shear force and bending moment diagram of inclined load with udl

draw shear force and bending moment diagram of given figure. 

SIGN CONVENTION

STEPS FOR FINDING REACTIONS

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

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 +RE and Three Vertical Loads are acting downward

i.e., -7KN of nodal load, 4KN/m of UDL &  Inclined load of -3KN and the angle inclination is 60 degree.

Divide the inclined load into two components.

Horizontal force= 3cos30 and vertical force= 3sin30.     

Take The Summation Of All Vertical Forces Acting On The Beam (Both Upward And Downward)

In the inclined load case, we have to find Horizontal force (FX) also

Take The Summation Of Horizontal Force Acting On The Beam (Both left And right).

Force acting towards right is taken as positive (+) and force acting towards left is taken as negative (-).

2.   Apply Second Equilibrium Equation

∑M(Summation Of Moments)= 0

Formula Of Moment= Force*Distance.

e.g. ∑MD= +(RA*7)-(5*5)-(4*3*1.5)

Here We Have Taken ∑M At D.

Take The Summation Of All Moments Acting On The Beam (Both Clock Wise And Anti clock 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 REACTION

Horizontal Reaction

∑FX=0

RAX-3COS60=0

RAX= 1.5 KN

Vertical Reaction

∑FY=0

RA-7-(4*3)-3SIN60+RE=0

RA+RE= 21.598KN – (1)

∑MA=0

(RA*0)+(7*2)+(4*3*5.5)+(3SIN60*7)-(RE*10)= 0

RE= 9.819KN

FROM Eqn (1)

RA=21.598-9.819= 11.779KN

Or,

∑ME=0

(RE*0)-(3SIN60*3)-(4*3*4.5)-(7*8)+(RA*10)=0

RA= 11.779KN

To Calculate The Axial Force

 Take summation of forces acting on node or member.

CALCULATION OF Axial Force

FA=RAX= 1.5KN

FB=1.5KN

FC=1.5KN

FD=1.5-(3COS60)= 0

To Calculate The End Shear Force

 Take summation of forces acting on node or member.

In the below calculation L indicates left and R indicates right.

CALCULATION OF END SHEAR FORCE

FA= 11.779KN

FB=11.779-7= 4.779KN

FC=11.779-7= 4.779KN

FDL=11.779-7-(4*3)= -7.221KN

FDR=11.779-7-(4*3)-3SIN60= -9.819

FE=11.779-7-(4*3)-3SIN60+9.819= 0

To Calculate The Mid Shear Force

 Take Summation of forces acting at the mid span of the member.

CALCULATION OF MID SHEAR FORCE

F(A-B)= 11.779KN

F(B-C)= 11.779-7= 4.779KN

F(C-D)= 11.779-7-(4*1.5)= -1.221KN

F(D-E)= 11.779-7-(4*3)-3SIN60= -9.819KN

To Calculate The End Bending Moment

 Calculate moments acting at the mid span of the member and add the remaining distance up to which node/member you want to calculate the moment.

CALCULATION OF END BENDING MOMENTS

MA= (11.779*0)= 0

MB= (11.779*2)= 23.559KNm

MC= (11.779*4)-(7*2)= 33.116KN

MD= (11.779*7)-(7*5)-(4*3*15)= 29.453KNm

ME= (11.779*10)-(7*8)-(4*3*4.5)-(3SIN60*3)= 0

To Calculate The Mid Bending Moment

 Calculate moments acting at the mid span of the member and add the remaining distance up to which node/member you want to calculate the moment.

CALCULATION OF MID BENDING MOMENTS

M(A-B)= (11.779*1)= 11.779KNm

M(B-C)=  (11.779*3)-(7*1)= 28.337KNm

M(C-D)= (11.779*5.5)-(7*3.5)-(4*1.5*0.75)= 35.785KNm

M(D-E)= (11.779*8.5)-(7*6.5)-(4*3*3)-(3SIN60*1.5)= 14.724KNm

shear force and bending moment diagram manually
shear force and bending moment diagram
software result

axial force

shear force

bending moment

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