Def numerical

Mechanics of Solids Deflection Numericals

Yatin Kumar Singh

1. A steel girder of I section supported over a span of

3.6 m has Ixx = 1,200 × 104 mm4. Determine the slope at

ends and deflection at the centre of the beam, if central

load is 16 kN, E = 200 GPa = 200 × 109 N/m2 = 200 ×

106 kN/m2.

Solution:

E = 200 × 106 kN/m2

I = 1,200 × 104 mm4

EI = 1,200 × 200 × 10-2 = 2,400 kNm2

W = 16 kN; L = 3.6m

2. A beam AB, 10 m long, carries point loads of 6 and 3

kN at C and D as shown in Fig. Determine support

reactions, deflection at C and D, and slope at

ends A and B, if EI is the flexural rigidity of the beam.

Solution:

Taking moments about A,

6 × 4 + 7 × 3 = 10RB

Reaction, RB = (24 + 21)/10 = 4.5 kN

Total load on beam = 6 + 3 = 9 kN

Reaction at A, RA = 9 − RB = 9 − 4.5 = 4.5 kN.

There are three portions, that is, AC, CD and DB in the

beam and if A is the origin then DB is the last portion.

Consider a section at a distance x from A in the

portion DB,

Bending moment, Mx = 4.5x − 6(x − 4) − 3(x − 7)

Integrating

Since the beam is not symmetrically loaded about its

centre, so we do not know where slope is zero.

By integrating eq (ii) we obtain,

EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 + C1x + C2

Boundary conditions:

x = 0 at end A, deflection y = 0

Moreover in portion AC, only the first term is valid and

the other two terms of equation are not valid.

So, 0 = 0.75 × 0, neglected term – neglected term + C1 ×

0 + C2

or, constant C2 = 0

So, EIy = 0.75x3 – (x – 4)3 – 0.5(x – 7)3 + C1x

At end B, x = 10 m, deflection y = 0, by substituting the

value

0 = 0.75 × 103 – 63 – 0.5 × 33 + 10C1

0 = 750 – 216 –13.5 + 10C1

Constant, C1 = –52.05

Finally, the equations are

EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 − 52.05x

Slope at end A, x = 0

EIiA = 2.25 × 0 – neglected terms –52.05

at end B, x = 10, so all the terms are valid

Deflection

At point C, x = 4 m, hence, the third term in the

equation is invalid.

EIyC = 0.75 × 43 − (4 − 4)3 − neglected term − 52.05 × 4

EIyC = 48 − 0 − 208.2 = −160.2

At point D, x = 7 m, and all the terms in the equation for

deflection are valid.

EIyD = 0.75 × 73 − (7 − 4)3 − 0.5(7 − 7)3 − 52.05 × 7 =

257.25 − 27 − 0 − 364.35 = −134 1

3. A beam ABCD, 6 m long hinged at end A and roller

supported at end D, is subjected to CCW moment of 10

kN m at point B and a point load of 10 kN at point C as

shown in Fig. Determine the deflection under load of

10 kN and slope at point B, by taking EI as flexural

rigidity of the beam.

Solution:

Taking moments at A,

10 + RD × 6 = 10 × 4

RD = 5kN ↑

Total load on the beam = 10 kN

Reaction at A, RA = 10 – 5 = 5kN ↑


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There are three portions: AB, BC and CD in beam,

with A as the origin portion and CD is the last. Take a

section YY in portion CD at a distance x from A as

shown in the figure.

The equation of bending moment is

Note that 10 kN m is a moment and we cannot take

moment of moment, but 10 kN m is applied at B at a

distance of (x – 2) m from section YY. Moreover (x – 2)°

= 1, so only to locate the position of moment, the term

(x – 2)° is taken, that is, (x – 2) raised to power zero, this

term locates the position of moment applied at B.

Integrating we get

where C1 is a constant of integration, we do not know

where the slope is zero as the beam is not

symmetrically loaded.

Also by integrating we obtain

where C2 is another constant of integration.

y = 0 at end A, x = 0, only first two terms in between are

valid in portion AB, so

or, constant, C2 = 0.

y = 0 at end D, where x = 6 m, all the terms in the

equation are valid, so

Finally, the equations for slope and deflection are

The slope at B and x = 2 m, the third term is not valid

Deflection at C, x = 4 m, all the terms in equation for

deflection are valid for x = 4 m

4. A beam ABCD, 7 m long hinged at A and roller

supported at D carries 7 kN load at B and 4 kN/m udl

over BC = 3 m. If EI = 14,000 kN m2 for the beam,

determine the slope at A and deflection at point C.

