Text book of Strength of Materials_P.H.Jain

Strength of Materials − P. H. Jain Shear Force and Bending Moment 2.47

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Strength of Materials − P. H. Jain Simple Stresses and Strains 3.21 23. A member ABCD is subjected to point loads P 1, P2, P3 and P4 as shown in fig. Calculate the forece

P2 necessary for equilibrium if P 1 = 10 kN, P3 = 40 kN and P 4 = 16 kN. Take E = 2.05 ×××× 105 N/mm2. Determine the total elongation of the member. S.R.T.M.U. [N 09]

Solution : Given : d1 = 25 mm, d2 = 50 mm, d3 = 30 mm.

L1 = 1000 mm, L2 = 600 mm, L3 = 800 mm.

P1 = 10 kN, P3 = 40 kN, P4 = 16 kN.

E = E1 = E2 = E3 = 2.05 × 105 N/mm2. Find P2 = ?, δl = ?

∴ A1 = 221 25

4

πd

4

π= = 156.25π mm2

A2 = 222 50

4

πd

4

π= = 625π mm2

A3 = 223 30

4

πd

4

π= = 225π mm2

For the equilibrium of the entire bar,

Σ Fx = 0 (considering → +ve and ← −ve)

∴ − 10 + P2 – 40 + 16 = 0

∴ P2 = + 34 kN (→)

Consider Free Body Diagram (F.B.D.) of each part as shown in fig.

Thus, forces on each part

P1 = + 10 kN = + 10 × 103 N (Tensile)

P2 = − 24 kN = − 24 × 103 N (Compressive)

P3 = + 16 kN = + 16 × 103 N (Tensile)

We know that, total change in length of bar

δl = δl1 + δl2 + δl3

∴ δl = 33

33

22

22

11

11

EA

LP

EA

LP

EA

LP ++

Here E1 = E2 = E3 = E

∴ δl =

++

3

33

2

22

1

11

A

LP

A

LP

A

LP

E

1

∴ δl =

××+××−××× 225π

0081016

625π

6001042

156.25π

10001010

102.05

1 333

5

∴ δl = + 0.1519 mm (Elongation)

10 kN 34 kN

1000 mm

40 kN 16 kN

10 kN

34 kN

600 mm

40 kN 16 kN 24 kN

10 kN 24 kN

1

2

800 mm

16 kN 10 kN 34 kN 40 kN

16 kN 3

P4

P1

Dia. 25 mm Dia. 30 mm

Dia. 50 mm

1000 mm 600 mm 800 mm

P2 P3 1

2 3

16 kN

10 kN

Dia. 25 mm Dia. 30 mm

Dia. 50 mm

1000 mm 600 mm 800 mm

P2 =

34 kN

40 kN

1 2

3

Copy Rights Reserved with Prof. P. H. Jain

Strength of Materials − P. H. Jain Simple Stresses and Strains 1.21


+ ve − ve

35. A simply supported beam carries an U.V.L. and a point load as shown in fig. Draw shear force and bending moment diagrams. Also, locate the posit ion and magnitude of the maximum bending moment. Solapur Univ. [M 08]

Solution : Step 1 : Support Reactions : ∑MA = 0 ( = +ve, = −ve)

∑V = 0 ( ↑ = +ve, ↓ = −ve) ∴ +(2

1× 20 × 3) × 3 + 30 × 5 − RB × 6 = 0

∴ RA + RB − (2

1× 20 × 3) − 30 = 0 ∴ 6RB = 240 kN

∴ RA + RB = 60 kN ∴ RB = 40 kN (↑↑↑↑) and RA = 60 −−−− 40 = 20 kN (↑↑↑↑)

Step 2 : S.F. Calculations :

SAL = 0 [L]

SAR = 20 kN [L]

SC = 20 kN [L]

SD = 20 − (2

1× 20 × 3) = −10 kN [L]

SEL = −10 kN [L]

SER = −10 − 30 = −40 kN [L]

SBL = −40 kN [L]

SBR = −40 + 40 = 0 [L]

Step 3 : B.M. Calculations :

MA = 0 [L]

MC = 20 × 1 = 20 kNm [L]

MD = 20 × 4 − (2

1× 20 × 3) × 1 = 50 kNm [L]

or −30 × 1 + 40 × 2 = 50 kNm [R]

ME = 40 × 1 = 40 kNm [R]

MB = 0 [R]

Step 4 : Maximum B.M. : In S.F.D., the S.F. is zero at point F. Let ‘x’ be the distance of point F from C.

