Shear Force and Bending Moment Diagram of Beam Span

This article explains how to calculate shear force and bending moment diagram of beam span with solved examples. It covers the concept of uniformly distributed load and point loads. It also discusses the slenderness ratio of I section beam.

Task: 1
Shear force and bending moment diagram of beam span.
∑ F x=0
HB =0
∑ M A =0
Taking anti close wise moment at positive and clockwise as negative.
RA + RB =170
P1 x 0.5−40 x 3 x 1.5+ RB x 3−P2 x 3.5=0
7.5−180+ RB x 3−122.5=¿
RB=98.33
RA =170−98.33=71.67
Shear force:
Force,
Q ( x 1=0.5 )=−P 1
Q1 ( x 1=0 ) =−15 kN
Q1 ( x 1=0.5 ) =−15 kN
Bending moment at CA span,
M ( x1 ) =−Load x perpendicular distance
M ( xc ) =−15 x 0=0
M ( xA ) =−15 x 0.5=−7.5
Shear force and bending moment at AB span.
Q ( x 2 ) =−P 1+RA −q1 x ( x2 −0.5 )
¿−15+71.67−40 x ( 0.5−0.5 )
Q ( x 2=0.5 )=56.67 kN
Q1 ( x 1=3.5 ) =−15+71.67−40 x ( 3.5−0.5 )
Q1 ( x 1=3.5 ) =−63.33 kN
Bending moment at mid-span:
1

M ( x 2 )=−P 1 x x 2+ RA x ( x 2−0.5 )−q1
( x 2−0.5 )2
2
M ( x 2=0.5 ) =−15 x 0.5+71.67 x ( 0.5−0.5 ) −40 x ( 0.5−0.5 )2
2 =−7.5 kNm
M ( x 2=0.5 )=−15 x 3.5+71.67 x ( 3.5−0.5 )−40 x ( 3.5−0.5 )2
2 =−17.5 kNm
Extreme point,
M = W l2
8 = 40 x 3 x 3
8 =45 kNm
For BD span,
Shear force and bending moment.
Shear force,
Q ( x3 )=−P 1+RA−q1 x ( 3.5−0.5 ) +98.33
Q ( x3=3.5 ) =−15+71.67−40 x ( 3.5−0.5 ) +98.33=35
Q ( x3=4 ) =−15+71.67−40 x ( 4−0.5 ) +98.33=35
Bending moment,
M ( x3 ) =−P 1+ RA−q 1 x ( 3.5−0.5 ) x [ ( x3 −3.5 ) + ( 3.5−0.5 )
2 ] + RB x ( x 3−3.5 )
M (3.5 )=−15+3.5−71.67 x ( 3.5−0.5 )−40 x 3 x ( 1.5 ) +98.33 x ( 3.5−3.5 ) =−17.50 kNm
M ( 4 )=−15+ 4−71.67 x ( 4−0.5 )−40 x 3 x ( 2 ) +98.33 x ( 4−3.5 )=0
2

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Task: 2
As per the given condition, the serial size 203 x 203 x 52 which have specific dimension and
depth of section 206.2mm,width 204.3 mm, flange thickness 7.9mm and web thickness 12.5. A
uniformly distributed load is applied over the span which additional two point loads.
For the point loads the bending stress,
SF=25 kN
And the bending moment,
BM xx=225 x 2−25 x 2=0
BM yy =12.5 x X −25 X +50
Finding dM
dx ,
dM
dx =12.5−25=−12.5
Bending moment for shear force for the uniform distributed load,
BM xx=−W x X 2
2
Apply boundary condition and loading condition, x = 0 and x = 4m,
BM x=4=−W x 42
2 =−40 x 16
2 =−320 kN m2
Now, dm
dx =−Wx=−40 x 4=−160
Bending moment at uniformly distributed load,
BM x=WL
2 x− W x2
2 = 40 Lx
2 − 40 x2
2 =20 ( 4 x−x2 )
d BM xx
dx =20 ( 4−2 x )
20 ( 4−2 x ) =0
4

x=2 m
The point of contra-flexure obtained by,
−63.33−15 ( x +0.5 ) −35 x X x X
2
Differential above equation by x,
dm
dx =15 x−7.5−17.5 x2
15−35 x=0
x=2.33 m
As per shown in diagram,
Y = A 1 Y 1+ A 2 Y 2+ A 3 Y 3
A 1+ A 2+ A 3
¿
204.3 x 12.5 x ( 106.8+12.5+ 12.5
2 ) + ( 106.8 ) ( 7.9 ) ( 160.8
2 +12.5 ) + ( 204.3 x 12.5 x 6.25 )
2 x 204.3 x 12.5+160.8 x 12.5
Y =83.24 mm
I =INA + A h2
Iy 1=2 x [ 1
12 ( 204.3 ) ( 12.5 ) 3 + ( 204.3 x 12.5 ) ( 86.65 ) 3
] + 1
12 ( 160.8 ) ( 7.9 ) 3=¿
¿ 3322942056.0909375+6606.7226
¿ 3322948662.8135375
Iy 1=3.322 x 109 mm4
M
I = σ
Y
σ = MY
I = 45 x 83.24 x 106
3.322 x 109 =1127 N /mm2
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Task: 3
Slenderness ratio k = l
r =185.8
5.18 =35.86
Value of slenderness is less than 200 which is safe for the design point of view. If the slenderness
value greater than 200 so that there are higher chance to get failed.
This I section beam is short because of safe operation with respect to y-directional length.
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