89589540-Strength-of-Materials-by-S-K-Mondal-4.pdf

8/14/2019 89589540-Strength-of-Materials-by-S-K-Mondal-4.pdf

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4. Bending Moment and ShearForce Diagram

Theory at a Glance for IES, GATE, PSU)4.1 Shear Force and Bending Moment

At first we try to understand what shear force is and what is bending moment?

We will not introduce any other co-ordinate system.

We use general co-ordinate axis as shown in the

figure. This system will be followed in shear force and

bending moment diagram and in deflection of beam.Here downward direction will be negative i.e.

negative Y-axis. Therefore downward deflection of the

beam will be treated as negative.

We use above Co-ordinate system

Some books fix a co-ordinate axis as shown in the

following figure. Here downward direction will be

positive i.e. positive Y-axis. Therefore downward

deflection of the beam will be treated as positive. As

beam is generally deflected in downward directions

and this co-ordinate system treats downward

deflection is positive deflection.

Some books use above co-ordinate system

Consider a cantilever beam as shown subjected to

external load P. If we imagine this beam to be cut by

a section X-X, we see that the applied force tend to

displace the left-hand portion of the beam relative to

the right hand portion, which is fixed in the wall.

This tendency is resisted by internal forces between

the two parts of the beam. At the cut section a

resistance shear force (Vx) and a bending moment

(Mx) is induced. This resistance shear force and the

bending moment at the cut section is shown in the

left hand and right hand portion of the cut beam.

Using the three equations of equilibrium

0 , 0 0x y iF F and M

We find that xV P and .xM P x In this chapter we want to show pictorially thePage 125 of 429


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Chapter-4 Bending Moment and Shear Force Diagram S K Mondalsvariation of shear force and bending moment in a

beam as a function of x' measured from one end of

the beam.

Shear Force (V) equal in magnitude but opposite in direction

to the algebraic sum (resultant) of the components in the

direction perpendicular to the axis of the beam of all external

loads and support reactions acting on either side of the section

being considered.

Bending Moment (M) equal in magnitude but opposite in

direction to the algebraic sum of the moments about (the

centroid of the cross section of the beam) the section of all

external loads and support reactions acting on either side of

the section being considered.

What are the benefits of drawing shear force and bending moment diagram?

The benefits of drawing a variation of shear force and bending moment in a beam as a function of x'

measured from one end of the beam is that it becomes easier to determine the maximum absolute

value of shear force and bending moment. The shear force and bending moment diagram gives a

clear picture in our mind about the variation of SF and BM throughout the entire section of the

beam.

Further, the determination of value of bending moment as a function of x' becomes very important

so as to determine the value of deflection of beam subjected to a given loading where we will use the

formula,

2

2 x

d yEI M

dx .

4.2 Notation and sign convention

Shear force (V)

Positive Shear Force

A shearing force having a downward direction to the right hand side of a section or upwards

to the left hand of the section will be taken as positive. It is the usual sign conventions to be

followed for the shear force. In some book followed totally opposite sign convention.

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Chapter-4 Bending Moment and Shear Force Diagram S K Mondals

The upward direction shearing

force which is on the left hand

of the section XX is positive

shear force.

The downward direction

shearing force which is on the

right handof the section XX is

positiveshear force.

Negative Shear Force

A shearing force having an upward direction to the right hand side of a section or downwards

to the left hand of the section will be taken as negative.

The downward direction

shearing force which is on the

left hand of the section XX is

negativeshear force.

The upward direction shearing

force which is on the right

hand of the section XX is

negativeshear force.

Bending Moment (M)

Positive Bending Moment

A bending moment causing concavity upwards will be taken as positive and called as

sagging bending moment.

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Sagging

If the bending moment of

the left handof the section

XX is clockwisethen it is a

positivebending moment.


the right hand of the

section XX is anti-

clockwise then it is a


A bending moment causing

concavity upwards will be

taken as positive and

called as sagging bending

moment.

Negative Bending Moment

Hogging


the left hand of the

section XX is anti-

clockwise then it is a



the right hand of the

section XX is clockwise

then it is a positive

bending moment.

A bending moment causing

convexity upwards will be

taken as negative and called

as hogging bending moment.

Way to remember sign convention

Remember in the Cantilever beam both Shear force and BM are negative (ive).

