Shear Force
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Transcript of Shear Force
Shear Force
HNC In Engineering – Mechanical Science
Edexcel HN Unit: Engineering Science (NQF L4)
Author: Leicester CollegeDate created:Date revised: 2009
Abstract; A graphical method to determine the shear force and bending moment distribution along a simply supported beam is given. A suitable example is used to illustrate the major steps in the process.
A sample calculation for the determination of the maximum stress values is also given.
© Leicester College 2009. This work is licensed under a Creative Commons Attribution 2.0 License.
Contents• Shear force / bending moment diagrams• Find reactions R1 and R2• Shear force Diagram• To find point of max BM – Using similar triangles• Size of max bending moment• Bending moment diagram• Calculating max Stress in the beam• Credits
File Name Unit Outcome Key WordsStress introduction 1.1 Stress, strain, statics, young’s modulus
BM, shear force diagrams
1.1 Shear force, bending moment, stress
Selecting beams 1.2 Beams, columns, struts, slenderness ratio
Torsion introduction 1.3 Torsion, stiffness, twisting
Dynamics introduction
2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation
For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications
These files support the Edexcel HN unit – Engineering Science (Mechanical)
Shear force / bending moment diagrams
The problem; A B
20 kN/m UDL 10kN/m UDL
C
R1 6m 6m 13m R2
The beam is 25m long and 100mm square section.
Shear force / bending moment diagrams
The methodA) Determine the unknown reactions (Forces R1 and
R2)
B) Draw Shear force diagram
C) Find point of max BM (using similar triangle method)
D) Find BM at points along the beam
E) Draw BM diagram from points etc calculated
F) Calculate I for the beam
G) Use bending formula to find max stress value
Find reactions R1 and R2
Taking moments about R1 to find R2
(9 x 120) + ( 12.5 x 250) = 25 x R2
(1080 + 3125) / 25 = R2 and R2 = 168.2 kN
Taking moments about R2 to find R1
( 25 x R1) = (16 x 120) + (12.5 X 250 )
( 1920 + 3125) / 25 = R1 = 201.8 kN
Check – UP forces = Down Forces R1 + R2 = 120 + 250 = 370 kN OK
Shear force Diagram
Working from the left hand side;
Point of max BM
R1 201.8kN
10kN/m UDL
6m 20kN/m UDL
6m
R2 – 168.2kN
To find point of max BM – Using similar trianglesAB / AD = DF / EF
From dimensions and; AB = (DF x AD)/ EF = 6 x 141.8Shear forces given (141.8 + 38.2)
= 4.7267m
141.8 (ie 201.8 – (6 x 10)
D F
A B C
38.2 (ie 168.2 – 13 x 10) E
Max BM from R1 = 6 + 4.7267 = 10.7267m
Size of max bending momentAs with SF diagram – Working from the left (R1)
Max BM takes place when SF = 0 ie at 10.7267m from R1
Max BM = (wl – wl2 – wl2) (minus sign indicates ‘hogging’)
2 2 (divide by 2 because UDL acts at half length
and w = load(kN/m) and l = length)
Max BM = R1 x 10.73 – 10.732 x 10 – 4.472 x 20 = 1365.92 kNm
2 2
Max BM = 1365.92 kNm
And similarly;
BM at A = R1 x 6 - (6 x 6 x 10) /2 = 1030.8 kNm
For BM at C we can come from the other end;
BM at C = R2 x 13 – (13 x 13 x 10) / 2 = 1341.6kNm
Note – At A and C second UDL is not present
Bending moment diagram
1365.92kN/m
1030.8 kN/m 1341kN/m
Note BM diagram starts and finishes at zero
Ie no BM at the ends
Calculating max Stress in the beam
Maximum stress occurs at max BM point
To find max stress using ‘Bending Equation’ – pp 42
M = = E Where M = Max BM = 1365.92 x 103
I y R I = bd3 = (0.1 x 0.13) = 8.33 x 10-6
12 12
y = Distance from neutral axis0.1/2
Therefore, 1365.92 x 103 x 50 x 10-3 = 8195.5 MPa
8.33 x 10-6
This resource was created Leicester College and released as an open educational resource through the Open Engineering Resources project of the Higher Education Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.
© 2009 Leicester College
This work is licensed under a Creative Commons Attribution 2.0 License.
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