Solution:

Reactions

Total udl on beam = 4 × 3 = 12 kN

CG of this load lies at 2 + 1.5 = 3.5 m from end A

Taking moments about A,

7 × 2 + 4 × 3(3.5) = 7 RD

RA = 7 + 12 − 8 = 11kN

Last portion of the beam is CD. A section YY at a

distance x from end A in the portion CD of the beam is

taken to make the equation of bending moment valid

for all the three portions. The udl is extended to

section YY on both sides (upward and downward), so

that its net effect becomes zero.

Bending moment at section YY

where w is the rate of loading.

Note that the first term is valid for portion AB, the first

three terms are valid for portion BC, and all the four

terms are valid for portion CD of the beam.

Substituting the value of w = 4 kN/m,

Integrating


Yatin Kumar Singh

where C1 is a constant of integration.

Also integrating (2) we obtain


At end A, x = 0, y = 0 (in portion AB, x = 0, their three

terms are invalid)

Constant, C2 = 0.

At the end D, x = 7 m, y = 0, and all the terms in

equation are valid.

Constant, C1 = –54.5

The equations will become

EI = 14,000 kN m2

Slope at A, x = 0

Deflection at C, x = 5 m, and all the terms are valid

5. Cantilever AB, 5 m long, is simply supported

at A and fixed at B. If it carries a udl of 6 kN/m

over CB = 3 m, EI of cantilever is 3,600 kN m2.

Determine the reaction at A and slope at A and also find

out the deflection at C?

Solution:

Say reaction at A is RA. Taking a section at

distance x from A, in the portion CB,

Bending moment,

where w is rate of loading

or,

Integrating we get

where C1 is a constant of integration; dy/dx = 0 at x = 5,

by substituting this value, we obtain

So,

Also by integrating we obtain

At A, x = 0, y = 0

0 0 = − omitted term + C2

Constant, C2 = 0

Moreover, y = 0 at end B, x = 5 m, by substituting this

value,

RA = 2.754 kN

By substituting the value of RA, equations of slope and

deflection are

At A, x = 0

Deflection at C, x = 2 m


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6. A simply supported beam of span length L carries a

udl of intensity w throughout its length. Determine the

slope at A and deflection at C by moment area method.

Solution:

BM diagram is parabolic for this case with maximum

bending moment at centre

Area of the bending moment diagram from A to C

that is, area AC′C

Because of symmetrical loading, ic = 0, and slope at

centre is zero.

CG of area AC’C lies at a distance of (5/8 × L/2) from

A.

Now,

7. A beam, 6 m long, is simply supported at ends

carries a udl of intensity 4 kN/m throughout its length.

Draw BM diagram of the beam. Using conjugate beam

method, determine the slope at the ends and deflection

in the centre. EI is the flexural rigidity of the beam. EI =

10,500 kN m2.

Solution:

BM diagram is parabolic with Mmax =

Bending moment diagram supported on two supports

is shown as conjugate beam.

Area of BM diagram

Slope at A,

Slope at B,

M′C = moment at centre

8. A beam of length L carries a central load W as

shown in figure. Moment of inertia for quarter length

from ends is I1 and for the middle half length moment

of inertia is I2, such that I2 = 2I1, now draw the

conjugate beam diagram.

Solution:

If E is Young’s modulus of the material, diagram AE′B is

the bending moment diagram such that


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Conjugate beam diagram gives,

Beam is symmetrically loaded, therefore reactions, RA′

= RB′.

RA′ = area of conjugate beam diagram up to centre.

iE = slope at centre is zero

iE – iA = area of conjugate beam diagram up to centre.

Moment at centre, M′E

Deflection at centre =

9. A uniform beam of length L is supported

symmetrically over a span l. Determine the ratio

of l/L if the upward deflections at the ends is equal to

the downward deflection at the centre due to a

concentrated load at mid span.

Solution:

Fig. shows a beam AB of length L, supported

over CD = l, central load is W. Say EI is the flexural

rigidity of the beam. Reaction RC = RD = W/2 (on

account of symmetrical loading).

Let us analyze only half the portion of the beam and

slope at centre E will be zero. Take a section YY at a

distance of x from A, in portion CE.

Integrating

So,

Integrating

By substituting the value,


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Deflection at A, yA, x = 0

Deflection at E, x = L/2

But yA = –yE (as they are numerically equal)

10. A propped cantilever of length L is fixed at one end

and roller supported at the other end. Cantilever is

subjected to a couple M at L/4 from A as shown in fig.