4.3 Relation between S.F (Vx), B.M. (Mx) & Load (w)

xdV = -w (load)dx

The value of the distributed load at any point in the beam is

equal to the slope of the shear force curve. (Note that the sign of this rule may change

depending on the sign convention used for the external distributed load).

x

x

dM

= VdxThe value of the shear force at any point in the beam is equal to the slope

of the bending moment curve.Page 128 of 429


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4.4 Procedure for drawing shear force and bending moment diagram

Construction of shear force diagram

From the loading diagram of the beam constructed shear force diagram.

First determine the reactions.

Then the vertical components of forces and reactions are successively summed from the left

end of the beam to preserve the mathematical sign conventions adopted. The shear at a

section is simply equal to the sum of all the vertical forces to the left of the section.

The shear force curve is continuous unless there is a point force on the beam. The curve then

jumps by the magnitude of the point force (+ for upward force).

When the successive summation process is used, the shear force diagram should end up with

the previously calculated shear (reaction at right end of the beam). No shear force acts

through the beam just beyond the last vertical force or reaction. If the shear force diagram

closes in this fashion, then it gives an important check on mathematical calculations. i.e. The

shear force will be zero at each end of the beam unless a point force is applied at the end.

Construction of bending moment diagram

The bending moment diagram is obtained by proceeding continuously along the length ofbeam from the left hand end and summing up the areas of shear force diagrams using proper

sign convention.

The process of obtaining the moment diagram from the shear force diagram by summation is

exactly the same as that for drawing shear force diagram from load diagram.

The bending moment curve is continuous unless there is a point moment on the beam. Thecurve then jumps by the magnitude of the point moment (+ for CW moment).

We know that a constant shear force produces a uniform change in the bending moment,resulting in straight line in the moment diagram. If no shear force exists along a certain

portion of a beam, then it indicates that there is no change in moment takes place. We also

know that dM/dx= Vxtherefore, from the fundamental theorem of calculus the maximum or

minimum moment occurs where the shear is zero.

The bending moment will be zero at each free or pinned end of the beam. If the end is builtin, the moment computed by the summation must be equal to the one calculated initially for

the reaction.

4.5 Different types of Loading and their S.F & B.M Diagram

(i) A Cantilever beam with a concentrated load P at its free end.

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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsShear force:

At a section a distance x from free end consider the forces to

the left, then (Vx) = - P (for all values of x)negative in sign

i.e. the shear force to the left of the x-section are in downward

direction and therefore negative.

Bending Moment:

Taking moments about the section gives (obviously to the left

of the section) Mx = -P.x (negative sign means that the

moment on the left hand side of the portion is in the

anticlockwise direction and is therefore taken as negative

according to the sign convention) so that the maximum

bending moment occurs at the fixed end i.e. Mmax = - PL

(at x = L)

S.F and B.M diagram

(ii) A Cantilever beam with uniformly distributed load over the whole length

When a cantilever beam is subjected to a uniformly

distributed load whose intensity is given w /unit length.

Shear force:

Consider any cross-section XX which is at a distance of x from

the free end. If we just take the resultant of all the forces on

the left of the X-section, then

Vx= -w.x for all values of x'.

At x = 0, Vx = 0

At x = L, Vx= -wL (i.e. Maximum at fixed end)

Plotting the equation Vx = -w.x, we get a straight line

because it is a equation of a straight line y (Vx) = m(- w) .x

Bending Moment:

Bending Moment at XX is obtained by treating the load to the

left of XX as a concentrated load of the same value (w.x)

acting through the centre of gravity at x/2.

S.F and B.M diagram

Therefore, the bending moment at any cross-section XX is

2.

. .2 2

x

x w xM w x

Therefore the variation of bending moment is according toparabolic law.

The extreme values of B.M would be

at x = 0, Mx = 0

and x = L, Mx =

2

2

wL

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Maximum bending moment, 2

max

wL

2M at fixed end

Another way to describe a cantilever beam with uniformly distributed load (UDL) over its whole

length.

(iii) A Cantilever beam loaded as shown below draw its S.F and B.M diagram

In the region 0 < x < a

Following the same rule as followed previously, we get

x xV =- P; and M = - P.x

In the region a < x < L

x xV =- P+P=0; and M = - P.x +P .x a P a

S.F and B.M diagram

(iv) Let us take an example: Consider a cantilever bean of 5 m length. It carries a uniformly

distributed load 3 KN/m and a concentrated load of 7 kN at the free end and 10 kN at 3 meters from

the fixed end.

Draw SF and BM diagram.

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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsAnswer: In the region 0 < x < 2 m

Consider any cross section XX at a distance x from free end.

