Unit V Advanced topics in bending of beams

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unit V Advanced topics in bending of beams Contents • Unsymmetrical bending of beams of symmetrical sections • Unsymmetrical bending of beams of unsymmetrical sections • Shear Centre • curved beams – Winkler Bach formula.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Advanced topics in bending of beams References • Rajput R.K. "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2010.

• William A .Nash, “Theory and Problems of Strength of Materials”, Schaum‟s Outline Series, Tata McGraw Hill Publishing company, 2007. • Punmia B.C."Theory of Structures" (SMTS) Vol 1&II, Laxmi Publishing Pvt Ltd, New Delhi 2004.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical bending of beams Unsymmetrical bending of beams: 𝑀 𝑓

• While using the well known simple flexure formula = , it is 𝐼 𝑦 assumed that the neutral axis of the cross section of the beam is perpendicular to the plane of loading. • This condition implies that the plane of loading or plane of bending, is coincident with or parallel to a plane containing a principal centroidal axis of inertia of the cross section of the beam. • If however, the plane of loading or that of bending doesn’t lie in (or parallel to ) a plane that contains the principal centroidal axis of the cross section, the bending is called unsymmetrical bending. • In the case of unsymmetrical bending, the direction of neutral axis is not perpendicular to the plane of bending. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical bending of beams • Reasons for unsymmetrical bending: (i) The section is symmetrical (viz. rectangular, circular, I-section) but the load line is inclined to both the principal axes.

(ii) The section itself is unsymmetrical (viz. angle section or channel section vertical web) and the load line is along any centroidal axes.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Stresses due to unsymmetrical bending • Stresses due to unsymmetrical bending: 𝑀′′ =Mcos 𝜃 Y Figure 1 shows the cross V M section of a beam under the P (u,v) action of a bending moment M N dA u acting in plane YY. v G = Centroid of the section, XX, YY are coordinate axes passing through G. UU, VV are Principal axes inclined at an angle 𝜃 to XX and YY axes respectively.

𝑀′ = Msin 𝜃 U

𝜃

X

𝛼

G

X

U

A Y

V

Figure 1

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Stresses due to unsymmetrical bending • Let us determine the stress distribution over the section. The moment M in the plane YY can be resolved into its components in the planes UU and VV as follows. • Moment in the plane UU, 𝑀′ =Msin 𝜃, • Moment in the plane VV, 𝑀′′ =Mcos 𝜃 • The components 𝑀′ and 𝑀′′ have their axes along VV and UU respectively.

• The resultant bending stress at the point p(u,v) is given by, 𝑀′ 𝑢 𝑀′′ 𝑣 𝜎𝑏 = + 𝐼𝑉𝑉 𝐼𝑈𝑈 𝑀 sin 𝜃.𝑢 𝑀 cos 𝜃.𝑣 = + 𝐼𝑉𝑉 𝐼𝑈𝑈 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Stresses due to unsymmetrical bending • 𝜎𝑏 =M

𝑣 cos 𝜃 𝐼𝑈𝑈

𝑢 sin 𝜃 + 𝐼𝑉𝑉

• At any point the nature of 𝜎𝑏 will depend upon the quadrant in which it lies. In other words the signs of u and v will have to be taken into account while determining the resultant bending stress. • The equation of the neutral axis (N.A.) can be found by finding the locus of the points on which the resultant stress is zero. • Thus the points lying on neutral axis will satisfy the condition that 𝜎𝑏 =0, i.e., M

𝑣 cos 𝜃 𝐼𝑈𝑈

𝑢 sin 𝜃 + 𝐼𝑉𝑉

=0

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Stresses due to unsymmetrical bending • Or, • 𝑣=-

𝑣 cos 𝜃 𝐼𝑈𝑈 𝐼𝑈𝑈 𝐼𝑉𝑉

• Or, 𝑣 = -

×

+

sin 𝜃 cos 𝜃

𝐼𝑈𝑈 𝐼𝑉𝑉

𝑢 sin 𝜃 𝐼𝑉𝑉

=0

𝑢

× tan 𝜃 𝑢

• This is an equation of a straight line passing through the centroid G of the section and inclined at an angle 𝛼 with UU where, tan 𝛼 = -

𝐼𝑈𝑈 𝐼𝑉𝑉

× tan 𝜃

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Stresses due to unsymmetrical bending • Worth noting points: (i) The maximum stress will occur at a point which is at the greatest distance from the neutral axis. (ii) All the points of the section on one side of the neutral axis will carry stress of the same nature and on the other side of its axis, of opposite nature. (iii) In the case where there is direct stress in addition to the bending stress, the neutral axis will still be straight line but will not pass through G (Centroid of section). This is obvious from the fact that from finding the equation algebraic sum of direct and bending stresses will be equated to zero. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Deflection of beams due to unsymmetrical bending • Deflection due to unsymmetrical bending: • Figure 2 shows the transverse ′′ =Wcos 𝜃 Y 𝑀 section of the beam with centroid V W G. ′ = Wsin 𝜃 𝑀 • XX and YY are two rectangular coU N ordinate axes and UU and VV are the principal axesinclined at an 𝜃 angle 𝜃 to the xy set of coordinate X X axes. 𝛿𝑢 𝛽 𝛿𝑣 • W is the load acting along line YY U A 𝛿 on the section of the beam. The load W can be resolved into the following two components: V Y • (i) Wsin𝜃 ----along UG Figure 2 • (ii) Wcos 𝜃 ------ along VG Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Deflection of beams due to unsymmetrical bending • Let, 𝛿𝑢= Deflection caused by the component W sin 𝜃 along the line GU for its bending about VV axis, and • 𝛿𝑣 = Deflection caused by the component Wcos 𝜃 along the line GV due to bending about UU axis. • Then depending upon the end conditions of the beam, the values of 𝛿𝑢 and 𝛿𝑣 are given by, • 𝛿𝑢=

𝑘(𝑤 sin 𝜃)𝑙 3 𝐸 𝐼𝑉𝑉

• and 𝛿𝑣 =

𝑘(𝑤 cos 𝜃) 𝑙 3 𝐸 𝐼𝑈𝑈

• Where, k= A constant depending on the end condition of the beam and position of the load along the beam and 𝑙 = length of the beam. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Deflection of beams due to unsymmetrical bending • The total or resultant deflection 𝛿 can then be found as follows: • 𝛿=

𝛿𝑢

2

+ 𝛿𝑣

2

• 𝛿=

𝑘𝑙 3 𝐸

𝑤 sin 𝜃 2 𝐼𝑉𝑉

• 𝛿=

𝑘𝑤𝑙 3 𝐸

sin 𝜃 2 𝐼𝑉𝑉

+

+

𝑤 cos 𝜃 2 𝐼𝑈𝑈 cos 𝜃 2 𝐼𝑈𝑈

• The inclination 𝛽 of the deflection, with the line GU is given by tan 𝛽 =

𝛿𝑢 𝛿𝑣

=

𝐼𝑈𝑈 tan 𝜃 𝐼𝑉𝑉

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Deflection of beams due to unsymmetrical bending

𝑀′′ =Mcos 𝜃 V N

𝑀′′ =Wcos 𝜃 Y V W

Y

M P (u,v) dA u v

G

𝜃

X

𝛼

G

X

U

A Y

𝑀′ = Wsin 𝜃 U

𝑀′ = Msin 𝜃 U N

V

X

𝜃

X

𝛽 U

Figure 1

A Y

V

Figure 2

• The magnitudes of 𝛼 and 𝛽 are the same and are measured from perpendicular lines (GU and GV) in the same direction as shown in figure1 and figure 2. • Thus the deflection 𝛿 will be in a direction perpendicular to the neutral axis. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems • Problem: • A beam of T-section (flange: 60 mm x 10 mm , web 100 mm x 5 mm) is 3 m length and is simply supported at the ends. It carries a load of 4 kN inclined at 200 to the vertical and passing through centroid of section. If E= 200 GN/m2, calculate (i) Maximum tensile stress, (ii) Maximum compressive stress (iii) Maximum bending stress (iv) Deflection due to the load, 𝛿 (v) Position of neutral axis.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems 60 mm

