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 modulusBM, shear force diagrams

1.1 Shear force, bending moment, stress

Selecting beams 1.2 Beams, columns, struts, slenderness ratioTorsion introduction 1.3 Torsion, stiffness, twistingDynamics 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 diagramsThe problem;

A B

20 kN/m UDL 10kN/m UDLC

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 diagramC) Find point of max BM (using similar triangle method)D) Find BM at points along the beamE) Draw BM diagram from points etc calculatedF) Calculate I for the beamG) 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 kNTaking 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 BMR1 201.8kN

10kN/m UDL

6m 20kN/m UDL

6m

R2 – 168.2kN

To find point of max BM – Using similar trianglesAB / AD = DF / EFFrom 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 R1Max 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 kNmAnd similarly;BM at A = R1 x 6 - (6 x 6 x 10) /2 = 1030.8 kNmFor BM at C we can come from the other end;BM at C = R2 x 13 – (13 x 13 x 10) / 2 = 1341.6kNmNote – At A and C second UDL is not present

Bending moment diagram1365.92kN/m

1030.8 kN/m 1341kN/m

Note BM diagram starts and finishes at zeroIe no BM at the ends

Calculating max Stress in the beamMaximum 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 MPa8.33 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

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