Strength of Materials-Shear Force and Bending Moment Diagram.pptx

SFD and BMD of Cantilever and Simply Supported Beam

1. STRENGTH OF MATERIALS Dr.S.SURESH , Assistant Professor, Department of Mechanical Engineering. Unit-II Transverse loading on beams and stress in beam

2. Beams Types of beams Types of loads Shear force and bending moment SFD & BMD

3. Beams Structural element Resist loads which perpendicular to the their axis

4. Types of Beam

5. Cantilever beam Simply Supported beam

6. Over hanging Beam Continuous Beam

7. Types of Loads

8. Point Load UDL

9. Shear force The shear force at any point along a loaded beam may be defined as the algebraic sum of all vertical forces acting on either side of the point on the beam .

10. BENDING MOMENT Bending moment at any point along a loaded beam may be defined as the sum of the moments due to all vertical forces acting on either side of the point on the beam . 

11. Shear Force And Bending Moment Diagrams A shear force diagram (SFD) which shows the shear force at every section of the beam due to transverse loading on it.  A bending moment diagram (BMD) is a diagram which shows the bending moment at every section of the beam due to transverse loading on it.

12. Shear Force Calculation SF at B = 1.5 KN SF at C = 1.5 + 2 = 3.5 KN SF at A = 1.5 + 2 = 3.5 KN Bending Moment Calculation BM at B = 0 BM at C = -1.5 x 0.5 = -0.75 KN-m BM at A = (-1.5 x 1.5) + (-2 x 1) = -4.25 KN-m 1. Draw Shear force and Bending moment diagram for a cantilever beam of span 1.5 m carried point load as shown in Fig.

13. Example

14. Shear Force calculation SF at B = 0 SF at C = 1x1.5 =1.5 KN SF at A = 1.5 KN Bending Moment calculation BM at B = 0 BM at C = 1x1.5x(1.5/2) =1.125 KN BM at A = 1x1.5x(1.5/2 + 0.5) =1.875 KN A Cantilever of length 2.0 carries a uniformly distributed load of 1 KN/m run over a length of 1.5 m from the free end. Draw the shear force and bending moment diagrams for the cantilever.

15. Find the reactions at A & B 1.Taking moment at A 2. Sum of forces upwards = Sum of forces downwards

16. SF Calculation SF at B = -W/2 SF at C = -W/2 + W = +W/2 BM Calculation BM at B = 0 SF at C = W/2 x L/2 = WL/4 BM at A = 0 Note:

17. Shear Force Bending Moment

18. Reactions at A & B Shear Force SF at B = -0.5 wl SF at C = -0.5 wl + w (l/2) = 0 SF at A = +o.5wl Bending Moment BM at B = 0 SF at C = 0.5 wl (l) - w (l/2) (l/4) = wl 2 /8 SF at A = 0

19. Shear Force Calculation SF at B = - 50 KN SF at D = -50 + 40 = -10 KN SF at C = -50 + 40 + (10 x 4) = + 30 KN (Without Point Load) SF at C = 30 + 50 = + 80 KN SF at A = +80 KN SF at E = -R B + 40 + 10 (x-4) 0 = -50 + 40 + 10 (x-4) X = 5 m