Load intensity at F = h = 3

20x

=3

20

x

h Q

Since S.F. at F = 0.

∴ 20 − (2

1 × h × x) = 0

∴ 20 − (2

1 ×3

20x × x) = 0

∴ 20 − 3

10x2

= 0 ∴ 60 − 10x2 = 0 ∴ 10x2 = 60

∴ x = 6 = 2.449 m

∴B.M. at F = 20 (1 + x) − (2

1 × h × x) × 3

x = 20 + 20x − h

6

x2

= 20 + 20x − 3

20x

6

x2

= 20 + 20x − 18

20x3

= 20 + 20(2.449) − 18

20(2.449)3 = 52.66 kNm

∴ Maximum B.M. = B.M. at F = 52.66 kNm

Therefore, Maximum B.M. is at 1 + 2.449 = 3.449 m from A.

20 kN/m

1 m 3 m 1 m

A B D E

30 kN

C

1 m

0 0

20 kNm

50 kNm 40 kNm

52.66 kNm Cubic Curve (3rd degree curve)

RA = 20 kN RB = 40 kN

h

+

L.D.

20 kN

0 +

−

0

20 kN

10 kN 10 kN

40 kN 40 kN

x

Parabolic Curve (2nd degree curve)

F

S.F.D.

B.M.D.

20 kN/m

1 m 1 m

A B D E

30 kN

C

1 m 3 m

A C 1 m x

20 kN

h

F

20 kN/m

3 m

D

Strength of Materials − P. H. Jain S.F.B.M. 2.43


Strength of Materials − P. H. Jain Simple Stresses and Strains 3.21

Type 1

Formulae of Maximum Bending Moment (M) in Some Stan dard Cases :

Table below shows magnitude and position of maximum bending moment in some standard load conditions.

1) Simply supported beam with point load at mid span 2) Simply supported beam with u.d.l. on entire span

3) Cantilever with point load at free end 4) Cantilever with u.d.l. at free end

Problems on Beams having Symmetric Cross-Section 1. A steel cantilever beam of span 4 m is subjected to a point load of 2 kN at the free end. The cross -

section of the beam is 50 mm wide and 75 mm deep. D etermine the maximum bending stress in the beam. Dr.B.A.M.U.[M 07]

Solution : Given : L = 4 m, W = 2 kN, b = 50 mm, d = 75 mm. Find σmax = ?

Since, the beam is cantilever, tensile stress will develop in the top layer and compressive stress will develop in the bottom layer.

Maximum bending moment is at fixed end A

M = W.L = 2 × 4 = 8 kNm = 8 × 106 Nmm

Moment of Inertia of beam cross-section

I = 12

bd3

= 12

57 50 3× = 1757812.5 mm4

Distance of extreme layers from neutral axis N-A

ymax = yt = yc = 2

d =

2

75 = 37.5 mm

Using the relation I

M =

y

σ

∴ σmax = I

Mymax =

1757812.5

108 6× × 37.5 = 170.66 N/mm² (i.e. σt = σc = 170.66 N/mm²)

2 kN

A B

4 m

Load Diagram

σc = 170.66 N/mm²

σt = 170.66 N/mm²

Bending stress distribution

yt

yc

75 mm

50 mm

N

Cross-section

A

A B

L

W

L/2 L/2

W

A B

L

A B

L

w /unit length

A B

L

w /unit length

Mmax = 4

WL

Occurs at mid span.

Mmax = 8

2Lw

Occurs at mid span.

Mmax = WL

Occurs at fixed point.

Mmax = 2

2Lw

Occurs at fixed point.

