Text book of Strength of Materials_P.H.Jain

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Shear Force and Bending Moment 2.47

Strength StrengthofofMaterials Materials−−P.P.H.H.Jain Jain

Simple SimpleStresses Stressesand andStrains Strains 1.21 3.21

23. A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown in fig. Calculate the forece 5 2 P2 necessary for equilibrium if P1 = 10 kN, P3 = 40 kN and P4 = 16 kN. Take E = 2.05 × 10 N/mm . Determine the total elongation of the member. S.R.T.M.U. [N 09] Dia. 50 mm

Dia. 25 mm

1

d1 = 25 mm, L1 = 1000 mm, P1 = 10 kN,

600 mm

d2 = 50 mm, L2 = 600 mm, P3 = 40 kN,

P4

3

2

1000 mm

Solution : Given :

P3

P2

P1

Dia. 30 mm

800 mm

d3 = 30 mm. L3 = 800 mm. P4 = 16 kN.

E = E1 = E2 = E3 = 2.05 × 105 N/mm2. ∴

A1 =

π 2 π 2 d1 = 25 = 156.25π mm2 4 4

π 2 π 2 A2 = d 2 = 50 = 625π mm2 4 4 A3 =

Find P2 = ?, δl = ?

Dia. 25 mm

Dia. 50 mm P2 =

10 kN 1

34 kN

π 2 π 2 d 3 = 30 = 225π mm2 4 4

1000 mm

40 kN 3

2

600 mm

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. 10 kN

34 kN 40 kN 16 kN

1

10 kN

1000 mm 10 kN 34 kN

24 kN

2

40 kN 16 kN

24 kN

600 mm

16 kN

10 kN 34 kN 40 kN

3

16 kN

800 mm

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 = δl 1 + δl 2 + δl 3 δl =

PL P1L1 PL + 2 2 + 3 3 A1E1 A 2 E 2 A 3E 3

∴

δl =

1  P1L1 P2 L 2 P3L 3  + +   E  A1 A2 A3 

∴

δl =

10 × 103 × 1000 24 × 103 × 600 16 × 103 × 800  − +   625π 225π 2.05 × 105  156.25π 

∴

δl = + 0.1519 mm (Elongation)

∴

Dia. 30 mm

Here E1 = E2 = E3 = E

1

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800 mm

16 kN

Strength StrengthofofMaterials Materials−−P.P.H.H.Jain Jain

Shear Force and BendingS.F.B.M. Moment 2.43 2.47

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 position and magnitude of the maximum bending moment. Solapur Univ. [M 08] 20 kN/m

30 kN

C

A

B

D

E

3m

1m

1m

1m

Solution :

∑MA = 0 (

Step 1 : Support Reactions : ∑V = 0 ( ↑ = +ve, ↓ = −ve) ∴ RA + RB − (

1 × 20 × 3) − 30 = 0 2

∴ 6RB = 240 kN

∴ RA + RB = 60 kN Step 2 : S.F. Calculations :

SD

∴ RB = 40 kN (↑ ↑) and

RA = 60 − 40 = 20 kN (↑ ↑) 20 kN/m

− ve

+ ve

SAL = 0 [L] SAR = 20 kN [L] SC = 20 kN [L]

= −ve)

= +ve,

1 ∴ +( × 20 × 3) × 3 + 30 × 5 − RB × 6 = 0 2

h

C

A

B

D

RA = 20 kN 1m

1 = 20 − ( × 20 × 3) = −10 kN [L] 2

30 kN

3m

E RB = 40 kN 1m

1m

L.D.

SEL = −10 kN [L] SER = −10 − 30 = −40 kN [L] SBL = −40 kN [L]

Parabolic Curve nd (2 degree curve)

20 kN

20 kN +

0

F

SBR = −40 + 40 = 0 [L]

x

0 −

10 kN 10 kN

S.F.D.