• Solutioin: • To find the centroid: • 𝑦=

Y, V 10 mm

60×10×5 + 100×5×60 60×10 + 100×5

30 mm X, U

X, U

Now 𝐼𝑥𝑥 =

60×103 𝐼𝑈𝑈 = + 60 × 10 × 30 − 5 2 12 5×1003 + + 100 × 5 × 80 − 50 2 12 4 4

= (38 × 10 ) + (86.67 × 10 ) = 1.25 × 106 mm4 = 1.25 × 10−6 𝑚4 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

100 mm

𝑦 = 30 mm (from top)

Y, V

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Unsymmetrical beams - Problems 60 mm Y, V 10 mm 30 mm X, U

X, U

+

100×53 12

100 mm

Now 𝐼𝑌𝑌 = 𝐼𝑉𝑉 =

10×603 12

= 181041.67 mm4 = 0.181 × 10−6 𝑚4 Y, V

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems 10 mm

Y, V

A

W=4 kN

200

X, U

X, U

Bending moment: 𝑤𝑢 𝑙 1.368 × 3 𝑀𝑢 = = = 1.026 kNm 4 4 𝑤𝑣 𝑙 3.758 × 3 𝑀𝑣 = = = 7.82 kNm 4 4

B

100 mm

Components of W: 𝑊𝑢 = W sin20 = 4sin20 = 1.368 kN 𝑊𝑣 = Wcos20 = 4 cos20 = 3.759 kN

30 mm

60 mm

C

D Y, V

𝑀𝑢 will cause maximum compressive stress at B and D and tensile stress at A and C 𝑀𝑣 will cause maximum compressive stress at A and B and tensile stress at C and D Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems (i) Maximum tensile stress: 1.026×2.5×10−3 0.181×10−6

𝑀𝑣

𝐼𝑈𝑈

+

2.82×80×10−3 1.25×10−6

= 14171.27 + 180480 = 194651.27 𝑘𝑁/𝑚2 = 194.651 MN/𝑚2 (i) Maximum Compressive stress : At B, 𝜎𝐵 = =

𝑀𝑢 ×30×10−3 𝐼𝑉𝑉

+

1.026×30×10−3 0.181×10−6

Y, V

A

2.82×30×10−3 + 1.25×10−6

B

200

X, U

X, U

𝑀𝑣 ×30×10−3 𝐼𝑈𝑈

W=4 kN

100 mm

𝐼𝑉𝑉

+

10 mm

=

𝑀𝑢

×80×10−3

30 mm

At C, 𝜎𝑐 =

×2.5×10−3

60 mm

C

D Y, V

= 170055.25 + 67680 = 237735.25 𝑘𝑁/𝑚2 = 237.735 MN/𝑚2

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems 60 mm

𝐼𝑉𝑉

+

𝑐𝑜𝑠 2 𝜃 𝐼𝑈𝑈

Y, V

A

200

B

X, U

X, U

for a beam with simply supported ends

and carrying a point load at its centre. 𝛿=

W=4 kN

100 mm

1 k= 48

𝑠𝑖𝑛2 𝜃

30 mm

𝑘𝑤𝑙 𝛿= 𝐸

3

10 mm

(iii) Deflection due to the load, 𝛿:

𝑘𝑤𝑙3 𝐸𝐼𝑈𝑈

𝑠𝑖𝑛2 𝜃

𝐼𝑈𝑈 𝐼𝑣𝑣

2

+ 𝑐𝑜𝑠 2 𝜃 C

1 4 × 103 × 33 = × 48 200 × 109 × 1.25 × 10−6 ×

𝑠𝑖𝑛2 20 ×

10−6

D Y, V

2

1.25 × 2 20 + 𝑐𝑜𝑠 −6 0.181 × 10 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems 60 mm

30 mm

𝛿 =9× × 2.542 = 0.02287 𝑚 𝛿 = 22.9 𝑚𝑚

10 mm

10−3

Y, V

A

=

B

200

X, U

X, U

(iv) Position of neutral axis: 𝐼𝑈𝑈 tan 𝛽 = tan 𝜃 𝐼𝑉𝑉

𝛽 100 mm

1.25×10−6 0.181×10−6

W=4 kN

× tan 20 C

∴ 𝛽 = 68.30

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

D Y, V

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Unsymmetrical beams - Problems Problem: A Cantilever of I-section, 2.4 m long is subjected to a load of 200 N at the free end as shown in Figure. Determine the resulting bending stress at corners A and B, on the fixed section of the cantilevers. 600 N 200

2.5 mm

45 mm

2.5 mm 30 mm

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

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Unsymmetrical beams - Problems Solutiion: A Cantilever of I-section, 2.4 m long is subjected to a load of 200 N at the free end. 200 N Y, V 𝑙 = 1.8 𝑚. 2.5 mm Since I-section is symmetrical about XX 200 and YY axes, therefore XX and YY are the 2 mm principal axes UU and VV. Moment of inertia, X,U X,U 45 mm 30×503 𝐼𝑈𝑈 = 𝐼𝑋𝑋 = 12 −8 = 9.99 × 10 𝑚4

28×453 − 12

= 99875 𝑚𝑚4

2.5 × 303 45 × 23 𝐼𝑉𝑉 = 𝐼𝑌𝑌 = 2 × + 12 12 = 1.128 × 10−8 𝑚4 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

2.5 mm Y, V

30 mm

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Unsymmetrical beams - Problems Maximum bending moment, M=wl= 200× 2.4 = 480 𝑁𝑚. Components of M, 𝑀𝑈 = M sin20 = 480 × 𝑠𝑖𝑛20 = 164.17 𝑁𝑚 200 N Y, V 𝑀𝑉 = M cos 20 = 480 × 𝑐𝑜𝑠20 = 451 𝑁𝑚. A

𝑀𝑈 will cause tensile stresses at points A and C and compressive stresses at points B and D. X,U

B

2.5 mm

200 2 mm

X,U

45 mm

𝑀𝑉 will cause tensile stresses at points A and B and compressive stresses at points C and D. C

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

2.5 mm Y, V

30 mm

D

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Unsymmetrical beams - Problems Now, resultant bending stresses on A and B are as follows. 𝑀𝑈 × 15 × 10−3 𝑀𝑉 × 25 × 10−3 𝜎𝐴 = + 𝐼𝑉𝑉 𝐼𝑈𝑈 A =

164.17×15×10−3 1.128×10−8

+

451×25×10−3

200 N Y, V B 200

9.99×10−8

= 218.31 + 112.86 = 331.17 MN/𝑚2 X,U 𝑀𝑈 × 15 × 10−3 𝑀𝑉 × 25 × 10−3 𝜎𝐵 = − + 𝐼𝑈𝑈 𝐼𝑉𝑉

=

164.17×15×10−3 − 1.128×10−8

2.5 mm

451×25×10−3 + 9.99×10−8 C

= 218.31 + 112.86 = −105.4 MN/𝑚2

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

2 mm X,U

45 mm

2.5 mm Y, V

30 mm

D

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Unsymmetrical bending of beams - Problems

10 mm

80 mm

Problem: A 80 mm x 80 mm x 10 mm angle section shown in Figure is used as a simply supported beam over a span of 2.4 m. It carries a load of 400 N along the line YG, where G is the centroid of the section. Calculate: 10 mm U Y V (i) Stresses at the points A,B and C 𝑥 of the mid section of the beam. (i) Deflection of the beam at mid 450 section and its direction with the X X 𝑦 load line. (i) Position of the neutral axis. 80 mm V U Take E=200 GN/𝑚2 Y

Unsymmetrical bending of beams - Problems Solution: Let (𝑥, 𝑦) be the co-ordinates of centroid G, with respect to the rectangular axes B𝑌1 and B𝑋1 . = 23.66 mm A V

Moment of inertia about xx axis, 103

𝐼𝑋𝑋

80 × + 80 × 10 × 23.66 − 5 2 12 = 10 × 703 + + 70 × 10 × 45 − 23.66 12

= 889833 𝑚𝑚4

2

𝑌1 10 mm 𝑥

450

X

U

U

Y

G B

80 mm V Y

= 8.898 × 105 𝑚𝑚4 = 𝐼𝑦𝑦 (since it is an equal angle section)

𝑦

X

10 mm

80×10×40 + 70×10×5 80×10 + 70×10

80 mm

Now 𝑥 = 𝑦 =

𝑋1

Unsymmetrical bending of beams - Problems If θ is the inclination of principal axes with GX, passing through G then, tan 2𝜃 =

2𝐼𝑥𝑦 𝐼𝑦𝑦 −𝐼𝑥𝑥

= ∞=tan 90

2𝜃 = 90 i.e., 𝜃1 = 450 and 𝜃2 = 90 + 45 = 1350 are the inclination of the principal axes GU and GV respectively.