Strength of Materials − P. H. Jain Bending Stresses in Beams 3.13



32. A simply supported beam carries a u.d.l. of 80 kN/m over the entire span of 6 meter. The cross-section of the beam is a T-section having flange 15 0 mm × 50 mm and web 50 mm × 150 mm. Calculate the maximum shear stress for the section of the beam. Also draw shear stress distribution diagram. S.R.T.M.U.[D 11], Dr.B.A.M.U.[ [M 98]

Solution : Given : S.S.B. of L = 6 m, w = 80 kN/m, T-section as shown. Find τ1, τ2 and τmax = ?

From symmetry of load diagram,

Reaction RA = RB = 2

Lw =

2

6)(80× = 240 kN

∴ Maximum shear force = Shear force at A and B

∴S = 240 kN = 240 × 103 N

Distance of centroidal axis (N-A) from base of section

y = 21

2211

A A

yA yA

++

= )15050()50 (150

75)15050(175)50 (150

×+×××+××

= 125 mm

Moment of Inertia of beam cross-section about N-A

I = 1xxI +

2xxI = ]hAI[ 211G1

+ + ]hAI[ 222G2

+

=

−×+× 2

3

)125(175)50150(12

50501 +

−×+× 2

3

)75(125)15050(12

15050 = 53.125 × 106 mm4

Shear stresses at top and bottom of the section are zero.

Shear stress in top flange at junction of top flange and web (Section 1-1)

τ1 = bI

ySA =

6

3

10125.53501

)]50()50150[(10 240

××××××

= 11.29 N/mm²

Shear stress in web at junction of top flange and web (Section 2-2)

τ2 = bI

ySA =

6

3

10125.5350

)]50()50150[(10 240

××××××

= 33.88 N/mm²

or τ2 = τ1 × widthWeb

widthFlange = 11.29 ×

50

150 = 33.88 N/mm²

Maximum shear stress = Shear stress at N-A. Consider area above N-A.

τmax = τN−A = bI

ySA =

6

3

10125.5350

)]5.12()2550()50()50150[(10 240

××××+××××

= 35.29 N/mm²

or alternatively, by considering area below N-A

τmax = τN−A = bI

ySA =

6

3

10125.5350

)]5.62()50125[(10 240

××××××

= 35.29 N/mm²

A B 6 m

80 kN/m

RA = 240 kN RB = 240 kN

50 mm

A N

50 mm

150 mm

150 mm

y = 125 mm

75 mm 50 mm

12.5 mm 25 mm

τ1 τ2

τmax = 35.29

11.29 33.88

Shear stress in N/mm²

Shear stress distribution Cross-section

62.5 mm

0

0

1 1 2 2

Strength of Materials − P. H. Jain Shear Stresses in Beams 4.29

Strength of Materials − P. H. Jain Simple Stresses and Strains 3.21 12. A rectangular column 200 mm wide and 150 mm thi ck is carrying a vertical load of 15 kN at an

eccentricity of 50 mm in a plane bisecting the thic kness. Determine the maximum and minimum intensities of stress in the section. Dr. B.A.M.U.[M 08], Amravati Univ.[M 09]

Solution : Given : b = 200 mm, d = 150 mm, P = 15 kN = 15 × 103 N, e = 50 mm. Find σmax and σmin = ?

Area of section A = b × d = 200 × 150 = 30000 mm2

Moment of Inertia about bending axis (Y-Y axis)

Iyy = 12

db3

= 12

200150 3× = 100 × 106 mm4

Distance of extreme layers of section from Y-Y axis

x = 2

b =

2

200 = 100 mm

Direct stress

σd = A

P =

30000

10 15 3× = 0.5 N/mm² (Compressive)

Bending stress

σb = Z

M =

yyI

eP×× x =

6

3

10100

5010 15

×××

× 100 = 0.75 N/mm²

Maximum stress σmax = σd + σb = 0.5 + 0.75 = 1.25 N/mm² (Comp. on side CD)

Minimum stress σmin = σd − σb = 0.5 − 0.75 = −−−−0.25 N/mm² (Tensile on side AB)

Stress distribution at base section is shown in fig.

