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

2. Introduction  SFD and BMD used for analysis of Beam.  SFD- Variation of the shear force along the length.  BMD- Variation of the bending moment along the length.  Value of Shear Force use to calculate shear stress  Value of Bending moment use to calculate Bending Stress

3. Types of Load Concentrated LoadUniform Load Uniform Varying Load Applied Couple

4. Types of Support Fixed or Build in RollerPinned or Hingged

5. Types of Beam Supports Simply Supported beamOverhanging beam Cantilever beam Continuous beam Beam fixed at one end and simply supported at other end Fixed beam at both ends

6. Sign Convention for Shear Force Positive Shear force Negative Shear force Direction of Shear force When moves from left to right

7. Relation between Load, SFD and BMD

8. Variation of Shear Force and Bending Moment Types of Load SFDBMD Point Load Horizontal Line Inclined Line (1 Degree) UDL Inclined Line (1 Degree) Parabolic Curve (2 Degree) UVL Parabolic Curve (2 Degree) Cubic Curve (3 Degree)

9. Example 1 Find SFD and BMD for following Beam

10. Reaction at A is sum of all forces at right side of A RA =400+300+800+500 = 2000 N Shear Force To find shear force start from left side Shear force at A (Just left) = 0 N Shear force at A (Just right) = 2000 N Shear force at B (Just left) = 2000 N Shear force at B(Just right) = 2000-400= 1600 N Shear force at C (Just left) = 1600 N Shear force at C (Just right) = 1600-300= 1300 N Shear force at D (Just left) = 1300 N Shear force at D (Just right) = 1300-800= 500 N Shear force at E (Just left) = 500 N Shear force at E (Just right) = 500-500 N= 0 N

11. Bending moment Starting from free end BM E = 0 Nm BM D = -500*0.5= -250Nm BM C = -800*0.5-500*1= -900Nm BM B = -300*0.5-800*1-500*1.5= -1700Nm BM A = -400*0.5-300*1-800*1.5-500*2= -2000Nm

12. Example 2 Find SFD and BMD for following loaded Beam

13. HINTS Take the entire beam as a free body. Determine the reactions at C. Apply the relationship between shear and load Develop the shear diagram.. Apply the relationship between bending moment and shear Develop the bending moment diagram.

14. Taking the entire beam as a free body, determine the reactions at C. Result from integration of the load and shear distributions should be equivalent.

15. Apply the relationship between shear and load to develop the shear diagram. No change in shear between Band C. Compatible with free body analysis

16. Apply the relationship between bending moment and shear to develop the bending moment diagram. Results at Care compatible with free-body analysis

17. Example 3 Find SFD and BMD for following loaded Beam

18. Reaction is total load on span RA =3+1*2+2.5 = 7.5 kN Shear Force To find shear force start from left side Shear force at A (Just left) = 0 N Shear force at A (Just right) = 7.5 kN Shear force at B (Just left) = 7.5 kN Shear force at B(Just right) = 7.5-3= 4.5 kN Shear force at C =4.5 kN Shear force at D = 4.5- 1*2= 2.5 kN Shear force at E (Just left) = 2.5 kN Shear force at E (Just right) = 2.5-2.5= 0 kN

19. Bending Moment To find bending moment for cantilever beam, start from free end Bending Moment at free end = 0 Nm Bending Moment at E= 0 kN-m Bending Moment at D= -2.5*0.5=-1.25 kN-m Bending Moment at C= -2.5*2.5-1*2*1=-8.25 kN-m Bending Moment at B= -2 kN-m Bending Moment at A= -2 kN-m

20. Draw SFD and BMD for the beam shown and calculate BM max Ans: BM max = 67.6 kNm Workout Numerical