1 2

1 2

𝐼𝑥𝑥 + 𝐼𝑦𝑦 + (𝐼𝑥𝑥 − 𝐼𝑦𝑦 )cos 90

−𝐼𝑥𝑦 sin 90 (At 𝜃1 = 450 )

1 = 8.898 + 8.898 × 105 + 0 2 −5.2266 × 105 = 14.1246 × 105 𝑚𝑚4

10 mm Y 𝑥

𝐺2 G

X

U

450

𝑦 𝐺1

U

B

80 mm V Y

X

10 mm

𝐼𝑈𝑈 =

A V 80 mm

Principal moment of inertia,

𝑌1

𝑋1

Unsymmetrical bending of beams - Problems Co-0rdinates of 𝐺1 = 40 − 23.66, − 23.66 − 5

= 16.34, −18.66

Centroid of 𝐺2 = − 23.66 − 5 , 45 − 23.66

= −18.66, 21.34

Product of inertia, 𝐼𝑥𝑦 = [80 × 10 × 16.34 × (−18.66)]+

(Product of inertia about the centroidal axes is zero because portions 1 and 2 are X rectangular strips) U

𝐺2 G

U

450

𝑦 𝐺1

B

80 mm V Y

X

10 mm

80 mm

[70×10× (−18.66) × (21.34)] 𝑌1 = −243923.5 − 278743 = −522666 𝑚𝑚4 10 mm 5 4 A Y = −5.2266× 10 𝑚𝑚 V 𝑥

𝑋1

Unsymmetrical bending of beams - Problems Also 𝐼𝑈𝑈 + 𝐼𝑉𝑉 = 𝐼𝑋𝑋 + 𝐼𝑌𝑌 ∴ 𝐼𝑉𝑉 = 𝐼𝑋𝑋 + 𝐼𝑌𝑌 − 𝐼𝑈𝑈 = 2 × 8.898 × 105 − 14.124 × 105 = 3.67 × 105 𝑚𝑚4

𝑊𝐿 400 × 2.4 × 𝑀= = = 4 4 The components of bending moment are: 𝑀′ = M sin 𝜃 = 2.4 × 105 sin 45 = 1.697 × 105 Nmm 𝑀′′ = M cos 𝜃 = 2.4 × 105 cos 45 = 1.697 × 105 Nmm

A 5 2.4 × 10 𝑁𝑚𝑚 V 80 mm

103

𝑌1

10 mm Y 𝑥

𝐺2 G

X

U

450

𝑦 𝐺1

U

B

80 mm V Y

X

10 mm

Stresses at the points A, B and C: Bending moment at the mid section,

𝑋1

Unsymmetrical bending of beams - Problems u,v co-ordinates: Point A, 𝑥 = −23.66, 𝑦 = 80 − 23.66 = 56.34 𝑢 = 𝑥 cos 𝜃 + 𝑦 sin 𝜃 = −23.66 cos 45 + 56.34 sin 45 = 23.1 𝑚𝑚 𝑣 = 𝑦 cos 𝜃 − 𝑥 sin 𝜃 = 56.34 cos 45 − (−23.66 sin 45) = 56.56 𝑚𝑚 Point B, 𝑥 = −23.66, 𝑦 = −23.66 𝑢 = 𝑥 cos 𝜃 + 𝑦 sin 𝜃 = −23.66 cos 45 + (−23.66 sin 45) = −33.45 𝑚𝑚 𝑌1 𝑣 = −23.66 cos 45 − (−23.66 sin 45) = 0

𝐺2 G

X

U

450

𝑦 𝐺1

U

B

80 mm V Y

X

10 mm

80 mm

Point C, 𝑥 = 80 − 23.66 = 56.34, 𝑦 = −23.66 𝑢 = 56.34 cos 45 + (−23.66 sin 45) = 23.1 𝑚𝑚 𝑣 = −23.66 cos 45 − 56.34 sin 45 = −56.56 𝑚𝑚

10 mm Y 𝑥

A V

C 𝑋1

Unsymmetrical bending of beams - Problems Stresses at Various points: Point A, 𝜎𝐴 =

𝑀′ 𝑢 𝐼𝑉𝑉

+

𝑀′′ 𝑣 1.697×105 ×23.1 = 𝐼𝑈𝑈 3.67×105

1.697×105 ×56.56 + 14.1246×105

∴ 𝜎𝐴 = 17.47 N/𝑚𝑚2

1.697×105 ×0 + 14.1246×105 A V

Point C, 𝜎𝐶 =

1.697×105 ×(23.1) 3.67×105

∴ 𝜎𝐶 = 3.88 N/𝑚𝑚2

80 mm

∴ 𝜎𝐵 = −15.47 N/𝑚𝑚2 +

1.697×105 ×(−56.56) 14.1246×105

𝑌1

10 mm Y 𝑥

𝐺2 G

X

U

450

𝑦 𝐺1

U

B

80 mm V Y

X

10 mm

Point B, 𝜎𝐵 =

1.697×105 ×(−33.45) 3.67×105

C 𝑋1

Unsymmetrical bending of beams - Problems Deflection of the beam 𝛿:

𝑘𝑤𝑙3 𝑠𝑖𝑛2 𝜃 𝑐𝑜𝑠 2 𝜃 𝛿= + 2 𝐸 𝐼𝑉𝑉 𝐼𝑈𝑈 2 k=

1 48

for a simply supported beam with a point load at its centre.

W=400N, l=2.4 m, E=2.4× 103 N/𝑚𝑚2 ,𝐼𝑈𝑈 =14.1246 × 106 𝑚𝑚4 , 𝐼𝑉𝑉 = 3.67 × 105 𝑚𝑚4 𝑌

sin 452 3.67 × 105

×

cos 452 + 2 14.1246 × 106

∴ 𝛿 = 1.146 𝑚𝑚

A V

2

80 mm

1 400 × 2.4 × 10 𝛿= × 48 200 × 103

10 mm Y 𝑥

𝐺2 G

X

U

450

𝑦 𝐺1

U

B

80 mm V Y

X

10 mm

1

3 3

C 𝑋1

Unsymmetrical bending of beams - Problems

𝐺1

U

B

80 mm V Y’

10 mm

80 mm

The deflection 𝛿 will be inclined at an angle 𝛽 clock wise with the line GV, given by 𝐼𝑈𝑈 14.1246 × 106 tan 𝛽 = tan 𝜃 = tan 45 = 3.848 5 𝐼𝑉𝑉 3.67 × 10 𝛽 = 75.430 Thus the deflection is at 75.43 - 45=30.430 clockwise with the load line Gy’ 𝑌1 10 mm Position of the neutral axis: A U Y V 𝑥 The neutral axis will be at 90- 30.43 =59.570 Anticlockwise with the load line, because the 𝐺2 Neutral axis is perpendicular to the line of 450 G X X 𝑦 deflection.

C 𝑋1

Unsymmetrical beams - Problems Problem: A beam of rectangular section, 80 mm wide 120 mm deep is subjected to a bending moment of 20 kNm. The trace of the plane of loading is inclined at 450 to the YY axis of the section. Locate the neutral axis of the section and calculate the bending stress induced at each corner of the beam section.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

35

Curved beams Curved beams – Introduction: • The bending equation,

𝑀 𝐼

=

𝑓 𝑦

=

𝐸 𝑅

was derived assuming the beam

to be initially straight. • The simple flexure formula may be used for curved beams for which the radius of curvature is more than five times the depth of the beam. • The simple bending formula, however is not applicable for deeply curved beams where the neutral and centroidal axes do not coincide. To deal with such cases Winkler-Bach theory is used.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

36

Curved beams Stresses in curved bars (Winkler – Bach theory): M A

dA

E

B

M

H D

G C G’

R

dy

y G

F

B’ F’

C’ R’

O’

𝜃′ 𝜃

O

Fig. Bending of a curved bar Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

37

Curved beams Stresses in curved bars (Winkler – Bach theory): M A • Figure shows a bar ABCD initially in E Its unstrained state. H • Let AB’C’D be the strained position of the D bar.