13. A hollow rectangular column is having external and internal dimensions as 1200 mm deep × 800 mm wide and 900 mm deep × 500 mm wide respectively. A vertical load of 200 kN is transmitted in the vertical plane bisecting 1200 mm side and at an eccentricity of 110 mm from the geometric axis of the section. Calculate the maximum and minimum s tresses in the section. Dr. B.A.M.U. [N 09]

Solution : Given : B = 800 mm, D = 1200 mm, b = 500 mm, d = 900 mm, P = 200 kN = 200 × 103 N,

e = 110 mm. Find σmax and σmin = ?

Area of section

A = BD − bd = 800 × 1200 − 500 × 900 = 510000 mm2

Moment of Inertia about bending axis (Y-Y axis)

Iyy = 12

db

12

DB 33

−

= 12

8001200 3× −12

500900 3×

= 4.1825 × 1010 mm4

Distance of extreme layers of section from Y-Y axis

x = 2

B =

2

800 = 400 mm

Direct stress

σd = A

P =

510000

10 200 3× = 0.392 N/mm² (Compressive)

Bending stress

σb = Z

M =

yyI

eP×× x =

10

3

101825.4

11010 200

×××

× 400 = 0.210 N/mm²

Maximum stress σmax = σd + σb = 0.392 + 0.210 = 0.602 N/mm² (Compressive on side CD)

Minimum stress σmin = σd − σb = 0.392 − 0.210 = 0.182 N/mm² (Compressive on side AB)

Stress distribution at base section is shown in fig.

50 mm

15 kN

150 mm

200 mm Y

Y

X X

A

B

C

D

x x

σmin

= −0.25 σmax

= 1.25

50 mm

σmin

σmax

Stress Distribution at base

800 mm

12

00 m

m

Y

X X

A

B C

D 500 mm

P

90

0 m

m

11

0 m

m

Y

x x

Base Section

Strength of Materials − P. H. Jain Direct and bending Stresses 5.9



Type 3.2

Problems on Composite Shafts −−−− In Series (with different torques)

34. A stepped shaft is subjected to torques as sho wn in fig. (a). The length of each section is 0.5 m and the diameters are 80 mm, 60 mm and 40 mm as sho wn in fig. If G = 80 GPa, what is the angle of twist in degrees at the free end ? Dr. B.A.M.U. B.Tech.C.[N 04], Dr. B.A.M.U.[N 99]

Solution : Compound shaft in series as shown. L1= L2 = L3 = 0.5 m = 500 mm, D1 = 40 mm, D2 = 60 mm, D3 = 80 mm, G1 = G2 = G3 = G = 80 GPa = 80 × 103 N/mm².

Find θ = ?

Consider F.B.D. of each shaft separately as shown in fig. (b) starting from shaft 1

Thus, torques on each shaft

T1 = 1 kNm = 1 × 106 Nmm

T2 = 3 kNm = 3 × 106 Nmm

T3 = 6 kNm = 6 × 106 Nmm

Therefore, angle of twist at free end

θ = θ1 + θ2 + θ3 (in anticlockwise direction)

= P33

33

P22

22

P11

11

IG

LT

IG

LT

IG

LT ++

=

++

P3

3

P2

2

P1

1

I

T

I

T

I

T

G

L [Q L1= L2 = L3 & G1 = G2 = G3]

=

π×+

π×+

π×

× 4

6

4

6

4

6

380

32

106

6032

103

4032

101

1080

500

= 0.4893 rad. = 0.4893 × π

180 = 2.80O

35. The stepped steel shaft shown in fig. is subjec ted to a torque ‘T’ at the free end and a torque of ‘2T’ in the opposite direction at the junction of t he two sizes. What is the total angle of twist at t he free end, if the maximum shear stress in the shaft is limited to 70 MN/m². Take the modulus of rigidity as 84 GN/m². Dr. B.A.M.U. [D 00]

