Sf and bm

BENDING

MOMENT AND

SHEAR FORCE [16 MARKS]

Chapter 2

Chapter Details 2.1 Introduction

2.2 Shear Force Diagram (SFD) and Bending Moment Diagram (BMD)

2.3 Reaction Calculation

2-2 Strength of Materials

Graphical and Statistical representation of questions asked from this chapter in previous years MSBTE Question Papers

16

14

12

0

2

4

6

8

10

12

14

16

To

tal M

ark

s

W-2009 S-2010 W-2010

Years

Graphical Representation

Statistical Analysis

MSBTE paper Total marks questions asked in this chapter

W-2009 16

S-2010 14

W-2010 12

2-3 Bending Moment & Shear Force

2.1 INTRODUCTION

Beam is a horizontal member in the structure. Due to

load on it, beam suffers bending, shear torsion etc.

Beam is generally designed for bending and checks for shear. We have already discussed concept of bending

stresses and shear stresses.

Due to bending, beam suffers tension as well as

compression at extreme fibres according to the type of

beam.

Shear failure is sudden failure. It does not give any

prior warning before failure.

Shear force acts tangential to the cross-section.

Shearing action and bending action occurs due to load at the same time.

Hence there is a relation between loading rate,

shearing force and bending moment.

Types of Load or Loading on a Beam

Practically beam suffers from various types of load:

a. concentrated load/point load

b. uniformly distributed load (udl)

c. uniformly varying load (uvl)

Among these, we are learning concentrated loads and

uniformly distributed loads.

a. Concentrated load / point load: A load acting at

a point on the beam is known as a concentrated

load or point load.

In practice, a load distributed over small area is

taken as concentrated load.

e.g., column

RA RB

W1 W2

L

C DA B

Figure 2.1: Simply supported Beam subjected to point loads

S.S.BLoad

S.S. beam: Simply supported beam.

(Applied mechanics → equilibrium)

Compression

Tension

LoadingCantilever

Tension

Compression W

Shear on beam

acting

Bending actionon beam

Wall

Beam

Even this portion of wall on beam is considered as concentrated load.


Conversion of udl into point load

L

w/m

L

L/2

W L×

W1 W2 W3 W4

C

RA RB

C CLeftpartof beam

Rightpart

of beam

RA RB

w1 w2 w3 w4

SFc = ΣVLc = ΣVRc

SFc = RA – W1 – W2

SFc = –RB – W3 + W4

ΣVL = Summation of vertical forces on left hand part of beam

ΣVR = Summation of vertical forces on right part of beam.

b. Uniformly distributed load (udl): A load which is

spread up uniformly on the beam is known as a

uniformly distributed load or rectangular load and is

written as u.d.l.

e.g., self weight of a beam, or load from the floor or a

slab on the beam.

For convenience u.d.l. is converted into its equivalent

point load which is assumed to act at the centre of

gravity of the load.

Shear Force (SF)

Shear force at any cross-section of the beam is the

algebraic sum of all vertical forces on the beam, acting on

the right or left side of the section.

Sign convention for the shear force

L R

X

XPositiveshear

L R

X

XNegative

shear

Figure 2.2

i.e. If we consider only left part of a beam upward force is

to be considered positive (+ve) and downward force is to

be considered negative (–ve). While if right part of beam is considered, downward force is to be considered positive

(+ve) while upward force is to be considered negative

(–ve).


Bending Moment (BM)

Bending moment at any cross-section of the beam, is the

algebraic sum of the moments of all the forces acting on the

right or left side of the section.

Sign convention for the beam:

L R

X

XPositive

BM

L R

X

XNegative

BM

Figure 2.3

Clockwise moment to the left of the section and

anticlockwise moment on the right of the section, is taken as

positive. (i.e. Sagging BM is considered positive).

Sagging from both sides i.e. L or R; upward force will

produce +ve BM.

Anticlockwise moment to the left of the section and

clockwise moment on the right of the section is taken as negative. (i.e. Hogging BM is consider negative).

From both side i.e. L or R down word force will produce

–ve BM.

BM = ΣWx

x2

x1

RA

Left part of beam

W

BM = R x – W xC A 1 1 2

C

L R

+ve

Sagging

From both sides i.e. L or R; upward force will produce +ve

BM and from both sides i.e. L or R downward force will

produce –ve BM.

L R

–ve

Hogging

Relation between Loading Shear force and Bending Moment

Relation between Rate of Loading and Shear Force

Shear force is a function of load. Hence rate of change of

shear force with respect to the distance, is equal to the

intensity of loading.

dF

dx = W

��

BTE W.2010 – 2M

State the relation

between rate of loading,

shear force and bending

moment.


Relation between Bending Moment and Shear Force

dM

dx= F

The rate of change of bending moment at any section, is equal to the shear force at that

section.