• Let, R= Radius of the curvature of the centroidal axis HG. • y= Distance of the fibre EF from the centroidal layer HG,

B G C R R’

O’ O

C’

G’

M

F B’ F’

𝜃′ 𝜃

dA

dy

y G

• R’=Radius of curvature of HG’ • y’=Distance between EF’ and HG’ after straining. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

38

Curved beams Stresses in curved bars (Winkler – Bach theory): • M=Uniform bending moment applied M A E to the beam (assumed positive when H tending to increase the curvature)

B

D

G C

• 𝜃= Original angle subtended by the centroidal axis HG at its centre of curvature O.

R R’ O’ O

C’

G’

M

F B’ F’

𝜃′ 𝜃

• 𝜃 ′ = Angle subtended by HG’ (after bending) at the centre of curvature O’.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

39

Curved beams Assumptions made in the analysis: 1. Plane sections (transverse sections) remain plane during bending.

M

A E

B

M

H

2. The material obeys Hooke’s law (limit of proportinality is not exceeded) 3. Radial strain is negligible.

D

G C R R’

O’ O

C’

G’

F B’ F’

𝜃′ 𝜃

4. The fibres are free to expand or contract without any constraining effect from the adjacent fibres. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

40

Curved beams • For finding the strain and stress normal to the section, consider the fibre EF at a M distance y from the centroidal axis.

A E

B

M

H

• Let 𝜎 be the stress in the strained layer EF’ under the bending moment M and e is the strain in the same layer.

D C R R’ O’

Strain, 𝑒 = or 𝑒 =

𝐸𝐹 ′ −𝐸𝐹 𝐸𝐹

(𝑅′ +𝑦 ′ )𝜃′ (𝑅+𝑦)𝜃

1+𝑒 =

=

G

(𝑅′ +𝑦 ′ )𝜃′ −(𝑅+𝑦)𝜃 , (𝑅+𝑦)𝜃

O

C’

G’

F B’ F’

𝜃′ 𝜃

−1

(𝑅′ +𝑦 ′ )𝜃′ (𝑅+𝑦)𝜃

--------------(1) Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

41

Curved beams • Also strain in the centroidal layer i.e., when y=0 𝑒0 =

𝑅′ 𝜃′ 𝑅𝜃

1 + 𝑒0 =

M

−1

𝑅′ 𝜃′ 𝑅𝜃

D

(𝑅 ′ +𝑦 ′ )𝜃 ′ 𝑦′ ′ 1 + 1+𝑒 𝑅 + 𝑦′ 𝑅 (𝑅 + 𝑦)𝜃 𝑅′ = = × = 𝑅′ 𝜃 ′ 1 + 𝑒0 𝑅 + 𝑦 𝑅′ 1 + 𝑦 𝑅 𝑅𝜃

• 𝑒 = 1 + 𝑒0

B

H

-------(2)

1+

A E

𝑦′

−1

𝑅′ 𝑦

1+

G C R R’

O’ O

C’

G’

F B’ F’

𝜃′ 𝜃

𝑅

𝑦′

• or, 𝑒 =

𝑦′

𝑦

𝑒0 𝑅′ + 𝑅′ + 𝑒0 − 𝑅 1+

𝑦 𝑅

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

42

Curved beams • According to assumption (3) i.e., radial strain is zero. M A • ∴ 𝑦 = 𝑦′ • 𝑒=

𝑦′ 𝑦 𝑒0 𝑅′ + 𝑅′ + 𝑒0 𝑦 1+𝑅

E

𝑦 −𝑅

H

• Adding and subtracting the • 𝑒=

𝑦 term 𝑒0 , 𝑅

D

G C

we get

𝑦 𝑦 𝑦 − +𝑒0 −𝑒0 𝑅 𝑅 𝑅 𝑦 1+𝑅

R R’

𝑦′ 𝑦′ 𝑦 𝑒0 +𝑒0 + 𝑒0 𝑅′ 𝑅′𝑅′

𝑦

=

B

1

1

1

1

𝑒0 1+𝑅 +𝑦 𝑅′−𝑅 + 𝑒0 𝑦 𝑅′−𝑅 𝑦 1+𝑅

𝑒 = 𝑒0 +

1 1 − 𝑅′ 𝑅 𝑦 1+ 𝑅

1+𝑒0

𝑦

O’ O

C’

G’

F B’ F’

𝜃′ 𝜃

-------------(3) Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

43

Curved beams • From Figure, it is obvious that for the given bending moment the M layers above the centroidal layer are A in tension and layers below the centroidal E B M H layer are in compression. F D

1 1 1 + 𝑒0 − 𝑦 𝑅 𝑅′ 𝜎 = 𝐸𝑒 = 𝐸 𝑒0 + 𝑦 1+ 𝑅

C R R’

− −(4)

(Where, E=Young’s modulus of the material)

Total force on the section, 𝐹 =

G

𝜎 𝑑𝐴.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

O’ O

C’

G’

B’ F’

𝜃′ 𝜃

44

Curved beams Consider a small strip of elementary area dA, at a distance y from the M centroidal layer HG, we have A E

𝐹=𝐸 =𝐸

𝑒0 dA + E

1 1 − 𝑦 𝑅′ 𝑅 dA 𝑦 1+ 𝑅 1 1 𝑦 − dA 𝑦 ′ 𝑅 𝑅 1+ 𝑅

1 + 𝑒0

𝑒0 dA + E 1 + 𝑒0

H D

= 𝐸𝑒0 A + E 1 + 𝑒0

1 − 𝑅

𝑦

𝑦

1+𝑅

dA ----(5)

G C R R’

O’ 1 𝑅′

B

O

C’

G’

F B’ F’

𝜃′ 𝜃

Where, A=Area of cross section of the bar,

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

45

Curved beams The total resisting moment is given by, 𝑀=

M

𝜎𝑦 𝑑𝐴

A E

B

H

D

1 1 2 1 + 𝑒0 − 𝑦 𝑑𝐴 𝑅 𝑅′ = 𝐸 𝑒0 𝑦 𝑑𝐴 + 𝐸 𝑦 1+ 𝑅 1 1 𝑦 2 𝑑𝐴 O’ 𝜃′ = 𝐸𝑒0 × 0 + 𝐸 1 + 𝑒0 − 𝑦 𝜃 𝑅′ 𝑅 1+ 𝑅 O (since

G C R R’

C’

G’

F B’ F’

𝑦 𝑑𝐴 = 0)

𝑀 = 𝐸 1 + 𝑒0

1 𝑅′

1 − 𝑅

𝑦 2 𝑑𝐴 𝑦

1+𝑅 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

46

Curved beams 𝑦 2 𝑑𝐴

𝐿𝑒𝑡,

𝑦 1+𝑅

= Aℎ

2

M

-------------(6)

Where ℎ2 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟

𝑀 = 𝐸 1 + 𝑒0 𝑁𝑜𝑤, =

1 − 𝑅

B

H

D

G C

𝐴ℎ2 ----------(7)

𝑅𝑦 dA 𝑅+𝑦

𝑦 𝑑𝐴 = 1+ 𝑅 𝑦2 𝑦− dA = 𝑅+𝑦

1 =0− 𝑅 ∴

𝑦

1 𝑅′

A E

𝑦𝑑𝐴 −

R R’ O’

𝑦 2 𝑑𝐴 𝑅+𝑦

O

C’

G’

F B’ F’

𝜃′ 𝜃

𝑦2

𝑦 dA 1+ 𝑅 𝑦 1 𝑦 𝑑𝐴 = − 𝑅 1+ 𝑅

𝑦2

1 2 − Aℎ −−−−− −(8) 𝑦 dA = Dr.P.Venkateswara rao, Associate 1 + Dept. of Civil Engg.,𝑅SVCEProfessor, 𝑅

M

47

Curved beams M

𝑦

1 2 𝑦 𝑑𝐴 = − 𝑅 Aℎ −−−−− −(8) 1+ 𝑅 𝐹 = 𝐸𝑒0 A + E 1 + 𝑒0

1 𝑅′

1 − 𝑅

𝑦

𝑦 1+ 𝑅

dA

A E

B

H

---(5)D

G C R R’

Hence equation (5) becomes 𝐹 = 𝐸𝑒0 A − E 1 + 𝑒0

1 𝑅′

1 𝐴ℎ2 − 𝑅 𝑅

−− −-(9)

Since transverse plane section remain plane during bending , ∴ 𝐹 = 0. or 0 = 𝐸𝑒0 A − E 1 + 𝑒0 1 𝑅′

or, 𝐸𝑒0 A = E 1 + 𝑒0

or, 𝑒0 = 1 + 𝑒0

1 𝑅′

−

1 𝑅′

−

O’ O

C’

G’

F B’ F’

𝜃′ 𝜃

1 𝐴ℎ2 𝑅 𝑅

1 𝐴ℎ2 − 𝑅 𝑅

1 ℎ2 rao, Associate Professor, 𝑅Dr.P.Venkateswara 𝑅 Dept. of Civil Engg., SVCE

M

48

Curved beams or,

𝑅 𝑒0 2 ℎ

1 𝑅′

= 1 + 𝑒0

1 − 𝑅

---------(10).

