Strength of Materials

UNIT - II Shear Force Diagrams

and Bending Moment Diagrams

Lecture Number -1 Prof. M. J. Naidu

Mechanical Engineering DepartmentSmt. Kashibai Navale College of Engineering, Pune-41

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

Types of Load

Concentrated Load Uniform Load

Uniform Varying Load Applied Couple

Types of Support

Fixed or Build in Roller Pinned or Hingged

Types of Beam Supports

Simply Supported beam Overhanging beam

Cantilever beam Continuous beam

Beam fixed at one end and simply supported at other end

Fixed beam at both ends

Sign Convention for Shear Force

Positive Shear force Negative Shear force

Direction of Shear force

When moves from left to right

Relation between Load, SFD and BMD

Variation of Shear Force and Bending Moment

Types of Load SFD BMD

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)

Example 1

Find SFD and BMD for following Beam

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

Bending momentStarting from free endBME= 0 NmBMD= -500*0.5= -250NmBMC= -800*0.5-500*1= -900NmBMB= -300*0.5-800*1-500*1.5= -1700NmBMA= -400*0.5-300*1-800*1.5-500*2= -2000Nm

Example 2

Find SFD and BMD for following loaded Beam

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.

• 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.

• Apply the relationship between shear and load to develop the shear diagram.

• No change in shear between Band C.• Compatible with free body analysis

• Apply the relationship between bending moment and shear to develop the bending moment diagram.

• Results at Care compatible with free-body analysis

Example 3

Find SFD and BMD for following loaded Beam

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

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

• Draw SFD and BMD for the beam shown and calculate BMmax

• Ans: BMmax = 67.6 kNm

Workout Numerical