Shear force (Vx)= -7- 3x

So, the variation of shear force is linear.

at x = 0, Vx = -7 kN

at x = 2 m , Vx= -7 - 3 2 = -13 kNat point Z Vx= -7 -3 2-10 = -23 kN

Bending moment (Mx)= -7x - (3x).

2x 3x7x

2 2

So, the variation of bending force is parabolic.

at x = 0, Mx = 0

at x = 2 m, Mx= -7 2 (3 2) 2

2= - 20 kNm

In the region 2 m < x < 5 m

Consider any cross section YY at a distance x from free

end

Shear force (Vx) = -7 - 3x 10 = -17- 3x

So, the variation of shear force is linear.

at x = 2 m, Vx = - 23 kN

at x = 5 m, Vx= - 32 kN

Bending moment (Mx) = - 7x (3x)

x

2

- 10 (x - 2)

23 x 17 202

x

So, the variation of bending force is parabolic.

at x = 2 m, Mx23 2 17 2 20

2 = - 20 kNm

at x = 5 m, Mx = - 102.5 kNm

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(v) A Cantilever beam carrying uniformly varying load from zero at free end and w/unit

length at the fixed end

Consider any cross-section XX which is at a distance of x from the free end.

At this point load (wx) =w

.xL

Therefore total load (W)

L L

x

0 0

wLw dx .xdx =

L 2

w

2

max

area of ABC (load triangle)

1 w wx . x .x

2 2L

The shear force variation is parabolic.

at x = 0, V 0

WL WLat x = L, V i.e. Maximum Shear force (V ) at fi

2 2

x

x

L

xShear force V

xed end

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2 3

2 2

max

load distance from centroid of triangle ABC

wx 2x wx.

2L 3 6L

The bending moment variation is cubic.

at x= 0, M 0

wL wLat x = L, M i.e. Maximum Bending moment (M ) at fi6 6

x

x

xBending moment M

xed end.

x

Integration method

d V wWe know that load .x

dx L

wor d(V ) .x .dx

Lx

Alternative way :

V x

0 0

2

Integrating both side

w

d V . x .dxL

w xor V .

L 2

x

x

x

2x

x

2

x

Again we know that

d M wx V -

dx 2L

wxor d M - dx

2L

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xM 2

x

0 0

3 3

x

Integrating both side we get at x=0,M =0

wxd(M ) .dx

2L

w x wxor M - -

2L 3 6L

x x

(vi) A Cantilever beam carrying gradually varying load from zero at fixed end and

w/unit length at the free end

Considering equilibrium we get, 2

A A

wL wLM and Reaction R

3 2

Considering any cross-section XX which is at a distance of x from the fixed end.

At this point loadW

(W ) .xL

x

Shear force AR area of triangle ANMxV

2

x max

x

wL 1 w wL wx- . .x .x = + -

2 2 L 2 2L

The shear force variation is parabolic.

wL wL

at x 0, V i.e. Maximum shear force, V2 2

at x L, V 0

Bending moment 2

A A

wx 2x=R .x - . - M

2L 3x

M

3 2

2 2

max

x

wL wx wL= .x - -

2 6L 3

The bending moment variation is cubic

wL wLat x = 0, M i.e.Maximum B.M. M .

3 3

at x L, M 0

x

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(vii) A Cantilever beam carrying a moment M at free end

Consider any cross-section XX which is at a distance of x from the free end.

Shear force:Vx= 0 at any point.

Bending moment (Mx) = -M at any point, i.e. Bending moment is constant throughout the

length.

(viii) A Simply supported beam with a concentrated load P at its mid span.

Considering equilibrium we get, A BP

R = R =2

Now consider any cross-section XX which is at a distance of x from left end A and section YY ata distance from left end A, as shown in figure below.

Shear force: In the region 0 < x < L/2Page 136 of 429


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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsVx= RA= + P/2 (it is constant)

In the region L/2 < x < L

Vx= RA P =2

P- P = - P/2 (it is constant)

Bending moment: In the region 0 < x < L/2

Mx=

P

2 .x (its variation is linear)

at x = 0, Mx= 0 and at x = L/2 Mx=PL

4i.e. maximum

Maximum bending moment, maxPL

4M at x = L/2 (at mid-point)


Mx=2

P.x P(x - L/2) =

PL

2

P

2.x (its variation is linear)

at x = L/2 , Mx=PL

4 and at x = L, Mx= 0

(ix) A Simply supported beam with a concentrated load P is notat its mid span.