Determine reaction at propped end, deflection at C.

Solution:

The reaction at propped end is RA. Consider a

section YY at a distance x from A (on the portion CB).

Integrating


Now dy/dx = 0 at x = L, so

So,

Also by integrating

C2 is another constant of integration,

0 = 0 – write then omitted term – 0 – 0 – C2

Now y = 0 at x = 0,

Constant, C2 = 0

Finally,

At fixed end, y = 0, x = L, by substituting this value

Deflection at C, x = L/4 by substituting

Substituting the value of RA

11. A long steel strip of uniform width and thickness

2.5 mm is lying on a level floor. Its one end is passing

over a roller of 60 mm in diameter lying on the floor at

one point. At what distance on either side of the roller

will the strip be clear of the ground? What is the

maximum bending moment in strip? ρ for steel = 76.44

× 10–6 N/mm3. Take b = 50 mm and E = 210 kN/mm2.


Yatin Kumar Singh

Solution:

b = 60 mm ; t = 2.5 mm;

w = b × t × ρ N/mm = weight per unit length

= 60 × 2.5 × 76.44 × 10−6 = 0.011466 N/mm.

Say L is the length of the strip which is clear from the

floor. At A, strip just leaves the ground. Say RC is

reaction at C and RA is reaction at A. Consider a

sections YY at a distance x from C.

Bending moment,

or

Integrating , we get


0 = 0 – 0 + C1

Constant, C1 = 0

where w is rate of loading.

By integrating


Now, x = L, y = 0, by substituting this value, we get,

Finally,

Moreover at end A, x = L, dy/dx = 0,

Substituting these values

by substituting the value of RC

At point C, y = –60 mm; x = 0. (we have taken x positive

towards right. The y is negative upwards).

Maximum bending moment occurs as a

E = 210 × 1,000 N.mm; w = 0.011466 N/mm

Substituting these values

L = 1894.2 mm = 1.894 m

12. A beam AB of length 4L is being lifted by a crane

and is bolted to the left and right hand cantilevers at A

and B. In raising this beam into position and before it is

properly aligned, it fouls at A and B carrying one

upward force P exerted by crane as shown in figure.

Show that the deflection of the beam at the crane hook

if E and I are same for each of the beam section is

1.328PL3/EI.

Solution:

Reaction at A, RA = 0.375 P

Reaction at B, RB = 0.625 P

Deflection at A,

Deflection at B,

δE due to P at E,

Total deflection at E


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13. A beam AB is simply supported at ends of length 8

m, which carries a udl of 4 kN/m throughout its length.

It is propped at the centre by a force P such that the

level of the centre is the same as the level of ends.

Determine the magnitude of P. What are reactions at

support and bending moment at centre? If EI is the

flexural rigidity, what is the slope at ends?

Solution:

For a simply supported beam, span length L, carrying a

udl of intensity w, the central deflection is

For a simply supported beam, span length L with

central load P, the central deflection is

But net deflection at C is zero.

Substituting the values w = 4 kN/m, L = 8 m, we obtain

Propping force, P = 20 kN

Total load on the beam = 4 × 8 = 32 kN

Reaction, RA = RB is due to symmetric loading

RA + RB + P = 32 kN

RA + RB = 32 − 20 = 12 kN,

RA = RB = 6 kN

Bending moment at centre C

Taking moments about C, for portion AC

= -8 kN.m

Consider a section at a distance x from A,

y = 0 , x = 0, constant, C2 = 0

y = 0, x = 4 m

Slope at A, x = 0

Due to symmetry, slope at B,

14. A beam AB, 2 m long, carries a uniformly

distributed load of 10 kN/m is resting over a similar

beam (of same cross section) CD, 1 m long as shown

in Fig. Determine the reaction at C.

Solution:

Say the reaction at C = P′

Consider a section YY at a distance x from B,

Integrating

By integrating

y = 0 at x = 2 m


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Now,

at x = 0 and deflection on CD

15. A beam of length L is hinged at both the ends. An

anticlockwise turning moment M is applied at point C.

Point C is at a distance of L/4 from end A. Determine

the slope and deflection at A. Given EI is the flexural

rigidity of the beam.

Solution:

For reactions, let us take moments about A

(e cw) M = RB × L (cw)

To balance RB, reaction

Consider the section YY at a distance of x from A in

portion CB of beam.