Substituting the value of 1 + 𝑒0

1 𝑅′

−

1 𝑅

Equation (7), we get 𝑅 𝑀 = 𝐸 𝑒0 2 ℎ 𝑀 or, 𝑒0 = 𝐸𝐴𝑅

×

𝐴ℎ2

M

in

A E

B

H

D

G C

= 𝑒0 EAR

−−− − 11 1 1 1 + 𝑒0 − 𝑦 𝑅 𝑅′ 𝜎 = 𝐸 𝑒0 + 𝑦 1+ 𝑅

R R’ O’

− −(4)

O

F B’ F’

𝜃′ 𝜃

Substituting the value of 𝑒0 in equation (4), we get 𝑀 𝑦 𝑅 𝜎=𝐸 + 𝑒0 2 𝑦 𝐸𝐴𝑅 1 + ℎ 𝑅 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

C’

G’

M

49

Curved beams M 𝑀 𝑦 𝑅 A 𝜎= +𝐸 𝑒0 2 𝑦 E 𝐴𝑅 ℎ 1+ 𝑅 H 𝑀 𝑦 𝑅 D = +𝐸 𝑒 0 𝑦 ℎ2 𝐴𝑅 1+ 𝑅 𝑀 𝑦 𝑀 𝑅 = +𝐸 × × 2 𝑦 𝐴𝑅 𝐸𝐴𝑅 ℎ 1+ 𝜃′ 𝑅 O’ 𝑀 𝑀 𝑅𝑦 1 𝜃 = + ` × O 𝐴𝑅 𝐴𝑅 1 + 𝑦 ℎ2 𝑅 𝑀 𝑅2 𝑦 𝜎= 1+ 2 × −−−− − 12 Tensile 𝐴𝑅 ℎ 𝑅+𝑦

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

B

G C R R’

C’

G’

M

F B’ F’

50

Curved beams On the other side of HG, y will be negative, and stress will be compressive 𝑀 𝑅2 𝑦 𝜎= 1− 2 × −−− − 13 𝐴𝑅 ℎ 𝑅−𝑦

M

A E

B

H

D

G C R R’

O’

C’

G’

F B’ F’

𝜃′

𝜃 When the bending moment is applied in such O a manner that it tends to decrease the curvature, then the equation (13), tensile.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

51

Curved beams M

Position of neutral axis: At the neutral axis, 𝜎 = 0 𝑀 𝑅2 𝑦 1+ 2 × =0 𝐴𝑅 ℎ 𝑅+𝑦 𝑅2 𝑦 × = −1 2 ℎ 𝑅+𝑦 𝑅2 y

−ℎ2

−Rℎ2

A E

B

H

D

G C R R’

O’

− ℎ2 y

C’

G’

F B’ F’

𝜃′ 𝜃

O = 𝑅+𝑦 = 𝑦 𝑅2 + ℎ2 = −Rℎ2 𝑹𝒉𝟐 𝒚=− 𝟐 𝑹 + 𝒉𝟐 Hence neutral axis is located below the centroidal axis. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

52

Curved beams M

Values of ℎ2 for various sections We know, ℎ2 =

𝑅 = 𝐴 𝑅 = 𝐴

𝑦2 𝑅+𝑦

1 𝐴

𝑦2

𝑦 1+𝑅

B

H

dA

D

G C R R’

dA 2

𝑦𝑑𝐴 −

𝑅 = 0 − 𝑅𝐴 + 𝐴 𝟑 𝑹 ∴ 𝒉𝟐 = 𝑨

A E

𝑅𝑑𝐴 +

𝑅 𝑑𝐴 𝑅+𝑦

O’ O

C’

G’

F B’ F’

𝜃′ 𝜃

𝑅2 𝑑𝐴 𝑅+𝑦

𝒅𝑨 − 𝑹𝟐 𝑹+𝒚 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

M

53

Curved beams Value of ℎ2 for Rectangular section: Figure shows the rectangular section with centre of curvature O lying on YY-axis and XX-axis is D the centroidal bending axis. X Consider an elementary strip of width B and depth dy at a distance y from the centroidal Layer. Area of the strip dA=Bdy Area of the section, A=BD 𝟑 𝑹 𝒅𝑨 𝟐 𝑊𝑒 𝑘𝑛𝑜𝑤, 𝒉 = − 𝑹𝟐 𝑨 𝑹+𝒚 3 𝑅 ∴ ℎ2 = 𝐵𝐷

B

Y

dA

dy

y G

X

Y

R

O

𝐷/2

𝐵𝑑𝑦 − 𝑅2 −𝐷/2 𝑅 + 𝑦

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

54

Curved beams 𝐷/2 3 𝑅 𝐵𝑑𝑦 2 ∴ℎ = − 𝑅2 𝐵𝐷 −𝐷/2 𝑅 + 𝑦 𝑅3 𝐷/2 = log 𝑒 𝑅 + 𝑦 − 𝑅2 −𝐷/2 𝐷 𝟑 𝑹 𝟐𝑹 + 𝑫 𝟐 ∴𝒉 = 𝐥𝐨𝐠 𝒆 − 𝑹𝟐 𝑫 𝟐𝑹 − 𝑫

B

Y

dA

dy

y

D X

G

X

Y

R

O

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

55

Curved beams - Problems Problem: Figure shows a frame subjected to a load of 2.4 kN. 18

120

2.4 kN

48 1

2

48

Dimensions in mm

2.4 kN

Find (i) The resultant stresses at points 1 and 2. (ii) Position of the neutral axis. Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

56

Curved beams - Problems Solution: Area of section at 1-2, 𝐴 = 48 × 18 × 10−6 = 8.64 × 10−4 𝑚2

18

120

2.4 kN

48 1

Bending moment, 𝑀 = −2.4 × 103 × 120 + 48 × 10−3 = −403.2 𝑁𝑚.

2

48

Dimensions in mm 2.4 kN

M is taken as –ve because it tends to decrease the curvature.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

57

Curved beams - Problems (i) Resultant stresses at points 1 and 2:

Direct stress, 𝜎𝑑 =

2.4×103 8.64×10−4

×

18

2.4 kN

48

10−6

= 2.77 MN/𝑚2 (tensile)

120

1

3 𝑅 2𝑅 + 𝐷 2 ℎ = log 𝑒 − 𝑅2 𝐷 2𝑅 − 𝐷 Here, 𝑅 = 48 𝑚𝑚 = 0.048 𝑚, 𝐷 = 48 𝑚𝑚 = 0.048 𝑚

2

48

Dimensions in mm 2.4 kN

3 0.048 2 × 0.048 + 0.048 2 ℎ = log 𝑒 − 0.0482 0.048 2 × 0.048 − 0.048

= 0.0482 log 𝑒 3 − 1 = 2.27 × 10−4 𝑚2 Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

58

Curved beams - Problems 18

120

2.4 kN

48 1

Bending stress due to M at point 2, 𝑀 𝑅2 𝑦 𝜎𝑏2 = 1− 2 𝐴𝑅 ℎ 𝑅−𝑦

2

48

Dimensions in mm 2.4 kN

−403.2 0.0482 0.024 = 1− 8.64 × 10−4 × 0.048 2.27 × 10−4 0.048 − 0.024

× 10−6 MN/𝑚2

∴ 𝜎𝑏2 = −9.722 1 − 10.149 = +88.95 MN/𝑚2 ( tensile) Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

59

Curved beams - Problems 18

120

2.4 kN

48 1

Bending stress due to M at point 1, 𝑀 𝑅2 𝑦 𝜎𝑏1 = 1+ 2 𝐴𝑅 ℎ 𝑅+𝑦

2

48

Dimensions in mm 2.4 kN

−403.2 0.0482 0.024 = 1+ 8.64 × 10−4 × 0.048 2.27 × 10−4 0.048 + 0.024

× 10−6 MN/𝑚2

= −42.61 𝑀𝑁/𝑚2 = 42.61 MN/𝑚2 (Comp.)