Considering equilibrium we get, RA= BPb Pa

and R =L L

Now consider any cross-section XX which is at a distance x from left end A and another section

YY at a distance x from end A as shown in figure below.Page 137 of 429


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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsShear force: In the range 0 < x < a

Vx= RA= +Pb

L (it is constant)

In the range a < x < L

Vx= RA- P = -Pa

L (it is constant)

Bending moment: In the range 0 < x < a

Mx= +RA.x =Pb

L.x (it is variation is linear)

at x = 0, Mx= 0 and at x = a, Mx=Pab

L (i.e. maximum)

In the range a < x < L

Mx= RA.x P(x- a) =Pb

L.x P.x + Pa (Put b = L - a)

= Pa (1 -x

1 LPa

)

at x = a, Mx=Pab

L and at x = L, Mx= 0

(x) A Simply supported beam with two concentrated load P from a distance a both end.

The loading is shown below diagram

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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsTake a section at a distance x from the left support. This section is applicable for any value of x just

to the left of the applied force P. The shear, remains constant and is +P. The bending moment varies

linearly from the support, reaching a maximum of +Pa.

A section applicable anywhere between the two applied forces. Shear force is not necessary to

maintain equilibrium of a segment in this part of the beam. Only a constant bending moment of +Pa

must be resisted by the beam in this zone.

Such a state of bending or flexure is calledpure bending.

Shear and bending-moment diagrams for this loading condition are shown below.

(xi) A Simply supported beam with a uniformly distributed load (UDL) through out its

length

We will solve this problem by following two alternative ways.

(a) By Method of Section

Considering equilibrium we get RA= RB=wL

2

Now Consider any cross-section XX which is at a distance x from left end A.

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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsThen the section view

Shear force:Vx=wL

wx2

(i.e. S.F. variation is linear)

at x = 0, Vx=wL

2

at x = L/2, Vx= 0

at x = L, Vx= -wL

2

Bending moment:2

.2 2

x

wL wx M x

(i.e. B.M. variation is parabolic)

at x = 0, Mx= 0

at x = L, Mx= 0

Now we have to determine maximum bending

moment and its position.

For maximum B.M:

0 . . 0x x

x x

d M d M i e V V

dx dx

or 02 2

wL Lwx or x

Therefore, maximum bending moment, 2

max8

wLM at x = L/2

(a) By Method of Integration

Shear force:

We know that, xd V

wdx

xor d V wdx

Integrating both side we get (at x =0, Vx=2

wL)

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0

2

2

2

xV x

x

wL

x

x

d V wdx

wLor V wx

wLor V wx

Bending moment:

We know that, x

x

d MV

dx

2

x x

wLor d M V dx wx dx

Integrating both side we get (at x =0, Vx=0)

02

2

.2 2

xM x

x

o

x

wLd M wx dx

wL wx or M x

Let us take an example:A loaded beam as shown below. Draw its S.F and B.M diagram.

Considering equilibrium we get

A

B

M 0 gives

- 200 4 2 3000 4 R 8 0

R 1700NB

or

A B

A

R R 200 4 3000

R 2100N

And

or

Now consider any cross-section which is at a distance 'x' from left end A and

as shown in figure

XX

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In the region 0 < x < 4m

Shear force (Vx) = RA 200x = 2100 200 x

Bending moment (Mx) = RA.x 200 x .x

2

= 2100 x -100 x2

at x = 0, Vx= 2100 N, Mx= 0

at x = 4m, Vx= 1300 N, Mx= 6800 N.m

In the region 4 m< x < 8 m

Shear force (Vx) = RA- 200 4 3000 = -1700

Bending moment (Mx) = RA. x - 2004 (x-2) 3000 (x- 4)

= 2100 x 800 x + 1600 3000x +12000 = 13600 -1700 x

at x = 4 m, Vx= -1700 N, Mx= 6800 Nm

at x = 8 m, Vx= -1700 N, Mx= 0

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(xii) A Simply supported beam with a gradually varying load (GVL) zero at one end and

w/unit length at other span.

Consider equilibrium of the beam =1

wL2

acting at a point C at a distance 2L/3 to the left end A.

B

A

A

BA

M 0 gives

wL LR .L - . 0

2 3

wLor R

6

wLSimilarly M 0 gives R

3

The free body diagram of section A - XX as shown below, Load at section XX, (wx) =w

xL

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The resulted of that part of the distributed load which acts on this free body is 21 w wx

x . x2 L 2L

applied at a point Z, distance x/3 from XX section.