If dM

dx= 0, ie. shear force = 0,

The bending moment (M) will be maximum i.e. The point at which Shear Force (SF)changes its value from +ve to –ve, at that point bending moment value is maximum;

and called as maximum bending moment point or contrashear point.

2.2 SHEAR FORCE DIAGRAM (SFD) AND BENDING MOMENT DIAGRAM (BMD)

A shear force diagram is that which shows the variation of shear force along the length of

the beam. A bending moment diagram shows the variation of bending moment along the

length of the beam.

There is a relation between rate of loading, shear force and bending moment. Hence, we

draw loaded beam, shear force diagram and bending moment diagram one below the

other as;

Loaded beam

then

Shear force diagram

then

Bending moment diagram

While drawing SFD and BMD, following points are to be

considered:

a. Base of SFD and BMD is equal to the span of beam.

(i.e. = L)

b. Positive values of SF and BM are plotted above the

base line and negative values of SF and BM are

plotted below the base line.

c. The SF and BM must be calculated for all the critical

points and written near the respective co-ordinates.

L

+ve SF

base of SFD

+ve BM

base of BMD

– ve SF

– ve BM


Critical points are as follows:

i. A point at which point load acts.

ii. A point at which udl starts or ends.

iii. A point at which SF value changes from positive to negative.( max. BM point)

i.e. +ve called –ve contrashear point or max BM point.

d. Location of zero bending moment point i.e. point of contraflexure must be found

out.

2.3 REACTION CALCULATION

While calculating SFE at various points – first step is to calculate reactions.

Reaction Calculation for Simply Supported Beam

We consider that the beam is statistically determinant, hence it follows conditions of

equilibrium.

Σ Fx = 0

Σ Fy = 0

Σ M = 0

Let us consider, simply supported beam

W1 W2 W3

x1

x2

L

x3RA RB

A B

Figure 2.4

Σ Fy = 0

RA + RB = W1 + W2 + W3 = ΣW

∴ RA = ΣW – RB ...............................................................................................................(1)

If we consider ΣFy = 0

i.e. Σ ↑ forces =Σ ↓ forces

RB = ΣWxL


As beam is in equilibrium ΣMA = 0

ΣMA = 0

+ve – ve

∴ 0 = W1x1 + W2x2 + W2x3 – RBL

∴ RBL = W1x1 + W2x2 + W3x3

+ve – ve

∴ RB = ΣWx

L .....................................................................................................................(2)

We can consider as formula to calculate RB. then by using.

RA = ΣW – RB

We can calculate value of RA.

Solved

Reaction Calculation Practice Example

1. Calculate reactions of following.

10N 20N 30N

RA RB

2m 2m 2m 2m

Solution

By using,

RB = ΣWx

L =

10 × 2 + 20 × 4 + 30 × 6

8

= 280

8 = 35 N

RA = ΣW – RB = (10 + 20 + 30) – 35

= 25 N

RA = 25 N

RB = 35 N

10N 20N 30N

2m

4m

8m

6m

RA RB


2. Calculate reaction for given beam.

10N/m

20N 5N

4m 2m 1m

DCA B

Solution

40N/m

20N5N

4m

6m

7m

2mDCRA RB

RB = ΣWx

L =

40 × 2 + 20 × 4 + 5 × 6

7

= 27.14 N

RA = (40 + 20 + 5) – RB = 65 – 27.14

= 37.86 N

10 4×

4 m

10 N/m

2 m 2 m

CG

udl is converted into

equivalent point load whose

rate is 10 N/m and considered

4m = 10 N/m × 4m = 50 N.

Action at center of udl pass.

3. Calculate reactions for given beam.

15N

5N/m

5N

1m 4m 1m

A BDC

Solution

RB = ΣWx

L =

15 × 1 + 30 × 3 – 5 × 5

6

= 13.33 N

RA = ΣW – RB = (15 + 30 + 5)–(13.33)

= 36.67 N

15N 30N 5N

1m

DC

3m

5m

6m

RA RB


Reaction Calculation for Cantilever

Only one support will bear total load on the beam hence in cantilever:

R = ΣW

as ΣFy = 0

↑ = ↓

4. Find reaction at cantilever support.

10N

2m 2m 2m

A

RA

20N 10N

Solution

↑ = ↓

RA = 10 + 20 + 10

∴ RA = 40N

5. Find reaction at cantilever support

60N

2m 2m 1.5 mRA

20N 5N

Solution

↑ = ↓

RA + 5 = 60 + 20

RA = 60 + 20 – 5

RA = 75 N

6. Find reaction at cantilever support

10N/m

5m

A B

10N

Solution

↑ = ↓

RA = 50 + 10

RA = 60 N

50N 10N

2.5mRA

5m

Engineering

Sf and bm