Bending moment,

or

By integrating

where C1 is constant of integration

Again integrating

at x = 0, y = 0, so

0 = 0 omitted term + C1 × 0 + C2

Constant, C2 = 0

At x = L, y = 0 at end B, by substituting this value

Expressions for slope and deflection are

at x = L/4, point C.

Slope,

Deflection

16. A simply supported beam of length L carries a

load W at point C, such that AC= a, CB = b, a + b = L.

Draw BM diagram for the beam and determine the

deflection under the load. Given EI as the flexural

rigidity of beam.

Solution:

BM diagram is shown with

Consider the portion AC, origin at A, x positive towards

right, first moment of area AC′C about origin A

Say slope at C is iC

Slope at A is iA.


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Then,

or

Consider portion BC, x origin at B, x positive towards

left

By adding we obtain;

But iC = –iC, because slope is same and taking x positive

towards left, iC = –iC′

17. A simply supported beam of length L carries two

loads W each at L/4 from each end. Draw BM diagram

of beam and determine slope at ends and deflection at

centre. EI is the flexural rigidity of beam.

Solution:

Beam with load W at L/4 from each end is shown in fig.

Reaction, RA = RB = W (because of symmetrical loading).

Bending moment,

Slope at ends

Area of BM diagram up to middle of the beam abcd.

Area,

Because of symmetrical loading, iE = 0.

(because of symmetrical loading)

First moment of area about A

But,

So,

Deflection at centre,

18. A cantilever AB of length L is of different section

with moment of inertia I1for L/2 and I2 for L/2 to L as

shown in fig. Determine the slope and deflection at A

if E is Young’s modulus of the material.

Solution:

By integrating both the sides

At fixed end, iB = 0

Multiply the equation of bending moment

by xdx throughout


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But slope iB = 0, deflection yB = 0 at fixed end.

So

19. A cantilever, 3 m long, carries a udl of 20 kN/m

throughout its length. Its free end is attached to a

vertical tie rod 2.4 m long and 20 mm in diameter. The

rod is initially straight. Determine the load taken by the

rod and deflection in cantilever at A. E = 210 kN/mm2

I = 1,000 × 104 mm4

Solution:

Tensile force P in rod produces reaction P in cantilever

as shown. Consider a section YY at a distance x for A.

Bending moment, Mx = Px – (wx2/2),

where w = 20 kN/m.

Integrating

where C1 is constant of integration.

dy/dx = 0 at B i.e., x = 3 m

By integrating

y = 0, at x = 3 m, end B

Deflection at A,

Extension in bar,

Change the length of the rod, dl = yA

Load taken by rod, P = 22.31 kN

Deflection in cantilever,

20. In a structure, a vertical member made of copper is

supported by two steel bars as shown in Fig. The

copper member is subjected to a temperature rise of

100°C. Calculate the maximum bending moment set up

in steel member which can be considered as simply

supported beam. Weight of all the members may be

neglected. Width of all the members is 20 mm.

For copper, EC = 105 kN/mm2, αC = 18 × 10–6/°C

For steel, ES = 200 kN/mm2


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Solution:

Steel member

Free expansion of copper member is prevented by

= 300 × α × T – 2δ = 300 × 18 × 10–6 – 2 × 0.211 W

= 0.54 – 0.422 W

Area of cross section of copper member = 20 × 5

= 100 mm2

Compressive force in copper member,

21. A beam, 9 m long, is simply supported at ends

carries a linearly varying load of zero intensity at C

increasing to 12 kN/m at B over a distance of 6 m.

If EI is the flexural rigidity of the beam, determine the

slope and deflection at point C.

Solution:

Total load on beam = (6×12)/2 = 36 kN

CG of load from A, 3 + 4 = 7 m.

Moment about A, 36 × 7 = 9RB.

Reaction, RB = 28 kN

RA = 36 – 28 = 8 kN.

Take a section YY at a distance x from A

wx = rate of loading at x

Mx = bending moment at x

By integrating

x = 0, y = 0, C2 = 0

x = 9m, y = 0

C1 = -93.6

At C, x = 3

22. A composite beam consists of a wooden section,

240 mm width × 260 mm depth. Two steel plates of

width 240 mm and thickness 20 mm are fixed on the

top and bottom of wooden section as shown in Fig.

11.48. Composite beam is simply supported over a

span of 5 m and a load of 30 kN is applied at the centre

of the beam. Determine the deflection at the centre of

composite beam. ES = 200 GPa, EW = 15 GPa.

Figure 11.48

Solution:

Section is symmetrical about the central axis XX.

Moment of inertia,

Moment of inertia, IS

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Yatin Kumar Singh

Def numerical

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