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.
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