Step 3 : B.M. Calculations :

40 kN

40 kN

MA = 0 [L] MC = 20 × 1 = 20 kNm [L] MD

1 = 20 × 4 − ( × 20 × 3) × 1 = 50 kNm [L] 2

Cubic Curve (3rd degree curve)

50 kNm 40 kNm

+

20 kNm

or −30 × 1 + 40 × 2 = 50 kNm [R] ME = 40 × 1 = 40 kNm [R]

52.66 kNm

0

0 B.M.D.

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 =

 h 20  Q x = 3   

20x 3

Since S.F. at F = 0. ∴ ∴ ∴

1 × h × x) = 0 2 1 20x 20 − ( × × x) = 0 2 3

1m

20 − (

20 −

10x 2 3

∴

= 0

x =

∴B.M. at F = 20 (1 + x) − (

20 kN

∴ 60 − 10x2 = 0

h F

C

A

x

D

3m

∴ 10x2 = 60

6 = 2.449 m

1 x x2 20x x 2 × h × x) × = 20 + 20x − h = 20 + 20x − 2 3 6 3 6

20(2.449)3 = 52.66 kNm 18 Maximum B.M. = B.M. at F = 52.66 kNm = 20 + 20(2.449) −

∴

20 kN/m

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

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= 20 + 20x −

20x 3 18

Strength Bending Stressesand in Beams StrengthofofMaterials Materials−−P.P.H.H.Jain Jain Simple Stresses Strains 3.13 3.21 Formulae of Maximum Bending Moment (M) in Some Standard 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 W A

Mmax =

B L/2

L/2

2) Simply supported beam with u.d.l. on entire span w /unit length

WL A

4

B

Occurs at mid span.

3) Cantilever with point load at free end

w /unit length Mmax =

Mmax = WL B

8

4) Cantilever with u.d.l. at free end

W A

wL2

Occurs at mid span.

L

L

Mmax =

A

B

Occurs at fixed point.

2

Occurs at fixed point.

L

L

2 wL

Type 1 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 crosssection of the beam is 50 mm wide and 75 mm deep. Determine 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 = ? σt = 170.66 N/mm²

50 mm 2 kN

A

B

75 mm

N

yt A yc

4m σc = 170.66 N/mm² Load Diagram

Cross-section

Bending stress distribution

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 =

50 × 753 bd 3 = = 1757812.5 mm4 12 12

Distance of extreme layers from neutral axis N-A

d 75 = = 37.5 mm 2 2 M σ Using the relation = I y ymax = yt = yc =

∴

σmax =

M 8 × 106 ymax = × 37.5 = 170.66 N/mm² I 1757812.5

(i.e. σt = σc = 170.66 N/mm²)

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Strength Shear inMoment Beams 4.29 Strengthof ofMaterials Materials−−P. P.H. H.Jain Jain Shear Force andStresses Bending 2.47 32. A simply supported beam carries a u.d.l. of 80 kN/m over the entire span of 6 meter. The crosssection of the beam is a T-section having flange 150 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 = ? 80 kN/m A

B 6m RB = 240 kN

RA = 240 kN

From symmetry of load diagram, wL (80 × 6) = = 240 kN Reaction RA = RB = 2 2 ∴ Maximum shear force = Shear force at A and B ∴S = 240 kN = 240 × 103 N 150 mm

0

50 mm 1 12.5 mm

1 50 mm75 mm 25 mm

2

2 N

τ1

τ2

11.29

33.88

τmax = 35.29

A 62.5 mm

150 mm

y = 125 mm

50 mm

0 Shear stress in N/mm²

Cross-section

Shear stress distribution

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

y =

A 1 y1 + A 2 y 2 A1 + A 2

=

(150 × 50) × 175 + (50 × 150) × 75 = 125 mm (150 × 50) + (50 × 150)

Moment of Inertia of beam cross-section about N-A I = I xx1 + I xx 2 = [ I G1 + A1h 12 ] + [ I G 2 + A 2 h 2 2 ]

150 × 50 3  =  + (150 × 50)(175 − 125) 2  +  12  Shear stresses at top and bottom of the section are zero.

 50 × 150 3  + (50 × 150)(125 − 75) 2  = 53.125 × 106 mm4   12 

Shear stress in top flange at junction of top flange and web (Section 1-1) τ1 =

240 × 10 3 × [(150 × 50) × (50)] SAy = = 11.29 N/mm² bI 150 × 53.125 × 10 6

Shear stress in web at junction of top flange and web (Section 2-2) τ2 = or

240 × 10 3 × [(150 × 50) × (50)] SAy = = 33.88 N/mm² bI 50 × 53.125 × 10 6

τ2 = τ1 ×

Flange width 150 = 11.29 × = 33.88 N/mm² Web width 50

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

240 × 10 3 × [(150 × 50) × (50) + (50 × 25) × (12.5)] SAy = = 35.29 N/mm² bI 50 × 53.125 × 10 6

or alternatively, by considering area below N-A τmax = τN−A =

240 × 10 3 × [(125 × 50) × (62.5)] SAy = = 35.29 N/mm² bI 50 × 53.125 × 10 6

Strength StrengthofofMaterials Materials−−P.P.H.H.Jain Jain

Direct bending Simpleand Stresses andStresses Strains 5.9 3.21

12. A rectangular column 200 mm wide and 150 mm thick is carrying a vertical load of 15 kN at an eccentricity of 50 mm in a plane bisecting the thickness. 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