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

60

Curved beams - Problems 18

120

2.4 kN

48

Resultant stress at point 2, 𝜎2 = 𝜎𝑑 + 𝜎𝑏2 = 2.77 + 88.95 = 91.72 MN/𝑚2 (tensile).

1

2

48

Dimensions in mm 2.4 kN

Resultant stress at point 1, 𝜎1 = 𝜎𝑑 + 𝜎𝑏1 = 2.77 − 42.61 = 39.84 MN/𝑚2 (Comp.)

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

61

Curved beams - Problems 18

120

2.4 kN

48 1

Position of neutral axis:

We know, 𝑦 = =−

𝑅ℎ2 − 2 2 𝑅 +ℎ

0.048×2.27×10−4 0.0482 +2.27×10−4

2

48

Dimensions in mm 2.4 kN

= −0.00435 𝑚 = −4.35 𝑚𝑚 Hence, neutral axis is at a radius of 4.35 mm below the centroidal axis.

Dr.P.Venkateswara rao, Associate Professor, Dept. of Civil Engg., SVCE

62

Curved beams 𝒉𝟐 for circular section: Figure shows the circular section of diameter d of a curved bar of radius of curvature of R from the centre of curvature O upto the centroid G of the section. Area of cross section,

b dy Y G

𝜋 2 A= 𝑑 4

Consider a strip of width b and a depth dy at a distance y from the centroidal layer as shown.

𝑏=2

𝑑 2

R

2

− 𝑦2

O

Curved beams 𝒉𝟐 for circular section (contd…):

Area of strip, 𝑑𝐴 = 𝑏. 𝑑𝑦 = 2 3 𝑅 ℎ2 = 𝐴

𝑅3 = 𝐴

𝑑𝐴 − 𝑅2 𝑅+𝑦

+𝑑/2 2

−𝑑/2

8𝑅3 = 𝜋𝑑2

𝑑 2 2

+𝑑/2

−𝑑/2

𝑑 2

2

b

− 𝑦 2 . dy

dy Y G

− 𝑦2

𝑅+𝑦

dy − 𝑅2

𝑑2 2 − 𝑦2 4 dy − 𝑅2 𝑅+𝑦

R

O

Curved beams 𝒉𝟐 for circular section (contd…): 𝑑2 +𝑑/2 2 3 2 − 𝑦 8𝑅 4 2 = dy − 𝑅 𝜋𝑑2 𝑅+𝑦

b dy Y

−𝑑/2

Equating the integral by binomial expression and then integrating, we get 𝟐 𝟒 𝒅 𝟏 𝒅 ∴ 𝒉𝟐 = + +⋯ 𝟐 𝟏𝟔 𝟏𝟐𝟖 𝑹

G

R

O

Curved beams 𝒉𝟐 for triangular section: Let 𝑅 + 𝑦 = 𝑎 𝑑𝑦 = 𝑑𝑎 Width of elementary strip, 𝑅2 − 𝑎 𝑏 𝑏′ = 𝑑 Area of elementary strip, 𝑑𝐴 = 𝑏 ′ 𝑑𝑦 = b′ . da

Now, ℎ2

=

𝑅3 𝐴

𝑅2

𝑅1

=

𝑅3 𝐴

𝑅2

𝑅1

𝑏

G

b

𝑅2 𝑅1

R

𝑑𝐴 − 𝑅2 𝑅+𝑦

𝑏 ′ 𝑑𝑎 − 𝑅2 𝑎

dy d Y

′

O d b 𝑅2 - a ? 𝑅2 − 𝑎 𝑏 𝑏 = 𝑑 ′

Curved beams 𝒉𝟐 for triangular section (contd…): ℎ2

=

𝑅3 𝐴

𝑅2

𝑅1

𝑅2 − 𝑎 𝑏 𝑑𝑎 − 𝑅2 𝑑 𝑎

𝑏

𝑅3 𝑅2 𝑏 𝑅2 𝑏 = log 𝑒 − 𝑅2 − 𝑅1 𝐴 𝑑 𝑅1 𝑑 Since, 𝑅2 = R

2 + d 3

; 𝑅1 = R −

𝑑 3

dy d Y

′

G

− 𝑅2

b

𝑅2 𝑅1

R

and 𝑅2 − 𝑅1 = d 3 𝑅 ℎ2 = 𝐴

2 2 R+ d 𝑏 R + 3d 3 log 𝑒 − 𝑏 − 𝑅2 𝑑 𝑑 R− 3

O

Curved beams 𝒉𝟐 for triangular section (contd…): 2 2 3 R + d 𝑏 R + 3d 𝑅 3 2 ℎ = log 𝑒 − 𝑏 − 𝑅2 𝑑 𝐴 𝑑 R− 3

G

b

𝑅2

2𝑅3 = 𝑏𝑑

3R + 2𝑑 𝑏 3R + 2𝑑 log 𝑒 − 𝑏 − 𝑅2 3𝑑 3R − 𝑑

𝟑 𝟐𝑹 ∴ 𝒉𝟐 = 𝒅

dy d Y

𝑏′

𝑅1

𝟑𝐑 + 𝟐𝒅 𝟑𝐑 + 𝟐𝒅 𝒍𝒐𝒈𝒆 − 𝟏 − 𝑹𝟐 𝟑𝒅 𝟑𝐑 − 𝒅

R

O

Curved beams 𝒉𝟐 for trapezoidal section : 𝑅2 − 𝑎 Consider an elementary strip of width b’ and depth dy at a distance y from the centroidal axis.

b 𝑑2 𝑑1

𝑏′

𝐵−𝑏 =b+ 𝑑1 + 𝑑2

𝐵−𝑏 2

𝑅2 − 𝑎 𝑑

G

B

𝑅2

Let 𝑅 + 𝑦 = 𝑎 𝑑𝑦 = 𝑑𝑎

𝑏

dy d Y

′

𝑅1

R

𝑅2 − 𝑎 O

Area of the strip, 𝑑𝐴 = 𝑏 ′ dy = b +

𝐵−𝑏 𝑑1 +𝑑2

𝑅2 − 𝑎 da

Curved beams 𝒉𝟐 for trapezoidal section (contd…): ℎ2

3

𝑅 = 𝐴

𝑅3 = 𝐴 =

=

𝑅3 𝐴 𝑅3 𝐴

𝑅2

𝑅1 𝑅1

𝑅1 𝑅1

𝑅1

b

𝑑𝐴 − 𝑅2 𝑅+𝑦

𝐵−𝑏 𝑏+ 𝑑1 + 𝑑2

𝑑𝑎 𝐵−𝑏 𝑏 + 𝑎 𝑑1 + 𝑑2

𝑏 log 𝑒

𝑑2

𝑏

𝑑1

𝑅2 𝑑𝑎 𝑅2 − 𝑎 − 𝑅2 𝑎 𝑅2

𝑅1

𝑅2 𝐵−𝑏 + 𝑅1 𝑑1 + 𝑑2

𝑅2 − 𝑎 𝑑𝑎 − 𝑅2 𝑎

𝑅2 log 𝑒

dy Y d

′

G

B 𝑅1

𝑅2 − 𝑅2 − 𝑅1 𝑅1

R

O

− 𝑅2

Curved beams 𝒉𝟐 for trapezoidal section (contd…): 3 𝑅 ℎ2 = 𝐴

𝑅 + 𝑑2 + 𝑅 − 𝑑1 − 𝑅2 𝑅 + 𝑑2 𝑅 + 𝑑2 log 𝑒 − 𝑅 − 𝑑1 𝐵−𝑏

b

𝑏 log 𝑒 𝐵−𝑏 𝑑

Where, 𝐵+𝑏 𝐴= d 2 𝑑 𝐵 + 2𝑏 𝑑1 = 3 𝐵+𝑏 𝑑2 = d − 𝑑1

𝑑2

𝑏

𝑑1

𝑅2

dy Y d

′

G

B 𝑅1

R

O

Curved beams - Problems Problem: The curved member shown in Figure has a solid circular cross-section 0.10 m in diameter. If the maximum tensile and compressive stresses in the member are not to exceed 150 Mpa and 200 Mpa respectively, determine the value of load P that can safely be carried by the member. 0.15 m

