Analysis of Bending Moment, Shear Force & Beam Design Elements

This report provides a detailed analysis of bending moment and shear force in simply supported beams under various loading conditions, including point loads and uniformly distributed loads. It includes calculations for reactions at supports, shear force and bending moment values at different points along the beam, and corresponding diagrams. The report also discusses statutory requirements for structural safety, the definitions of dead load, live load, and factor of safety, and their application in beam design. Furthermore, it covers the calculation of maximum deflection in beams and the factors influencing deflection, such as material properties and beam profile. The document concludes with a brief overview of different types of supports used in structural engineering. This solved assignment is available on Desklib, offering students a valuable resource for understanding beam design principles.

BENDING MOMENT
&
SHEAR FORCE
WITH
BEAM DESIGN CONSIDERATION
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TASK 1
TASK 1(A) P(1)
a) A simply supported beam of length is 14 meters and carries 150 kN load at its center.
The beam arrangement is shown below:-
Figure 1.1 Beam arrangement
Calculation for reaction at each support .
ΣMA = 0
The moments about point A is shown below:
- P1*7 + RB*14 = 0
ΣMB = 0:
The moments about point B is shown below:
- RA*14 + P1*7 = 0
Reaction at support B
RB = ( P1*7) / 14 = ( 150*7) / 14
RB = 75.00 (kN)
Reaction at support A
RA = ( P1*7) / 14 = ( 150*7) / 14 = 75.00 (kN)
RA = 75.00 (kN)
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As load is acting at centre of beam therefore, both support share equal reaction.
Calculations for shear force and bending moment diagram (Hibbeler, 2008)
For left half of beam
Equation for Shear force (Q)
Q(x1) = + RA
Shear force values at the edges of beam
Q1(0) = + 75 = 75 (kN)
Q1(7) = + 75 = 75 (kN)
Equation for bending moment (M)
M(x1) = + RA*(x1)
Bending moment values at the edges of beam
M1(0) = + 75*(0) = 0 (kN*m)
M1(7) = + 75*(7) = 525 (kN*m)
For right half of beam
Equation for Shear force (Q)
Q(x2) = + RA - P1
Q2(7) = + 75 - 150 = -75 (kN)
Q2(14) = + 75 - 150 = -75 (kN)
Equation for bending moment (M)
M(x2) = + RA*(x2) - P1*(x2 - 7)
Bending moment values at the edges of beam
M2(7) = + 75*(7) - 150*(7 - 7) = 525 (kN*m)
M2(14) = + 75*(14) - 150*(14 - 7) = 0 (kN*m)
From the shear force and bending moment values calculated, shear force and bending moment
diagram shown below :-
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Figure 1.2 SF & BM Diagram (Robert ,2012)
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b) A simply supported beam of length is 8 meters and carries 65 kN load at 2 m from the left of
support.
The beam arrangement is shown below :-
Figure 1.3 Beam arrangement
Calculation for reaction at each support .
ΣMA = 0
The moments about point A is shown below:
- P1*2 + RB*8 = 0
ΣMB = 0:
The moments about point B is shown below:
- RA*8 + P1*6 = 0
Reaction at support B
RB = ( P1*2) / 8 = ( 65*2) / 8 = 16.25 (kN)
RB = 16.25 (kN)
Reaction at support A
RA = ( P1*6) / 8 = ( 65*6) / 8 = 48.75 (kN)
RA = 48.75 (kN)
As load is acting at left side of beam therefore, left support having more reaction.
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Calculations for shear force and bending moment diagram
For left half of beam
Equation for Shear force (Q)
Q(x1) = + RA
Shear force values at the edges of beam
Q1(0) = + 48.75 = 48.75 (kN)
Q1(2) = + 48.75 = 48.75 (kN)
Equation for bending moment (M)
M(x1) = + RA*(x1)
Bending moment values at the edges of beam
M1(0) = + 48.75*(0) = 0 (kN*m)
M1(2) = + 48.75*(2) = 97.50 (kN*m)
For right half of beam
Equation for Shear force (Q)
Q(x2) = + RA - P1
Q2(2) = + 48.75 - 65 = -16.25 (kN)
Q2(8) = + 48.75 - 65 = -16.25 (kN)
Equation for bending moment (M)
M(x2) = + RA*(x2) - P1*(x2 - 2)
Bending moment values at the edges of beam
M2(2) = + 48.75*(2) - 65*(2 - 2) = 97.50 (kN*m)
M2(8) = + 48.75*(8) - 65*(8 - 2) = 0 (kN*m)
From the shear force and bending moment values calculated, shear force and bending moment
diagram shown below :-
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Figure 1.4 SF & BM Diagram
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c) A simply supported beam of length is 25 meters and carries uniform distributed load of 25
kN/m load .
The beam arrangement is shown below :-
Figure 1.5 Beam arrangement
Calculation for reaction at each support .
ΣMA = 0
The moments about point A is shown below:
- q1*25*(25/2) + RB*25 = 0
ΣMB = 0:
The moments about point B is shown below:
- RA*25 + q1*25*(25 - 25/2) = 0
Reaction at support B
RB = ( q1*25*(25/2)) / 25 = ( 25*25*(25/2)) / 25 = 312.50 (kN)
RB = 312.5 (kN)
Reaction at support A
RA = ( q1*25*(25 - 25/2)) / 25 = ( 25*25*(25 - 25/2)) / 25 = 312.50 (kN)
RA = 312.5 (kN)
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As load is acting uniform on the beam therefore, both support share equal reaction.
Calculations for shear force and bending moment diagram
For full beam
Equation for Shear force (Q)
Q(x1) = + RA - q1*(x1 - 0)
Shear force values at the edges of beam
Q1(0) = + 312.50 - 25*(0 - 0) = 312.50 (kN)
Q1(25) = + 312.50 - 25*(25 - 0) = -312.50 (kN)
Equation for bending moment (M)
M(x1) = + RA*(x1) - q1*(x1)2/2
Bending moment values at the edges of beam
M1(0) = + 312.50*(0) - 25*(0 - 0)2/2 = 0 (kN*m)
M1(25) = + 312.50*(25) - 25*(25 - 0)2/2 = 0 (kN*m)
Point at which shear force is zero having bending moment maximum is point of contraflexure
x = 12.50
Point of contraflexure point x = 12.50:
Maximum bending moment
M1(12.50) = + 312.50*(12.50) - 25*(12.50 - 0)2/2 = 1953.13 (kN*m)
From the shear force and bending moment values calculated, shear force and bending moment
diagram shown below :-
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Figure 1.6 SF & BM Diagram
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d) A simply supported beam of length is 14 meters and carries point load of 150 kN and also
uniform distributed load of 25 kN/m load .
Shear force and bending moment diagram shown below :-
Figure 1.7 SF & BM Diagram
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e) A simply supported beam of length is 25 meters and carries uniform distributed load of 25
kN/m load and 50kN at 10 m from right side side .
Shear force and bending moment diagram shown below :-
Figure 1.8 SF & BM Diagram
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