1. UNIT-II Shear Force Diagrams and Bending Moment Diagrams Lecture Number -6 Prof. M. J. Naidu Mechanical Engineering Department Smt. Kashibai Navale College of Engineering, Pune-41 Strength of Materials

2. Reverse Cases Points to remember to draw… SFD from BMD… 1.SFD is zero where BMD is horizontal straight line. 2.If BMD is inclined straight line then, V= Slope of BMD. 3.If BMD is a parabola, then consider M= ax 2 +bx+c. 4.Find values of a,b,c from values of M or from dM/dx for diff. points 5.SFD is inclined Straight line in this region. 6.Vertical step in BMD implies presence of external couple. Strength of Materials

3. Reverse Cases… Load diagram from SFD… 1.Straight horizontal line – No load region. 2. Inclined straight line in SFD rate of loading = Slope of SFD UDL on load diagram 3.Parabola in SFD consider V=ax 2 +bx+c Get a, b, c from V or from dV/dx at different points UVL in load diagram region. 4. Vertical step in SFD Point load on load diagram Strength of Materials

4. Reverse Cases… Strength of Materials Illustrative Problem 3: In the following problem, draw bending moment and load diagrams corresponding to the given shear diagram. Specify values at all change of load positions and at all points of zero shear.

5. Reverse Cases… To draw the Load Diagram Upward concentrated load at A is 10 kN. The shear in AB is a 2nd-degree curve, thus the load in AB is uniformly varying. In this case, it is zero at A to 2(10 + 2)/3 = 8 kN at B. No load in segment BC. A downward point force is acting at C in a magnitude of 8 - 2 = 6 kN. The shear in DE is uniformly increasing, thus the load in DE is uniformly distributed and upward. This load is spread over DE at a magnitude of 8/2 = 4 kN/m. Strength of Materials

6. Reverse Cases… To draw the Bending Moment Diagram To find the location of zero shear, F: x 2 /10 = 3 2 /(10 + 2) x = 2.74 m M A = 0 M F = M A + Area in shear diagram M F = 0 + 2/3 (2.74)(10) = 18.26 kN·m M B = M F + Area in shear diagram M B = 18.26 - [1/3 (10 + 2)(3) - 1/3 (2.74)(10) - 10(3 - 2.74)] M B = 18 kN·m Strength of Materials

7. Reverse Cases… M C = M B + Area in shear diagram M C = 18 - 2(1) = 16 kN·m M D = M C + Area in shear diagram M D = 16 - 8(1) = 8 kN·m M E = M D + Area in shear diagram M E = 8 - ½ (2)(8) = 0 The moment diagram in AB is a second degree curve, at BC and CD are linear and downward. For segment DE, the moment diagram is parabola open upward with vertex at E. Strength of Materials

8. Reverse Cases… Strength of Materials

9. Reverse Cases… Strength of Materials Workout numerical 1: In the following problem, draw bending moment and load diagrams corresponding to the given shear diagram. Specify values at all change of load positions and at all points of zero shear. [ consider the values in SI units directly without converting ]