P

0.05 m

P

Curved beams - Problems Solution: Given d=0.10 m, R=0.10 m, 𝜎1 = 150 MPa = 150 MN/𝑚2 (tensile) 𝜎2 = 200 MPa = 200MN/𝑚2 (Comp.) Refer to Figure, Area of cross section, 𝜋 2 𝜋 𝐴 = 𝑑 = × 0.102 = 7.854 × 10−3 𝑚2 4 4 Bending moment, M=P(0.15+0.10)=0.25P 2 4 𝑑 1 𝑑 ℎ2 = + 16 128 𝑅2 =

0.102 16

1 0.104 + 128 0.102

= 7.031 × 10−4 𝑚2

0.15 m

P

0.05 m

P

Curved beams - Problems Solution: Direct stress, 𝜎𝑑 =

0.15 m 𝑃 𝐴

P

(comp.)

Bending stress at point1 due to M, 0.05 m 0.10 m 2 1 𝑀 𝑅2 𝑦 𝜎𝑏1 = 1+ 2 tensile 𝐴𝑅 ℎ 𝑅+𝑦 Total stress at point 1, P 𝜎1 = 𝜎𝑑 + 𝜎𝑏1 𝑃 𝑀 𝑅2 𝑦 150 = − + 1+ 2 tensile 𝐴 𝐴𝑅 ℎ 𝑅+𝑦 𝑃 150 = − 7.854 × 10−3 0.25 𝑃 0.102 0.05 + 1+ −3 7.854 × 10 × 0.10 7.031 × 10−4 0.10 + 0.05 150 = −127.32𝑃 + 318.31𝑃 × 5.74 = 1699.78𝑃

Curved beams - Problems 150 150 × 103 ∴𝑃= 𝑀𝑁 = kN = 88.25 kN − (i) 1699.78 1699.78 0.15 m Bending stress at point 2 due to M, 𝑀 𝑅2 𝑦 𝜎𝑏2 = 1− 2 0.05 m 𝐴𝑅 ℎ 𝑅−𝑦 0.10 m 2 1 Total stress at point 2, 𝜎2 = 𝜎𝑑 + 𝜎𝑏2 𝑃 𝑀 𝑅2 𝑦 200 = − + 1− 2 𝐴 𝐴𝑅 ℎ 𝑅−𝑦 𝑃 200 = − 7.854 × 10−3 0.25 𝑃 0.102 0.05 + 1− −3 7.854 × 10 × 0.10 7.031 × 10−4 0.10 − 0.05 200 = −127.32𝑃 − 318.31𝑃 × 13.22 = −4335.38𝑃

P

P

Curved beams - Problems 200 200 × 103 𝑃=− 𝑀𝑁 = − kN = −46.13 kN 4335.38 4335.38 0.15 m ∴ 𝑃 = 46.13 𝑘𝑁 𝐶𝑜𝑚𝑝. − 𝑖𝑖 Comparing (i) and (ii) values, the safe load 0.05 m P will be lessor of these. 0.10 m 2 1 Hence, P= 46.13 kN.

P

P

Curved beams - Problems Problem: Figure shows a circular ring of rectangular section, with a slit and subjected to load P. (i) Calculate the magnitude of the force P if the maximum stress along the section 1-2 is not to 9 cm P 6 cm exceed 225 MN/𝑚2 . G (ii) Draw the stress distribution along 1-2. 1

2 G

Curved beams - Problems Solution: Area of section at 1-2, 𝐴 = 9 × 6 = 54𝑐𝑚2 = 0.0054 𝑚2 Permissible stress, 𝜎 = 225 𝑀𝑁/𝑚2 Bending moment, 𝑀 = 𝑃 × 16.5 × 10−2 = 0.165P Nm 9 cm 6 cm G

 M is taken as +ve because it tends to increase the curvature. (i) Magnitude of the force P: Direct stress, 𝜎𝑑 =

𝑃 − 𝐴

=

𝑃 − 0.0054

= 185.18 P N/𝑚2 (comp.)

1

G

2

P

Curved beams - Problems 3 𝑅 2𝑅 + 𝐷 2 ℎ = log 𝑒 − 𝑅2 𝐷 2𝑅 − 𝐷

9 cm 6 cm G

Here, 𝑅 = 16.5 × 10−2 m = 0.165 m 𝐷 = 9 𝑐𝑚 = 0.09 𝑚 1

3 0.165 2 × 0.165 + 0.09 2 ℎ = log 𝑒 − 0.1652 0.09 2 × 0.165 − 0.09

G

P

2

= 0.0499 × 0.5596 − 0.1652 = 6.99 × 10−4 Resultant stress at point 2,

𝜎𝑚𝑎𝑥 = 225×

𝑃 − 𝐴

106

+

𝑀 𝐴𝑅

1

𝑅2 𝑦 − 2 ℎ 𝑅−𝑦

= −185.18P +

0.165𝑃 0.0054×0.165

1

0.1652 0.045 − 6.99×10−4 0.165−0.045

𝑃 = −83189 𝑁 𝑜𝑟 𝑃 = 83.189 𝑘𝑁(𝑐𝑜𝑚𝑝. )

Curved beams - Problems 9 cm 6 cm G

(ii) Stress distribution along the section 1-2:

Stress distribution along G1: Resultant stress at point 1,

𝜎𝑚𝑎𝑥 =

𝑃 − 𝐴

+

83.189 − 0.0054

𝑀 𝐴𝑅

1

1 G

2

𝑅2 𝑦 + 2 ℎ 𝑅+𝑦

𝜎1 = 0.165 × 83.189 0.1652 0.045 + 1+ 0.0054 × 0.165 6.99 × 10−4 0.165 + 0.045 = 128.6 𝑀𝑁/𝑚2

P

Curved beams - Problems 9 cm 6 cm G

(ii) Stress distribution along the section 1-2: 128.6 MN/𝑚2

Hyperbolic

0

1 G

225 MN/𝑚2

2

P

Curved beams - Problems Problem: Figure shows a crane hook lifting a load of 150 kN. Determine the maximum compressive and tensile stresses in the critical section of the crane hook. d=180 𝑑1

𝑑2

b=45 A

A

G

B

Dimensions in mm

150 kN

B

B=135

Curved beams - Problems Solution: B=135 mm=0.135 m; b=45 mm=0.045 m; d=180 mm; P=150 kN Area of section, 𝐴 =

d=180 𝑑1 𝑑2 b=45 A

0.135+0.45 2

G

B

B=135

× 0.18 = 0.0162 𝑚2

𝑑 𝐵 + 2𝑏 0.18 0.135 + 2 × 0.045 𝑑1 = = = 0.075 m 3 𝐵+𝑏 3 0.135 + 0.045 A

B

𝑑2 = 0.18 − 0.075 = 0.105 m R=0.105+0.075=0.18 m

150 kN Dimensions in mm

Curved beams - Problems 3 𝑅 ℎ2 = 𝐴

𝑅 + 𝑑2 + 𝑅 − 𝑑1 − 𝑅2 𝑅 + 𝑑2 𝑅 + 𝑑2 log 𝑒 − 𝑅 − 𝑑1 b=45 A 𝐵−𝑏

d=180 𝑑1 𝑑2

𝑏 log 𝑒 𝐵−𝑏 𝑑

G

B

B=135

By substituting the values , ℎ2 = 0.00256 𝑚2 Bending moment , 𝑀 = −150 × 103 × 0.18 ∴ 𝑀 = −27000Nm A

Direct stress, 𝜎𝑑 = =

𝑃 𝐴

=

150×103 0.0162

B

× 10−6

−9.26 MN/𝑚2 (tensile)

150 kN Dimensions in mm

Curved beams - Problems Bending stress calculations, Bending stress at A, 𝜎𝑏𝐴

𝑀 𝑅2 𝑦 = 1+ 2 𝐴𝑅 ℎ 𝑅+𝑦

d=180 𝑑1 𝑑2

= −52.43 MN/𝑚2 b=45 A

G

𝜎𝑏𝐴 = 52.43 MN/𝑚2 (comp.) Bending stress at B, 𝑀 𝑅2 𝑦 𝜎𝑏𝐵 = 1− 2 𝐴𝑅 ℎ 𝑅−𝑦

B

B=135

= 74.45 MN/𝑚2

𝜎𝑏𝐵 = 74.45 MN/𝑚2 (Tensile.)