B A

1 kNm 1 kNm

C B C D

3 2 1

Fig. (b) F.B.D. of each shaft

2 + 1 = 3 kNm

2 + 1 = 3 kNm

3 + 2 + 1 = 6 kNm

3 + 2 + 1 = 6 kNm

0.5 m 0.5 m

D

C B

3 kNm

φ 60 mm φ 80 mm

2 kNm

0.5 m

A

1 kNm

φ 40 mm

Fig. (a)

3333 2222 1111

1.2 m 1.8 m

A

B C

2T T

φ 50 mm φ 100 mm

Strength of Materials − P. H. Jain Torsion 6.35


Strength of Materials − P. H. Jain Simple Stresses and Strains 3.21 44. A point in a strained material is subjected to st resses shown in fig. By using Mohr’s circle find :

1) The magnitude of principal stresses.

2) The direction of principal planes.

3) The magnitude of maximum shear stress. 4) The direction of planes of maximum shear stres s.

5) The normal stress on the planes carrying maxim um shear stress. Solapur Univ. [N 06]

Solution : Given : σx = 100 N/mm² (tensile), σy = 40 N/mm² (tensile), τ = 20 N/mm², Find σ1, σ2, θP1, θP2, τmax, θS1, θS2 , σn = ? Mohr’s Circle : Refer fig. (b) Lets choose scale : 1 cm = 10 N/mm2. 1) Mark origin ‘O’ and draw horizontal and vertical axes through O.

2) Draw OA = σx = 100 N/mm2 = 10 cm and OB = σy = 40 N/mm2 = 4 cm towards right from O. 3) At A and B, draw perpendicular lines AG and BH = τ = 20 N/mm2 = 2 cm as shown. 4) Mark mid-point of AB as C. Join G-H passing through C. With center C and diameter GH draw a circle. 5) From C, draw CR and CS perpendicular to OA.

By measurement,

Major Principal Stress

σ1 = Length OP × scale = 10.6 cm × 10 = 106 N/mm² (tensile)

Minor Principal Stress

σ2 = Length OQ × scale = 3.4 cm × 10 = 34 N/mm² (tensile)

Direction of Major Principal Plane

Q 2θP1 = Angle GCP (in anticlockwise direction) = 33.7O ∴∴∴∴ θP1 = 16.85O

Direction of Minor Principal Plane

θP2 = θP1 + 90O = 16.85O + 90O = 106.85O

Or Q 2θP2 = Angle GCQ (in anticlockwise direction) = 213.7O ∴∴∴∴ θP2 = 106.85O

Maximum Shear Stress

τmax = Radius CR × scale = 3.6 cm × 10 = 36 N/mm²

Direction of Planes of Maximum Shear Stress (+ ve)

Q 2θS1 = Angle GCR (in anticlockwise direction) = 123.7O ∴∴∴∴ θS1 = 61.85O

Direction of Planes of Maximum Shear Stress (− ve)

θS2 = θS1 + 90O = 61.85O + 90O = 151.85O

Or Q 2θS2 = Angle GCS (in anticlockwise direction) = 303.7O ∴∴∴∴ θS2 = 151.85O

Normal Stress on plane of maximum shear stress

σn = Length OC × scale = 7 cm × 10 = 70 N/mm² (tensile)

40 N/mm²

40 N/mm²

100 N/mm²

100 N/mm²

20 N/mm²

20 N/mm²

Fig. (a)

+σ

+τ

−τ

O A

B C

R

Fig. (b)

G

H

Q P

S

Scale : 1 cm = 10 N/mm2



CONTENTS OF THIS BOOK

1) Simple Stresses and Strains

2) Shear Force and Bending Moment

3) Bending Stresses in Beams

4) Shear Stresses in Beams

5) Direct and Bending Stresses

6) Torsion

7) Principal Stresses and Strains

8) Thin Cylinders

9) Strain Energy

10) Slope and Deflection

11) Axially Loaded Columns

Appendix : Solution of University Question Papers

A) Dr. B. A. M. U. Aurangabad

B) Solapur University, Solapur

C) S. R. T. M. U. Nanded

Maharashtra, (INDIA)

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Text book of Strength of Materials_P.H.Jain