Shear force (Vx) =2 2

A

wx wL wxR - -

2L 6 2L

Therefore the variation of shear force is parabolic

at x = 0, Vx=wL

6

at x = L, Vx= -wL

3

2 3wL wx x wL wx

and .x . .x6 2L 3 6 6LxBending Moment (M )

The variation of BM is cubic

at x = 0, Mx= 0

at x = L, Mx= 0

For maximum BM; x x

x x

d M d M0 i.e. V 0 V

dx dx

2

3 2

max

wL wx Lor - 0 or x

6 2L 3

wL L w L wLand M

6 6L3 3 9 3

i.e. 2

max

wLM

9 3L

at x3

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(xiii) A Simply supported beam with a gradually varying load (GVL) zero at each end and

w/unit length at mid span.

Consider equilibrium of the beam AB total load on the beam1 L wL

2 w2 2 2

A B

wLTherefore R R

4

The free body diagram of section A XX as shown below, load at section XX (wx)2w

.xL

The resultant of that part of the distributed load which acts on this free body is

21 2w wx.x. .x

2 L L

applied at a point, distance x/3 from section XX.

Shear force (Vx):Page 145 of 429


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Chapter-4 Bending Moment and Shear Force Diagram S K MondalsIn the region 0 < x < L/2

2 2

x A

wx wL wxV R

L 4 L

Therefore the variation of shear force is parabolic.

at x = 0, Vx=wL

4at x = L/4, Vx= 0

In the region of L/2 < x < L

The Diagram will be Mirror image of AC.

Bending moment (Mx):

In the region 0 < x < L/2

3

x

wL 1 2wx wL wxM .x .x. . x / 3 -

4 2 L 4 3L


at x = 0, Mx= 0

at x = L/2, Mx=

2wL

12


BM diagram will be mirror image of AC.

For maximum bending moment

x xx x

d M d M0 i.e. V 0 Vdx dx

2

2

max

wL wx Lor - 0 or x

4 L 2

wLand M

12

i.e. 2

max

wLM

12

Lat x

2

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(xiv) A Simply supported beam with a gradually varying load (GVL) zero at mid span and

w/unit length at each end.

We now superimpose two beams as(1) Simply supported beam with a UDL

through at its length

x 1

2

x 1

wLV wx

2

wL wxM .x

2 2

And (2) a simply supported beam with a gradually varying load (GVL) zero at each end and w/unit

length at mind span.

In the range 0 < x < L/2

2

x 2

3

x 2

wL wxV

4 L

wL wxM .x

4 3L

Now superimposing we get

Shear force (Vx):

In the region of 0< x < L/2

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2

x x x1 2

2

wL wL wxV V V -wx -

2 4 L

wx -L/2

L

Therefore the variation of shear force is parabolic

at x = 0, Vx= +

wL

4

at x = L/2, Vx= 0


The diagram will be mirror image of AC

Bending moment (Mx) = x 1M - x 2M =

2 3 3 2wL wx wL wx wx wx wL.x .x .x

2 2 4 3L 3L 2 4


x

2

x

at x 0, M 0

wxat x L / 2, M

24

(xv) A simply supported beam with a gradually varying load (GVL) w1/unit length at one

end and w2/unit length at other end.

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(xvi) A Simply supported beam carrying a continuously distributed load. The intensity of

the load at any point is,

sinx

xw w

L. Where x is the distance from each end of

the beam.

We will use Integration method as it is easier in this case.

We know that x x

x

d V d Mload and V

dx dx

x

x

d VTherefore sin

dx L

d V sin dxL

xw

xw

x x

Integrating both side we get

xw cos

x wL xLd V w sin dx or V A cos

L L

L

where, constant of Integration

A

A

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x

x x x

2

x 2

Again we know that

d M wL xV or d M V dx cos dx

dx L

Integrating both sidewe get

wL xsin

wL xLM x + B sin x + BL

L

A

A A

[Where B = constant of Integration]

Now apply boundary conditions

At x = 0, Mx= 0 and at x = L, Mx= 0

This gives A = 0 and B = 0

x max

2

x 2

2

max 2

wL x wLShear force V cos and V at x 0

L

wL xAnd M sinL

wLM at x = L/2

(xvii) A Simply supported beam with a couple or moment at a distance a from left end.