15 kN

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

50 mm

db 3 150 × 2003 Iyy = = = 100 × 106 mm4 12 12 Distance of extreme layers of section from Y-Y axis b 200 x= = = 100 mm 2 2 Direct stress σd =

200 mm Y

15 × 10 3 P = = 0.5 N/mm² (Compressive) A 30000

A

Bending stress

150 mm X

M 15 × 10 3 × 50 P×e σb = = ×x= × 100 = 0.75 N/mm² Z I yy 100 × 10 6

B

X D

Y x

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

C

50 mm

x

σmin = −0.25

σmax = 1.25

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 stresses 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 = ? 800 mm 500 mm Y

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

A

D

1200 × 800 900 × 500 − 12 12 3

=

3

P X

= 4.1825 × 1010 mm4 Distance of extreme layers of section from Y-Y axis B 800 x= = = 400 mm 2 2

X

110 mm

DB3 db 3 − 12 12

900 mm

Iyy =

1200 mm

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

B

C

Y x

x

Base Section

Direct stress σd =

P 200 × 10 3 = = 0.392 N/mm² (Compressive) A 510000

σmin σmax

Bending stress σb =

M P×e 200 × 103 × 110 = ×x= × 400 = 0.210 N/mm² Z I yy 4.1825 × 1010

Stress Distribution at base

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.

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Strength Strengthof ofMaterials Materials−−P. P.H. H.Jain Jain

Shear Force and Bending Torsion Moment 6.35 2.47 Type 3.2

Problems on Composite Shafts − In Series (with different torques) 34. A stepped shaft is subjected to torques as shown in fig. (a). The length of each section is 0.5 m and the diameters are 80 mm, 60 mm and 40 mm as shown 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. 2 kNm 1 kNm L1= L2 = L3 = 0.5 m = 500 mm, D1 = 40 mm, D2 = 60 mm, D3 = 80 mm, 2 1 3 G 1 = G2 = G3 = G = 80 GPa = 80 × 10 A B φ 40 mm N/mm². φ 60 mm Find θ = ?

3 kNm

3

D

C φ 80 mm

0.5 m

0.5 m

0.5 m

Fig. (a)

Consider F.B.D. of each shaft separately as shown in fig. (b) starting from shaft 1 3+2+1 = 6 kNm

3+2+1 = 6 kNm

2+1 = 3 kNm

3

1 kNm

1 kNm

1

2 B

B

C

C

D

2+1 = 3 kNm

A

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

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)

=

TL T1L1 T L + 2 2 + 3 3 G1I P1 G 2 I P2 G 3 I P3

=

L  T1 T2 T3  + +   G  I P1 I P2 I P3 

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

   1 × 10 6 3 × 10 6 6 × 10 6  + + =   π 4 π 4 80 × 10 3  π 40 4 60 80  32 32  32 180 = 0.4893 rad. = 0.4893 × = 2.80O π 500

35. The stepped steel shaft shown in fig. is subjected to a torque ‘T’ at the free end and a torque of ‘2T’ in the opposite direction at the junction of the two sizes. What is the total angle of twist at the 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] 2T T A φ 100 mm 1.2 m

B

φ 50 mm

C

1.8 m

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Strength of Materials − P. H. Jain Simple Stresses and Strains 3.21 44. A point in a strained material is subjected to stresses shown in fig. By using Mohr’s circle find : 1) 2) 3) 4)

The magnitude of principal stresses. The direction of principal planes. The magnitude of maximum shear stress. The direction of planes of maximum shear stress.

5) The normal stress on the planes carrying maximum 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 = ? +τ R

40 N/mm² 20 N/mm²

100 N/mm²

100 N/mm²

H

A O

Q

B

P

C

20 N/mm² G 40 N/mm² Fig. (a)

S −τ

Scale : 1 cm = 10 N/mm2

Fig. (b)

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)

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+σ

Strength of Materials − P. H. Jain

Force and Bending Moment CONTENTS OFShear THIS BOOK

2.47

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)