Shear Force
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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 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.

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Shear Force Presentation Transcript

  • 1. Shear Force HNC In Engineering – Mechanical Science Edexcel HN Unit: Engineering Science (NQF L4) Author: Leicester College Date 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 .
  • 2. 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
    For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications These files support the Edexcel HN unit – Engineering Science (Mechanical) File Name Unit Outcome Key Words Stress 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
  • 3. 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.
  • 4. Shear force / bending moment diagrams The method A) 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
  • 5. 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
  • 6. 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
  • 7. To find point of max BM – Using similar triangles AB / AD = DF / EF From dimensions and; AB = (DF x AD)/ EF = 6 x 141.8 Shear 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
  • 8. Size of max bending moment As with SF diagram – Working from the left (R1) Max BM takes place when SF = 0 ie at 10.7267m from R1 Max BM = (wl – wl 2 – wl 2 ) (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 – 10.73 2 x 10 – 4.47 2 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
  • 9. 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
  • 10. 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 10 3 I y R I = bd 3 = (0.1 x 0.1 3 ) = 8.33 x 10 -6 12 12 y = Distance from neutral axis0.1/2 Therefore,  1365.92 x 10 3 x 50 x 10 -3 = 8195.5 MPa 8.33 x 10 -6
  • 11. 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. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College.