Considering equilibrium we get

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A

B B

B

A A

M 0 gives

MR L+M 0 R

L

and M 0 gives

MR L+M 0 R

L

or

or

Now consider any cross-section XX which is at a distance x from left end A and another section YY

at a distance x from left end A as shown in figure.

In the region 0 < x < a

Shear force (Vx) = RA=M

L

Bending moment (Mx) = RA.x =M

L.x

In the region a< x < L

Shear force (Vx) = RA=M

L

Bending moment (Mx) = RA.x M =M

L.x - M

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(xviii) A Simply supported beam with an eccentric load

When the beam is subjected to an eccentric load, the eccentric load is to be changed into a couple =

Force (distance travel by force)= P.a (in this case) and a force = PTherefore equivalent load diagram will be

Considering equilibrium

AM 0 gives

-P.(L/2) + P.a + RB L = 0

or RB=P P.a

2 L and RA+ RB= P gives RA=

P P.a

2 L

Now consider any cross-section XX which is at a distance x from left end A and another section YY

at a distance x from left end A as shown in figure.

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In the region 0 < x < L/2

Shear force (Vx) =P P.a

2 L

Bending moment (Mx) = RA. x =P Pa

2 L

. x


Shear force (Vx) =P Pa P Pa

P = -2 L 2 L

Bending moment (Vx) = RA. x P.( x - L/2 ) M

=PL P Pa

.x - Pa2 2 L

4.6 Bending Moment diagram of Statically Indeterminate beam

Beams for which reaction forces and internal forces cannot be found out from static equilibrium

equations alone are called statically indeterminate beam. This type of beam requires deformation

equation in addition to static equilibrium equations to solve for unknown forces.

Statically determinate - Equilibrium conditions sufficient to compute reactions.

Statically indeterminate -Deflections (Compatibility conditions) along with equilibrium equations

should be used to find out reactions.

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Type of Loading & B.M Diagram Reaction Bending Moment

RA= RB =P

2 MA= MB =

PL-

8

RA= RB =wL

2 MA= MB =

2wL-

12

2

A 3

2

3

R (3 )

(3 )B

Pba b

L

PaR b a

L

MA= -

2

2

Pab

L

MB = -

2

2

Pa b

L

RA= RB =3

16

wL

Rc =5

8

wL

RA RB

+

--

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4.7 Load and Bending Moment diagram from Shear Force diagramOR

Load and Shear Force diagram from Bending Moment diagram

If S.F. Diagram for a beam is given, then

(i) If S.F. diagram consists of rectangle then the load will be point load

(ii) If S.F diagram consists of inclined line then the load will be UDL on that portion

(iii) If S.F diagram consists of parabolic curve then the load will be GVL

(iv) If S.F diagram consists of cubic curve then the load distribute is parabolic.

After finding load diagram we can draw B.M diagram easily.

If B.M Diagram for a beam is given, then

(i) If B.M diagram consists of inclined line then the load will be free point load

(ii) If B.M diagram consists of parabolic curve then the load will be U.D.L.

(iii) If B.M diagram consists of cubic curve then the load will be G.V.L.

(iv) If B.M diagram consists of fourth degree polynomial then the load distribution is

parabolic.

Let us take an example: Following is the S.F diagram of a beam is given. Find its loading

diagram.

Answer:From A-E inclined straight line so load will be UDL and in AB = 2 m length load = 6 kN if

UDL is w N/m then w.x = 6 or w 2 = 6 or w = 3 kN/m after that S.F is constant so no force is

there. At last a 6 kN for vertical force complete the diagram then the load diagram will be

As there is no support at left end it must be a cantilever beam.

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4.8 Point of Contraflexure

In a beam if the bending moment changes sign at a point, the point itself having zero bending

moment, the beam changes curvature at this point of zero bending moment and this point is called

the point of contra flexure.

Consider a loaded beam as shown below along with the B.M diagrams and deflection diagram.

In this diagram we noticed that for the beam loaded as in this case, the bending moment diagram is

partly positive and partly negative. In the deflected shape of the beam just below the bending

moment diagram shows that left hand side of the beam is sagging' while the right hand side of the

beam is hogging.

The point C on the beam where the curvature changes from sagging to hogging is a point of

contraflexure.

There can be more than one point of contraflexure in a beam.

4.9 General expression

EI4

2

d y

dx

3

3 x

d yEI V

dx

2

2 x

d yEI M

dx Page 157 of 429


34/34


dy

= = slopedx

y== Deflection

Flexural rigidity = EI

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