING MOMENTS 

MA= (18.423*0) = 0        (AS, THE DISTANCE IS ZERO) 

MB= (18.423*2) = 36.846KNm 

MC= (18.423*4)-(5*2) = 63.692KNm 

MD= (18.423*7)-(5*5)-(7*3*1.5) = 72.461KNm 

ME= (18.423*9)-(5*7)-(7*3*3.5) = 57.307KNm 

MF= (18.423*11)-(5*9)-(7*3*5.5)-(4*2*1) = 34.153KNm 

MG= (18.423*12)-(5*10)-(7*3*6.5)-(4*2*2)= 18.576KNm 

MH= (18.423*13)-(5*11)-(7*3*7.5)-(4*2*3)-(3*1) = 0 

MID – MOMENTS 

M(A-B)= (18.423*1) = 18.423KNm 

M(B-C)= (18.423*3)-(5*1) = 50.269KNm 

M(C-D)= (18.423*5.5)-(5*3.5)-(7*1.5*0.75) = 75.952KNm 

M(D-E)= (18.423*8)-(5*6)-(7*3*2.5) = 64.884KKNm 

M(E-F)= (18.423*10)-(5*8)-(7*3*4.5)-(4*1*0.5) = 47.73KNm 

M(F-G)= (18.423*11.5)-(5*9.5)-(7*3*6)-(4*2*1.5)  = 26.365KNm 

M(G-H)= 18.423*12.5)-(5*10.5)-(7*3*7)-(4*2*2.5) -(3*0.5) = 9.288KNm 

SFD & BMD MANUAL

STAAD OUTPUT

STAAD RESULT SUMMARY

FX = FORCE IN HORIZONTAL DIRECTION

FY = FORCE IN VERTICAL DIRECTION

MZ = BENDING MOMENT

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