A

B

150 kN Dimensions in mm

Curved beams - Problems d=180 𝑑1 𝑑2

Stress at A,

𝜎𝐴 = 𝜎𝑑 + 𝜎𝑏𝐴 = 9.26 − 52.43 = −43.17 MN/𝑚2

∴ 𝜎𝐴 = 43.17 MN/𝑚2 (Comp.)

b=45 A

G

B

B=135

Stress at B, 𝜎𝐵 = 𝜎𝑑 + 𝜎𝑏𝐵 = 9.26 + 74.45 = 83.71 MN/𝑚2 ∴ 𝜎𝐵 = 83.71 MN/𝑚2 (Tensile) A

B

150 kN Dimensions in mm

Curved beams Stresses in a ring: Figure shows a circular ring under the action of an axial pull P. P (A) Stresses on a section taken along the line of A action of P, (i) At outside ring, i.e., at point A, 𝑃 𝑅2 𝑅2 𝑦2 𝜎𝐴 = 1+ 2 𝜋𝐴 𝑅2 + ℎ2 ℎ 𝑅 + 𝑦2

(ii) At inside ring, i.e., at point B, 𝑃 𝑅2 𝑅2 𝑦1 𝜎𝐵 = 1− 2 2 2 𝜋𝐴 𝑅 + ℎ ℎ 𝑅 − 𝑦1

(tensile)

B D

(Comp. )

P

C

Curved beams Stresses in a ring (contd…): P (B) Stresses on a section perpendicular to the line of action of P, (i) At outside ring, i.e., at point C, 𝑃 𝑅2 𝑅2 2𝑅2 𝜎𝐶 = + 2 −1 𝐴 𝜋 𝑅 2 + ℎ2 2ℎ 𝜋 𝑅2 + ℎ2 𝑃 + (Comp. ) 2𝐴

𝑦2 𝑅 + 𝑦2

A

B D

(ii) At inside ring, i.e., at point D,

𝑃 𝑅2 𝑅2 2𝑅2 𝜎𝐷 = + 2 −1 𝐴 𝜋 𝑅 2 + ℎ2 2ℎ 𝜋 𝑅2 + ℎ2

−𝑦1 𝑅 − 𝑦1

+

𝑃 (Ten. ) 2𝐴

P

C

Curved beams - Problems Problem: A ring is made of round steel bar 30 mm diameter and the mean radius of ring is 180 mm. Calculate the maximum tensile and compressive stresses in the material of the ring if it is subjected to a pull of 12 kN.

Curved beams - Problems Solution: d=30 mm=0.03 m 𝜋 𝐴 = × 0.032 = 7.068 × 10−4 𝑚2 4 30 mm

P=12 kN

12 kN A

R=180 mm=0.18 m B

2 4 𝑑 1 𝑑 ℎ2 = + 16 128 𝑅2

=

0.032 16

+

1 128

×

0.034 0.182

D

= 5.64 × 10−5 𝑚2 12 kN

C

Curved beams - Problems Solution: Stresses: 𝑃 𝑅2 𝑅2 𝑦2 𝜎𝐴 = 1+ 2 2 2 𝜋𝐴 𝑅 + ℎ ℎ 𝑅 + 𝑦2 103

0.182

(tensile)

12 × 𝜎𝐴 = 𝜋 × 7.068 × 10−4 0.182 + 5.64 × 10−5 0.182 0.015 −6 1+ × 10 5.64 × 10−5 0.18 + 0.015

∴ 𝜎𝐴 = 243.7 MN/𝑚2 (tensile)

30 mm

12 kN A

B D

12 kN

C

Curved beams - Problems Stresses: 𝑃 𝑅2 𝑅2 𝑦1 𝜎𝐵 = 1− 2 2 2 𝜋𝐴 𝑅 + ℎ ℎ 𝑅 − 𝑦1 103

0.182

12 × 𝜎𝐵 = 𝜋 × 7.068 × 10−4 0.182 + 5.64 × 10−5 0.182 0.015 −6 1− × 10 5.64 × 10−5 0.18 − 0.015

∴ 𝜎𝐵 = −276.3 = 276.3 MN/𝑚2 (Comp. )

30 mm

12 kN A

B D

12 kN

C

Curved beams - Problems Stresses:

𝑃 𝑅2 𝑅2 2𝑅2 𝑦2 𝑃 𝜎𝐶 = + 2 −1 + (Comp. ) 𝐴 𝜋 𝑅 2 + ℎ2 2ℎ 𝜋 𝑅2 + ℎ2 𝑅 + 𝑦2 2𝐴 12 × 103 × 10−6 0.182 𝜎𝐶 = 7.068 × 10−4 𝜋 0.182 + 5.64 × 10−5 0.182 2 × 0.182 0.015 + −1 −5 2 −5 2 × 5.64 × 10 𝜋 0.18 + 5.64 × 10 0.18 + 0.015 12 kN 12 × 103 × 10−6 + (Comp. ) A −4 2 × 7.068 × 10 30 mm

𝜎𝑐 = −131.3 + 8.49 = −122.8 MN/𝑚2

B

D

∴ 𝜎𝑐 = 122.8 MN/𝑚2 (Comp.) 12 kN

C

Curved beams - Problems Stresses:

𝑃 𝑅2 𝑅2 2𝑅2 𝑦1 𝑃 𝜎𝐷 = − 2 −1 + (Tensile) 𝐴 𝜋 𝑅 2 + ℎ2 2ℎ 𝜋 𝑅2 + ℎ2 𝑅 − 𝑦1 2𝐴 12 × 103 × 10−6 0.182 𝜎𝐷 = 7.068 × 10−4 𝜋 0.182 + 5.64 × 10−5 0.182 2 × 0.182 0.015 − −1 −5 2 −5 2 × 5.64 × 10 𝜋 0.18 + 5.64 × 10 0.18 − 0.015 12 kN 12 × 103 × 10−6 + A 2 × 7.068 × 10−4 30 mm

𝜎𝐷 = 166.99 + 8.49 =

175.48 MN/𝑚2

(tensile)

∴Maximum tensile stress = 243.7 MN/𝑚2 Maximum compressive stress=276.3 MN/𝑚2

B

D

12 kN

C

Curved beams - Problems Problem: A steel ring has a rectangular cross-section, 75 mm in the radial direction and 45 mm perpendicular to the radial direction. If the mean radius of the ring is 150 mm and maximum tensile stress is limited to 180 MN/m2 . Calculate the tensile load the ring can carry.

Curved beams - Problems Solution: A steel ring has a rectangular cross-section, 75 mm in the radial direction and 45 mm perpendicular to the radial direction. If the mean radius of the ring is 150 mm 75 P A and maximum tensile 45 stress is limited to 180 MN/m2 .

Calculate the tensile load the ring can carry.

All dimensions in mm

P

Curved beams - Problems Area of cross section, 𝐴 = 0.075 × 0.045 = 0.003375 𝑚2 3 𝑅 2𝑅 + 𝐷 2 ℎ = log 𝑒 − 𝑅2 𝐷 2𝑅 − 𝐷

75

P

45

3 0.15 2 × 0.15 + 0.075 2 ℎ = log 𝑒 − 0.152 0.075 2 × 0.15 − 0.075

P ∴ ℎ2 = 4.87 × 10−4 𝑚2 All dimensions in mm

Curved beams - Problems Load P calculation: The maximum tensile stress will occur at A. ∴ 𝜎𝐴 = 180 × 106 N/𝑚2 𝑃 𝑅2 𝑅2 𝑦2 𝑁𝑜𝑤, 𝜎𝐴 = 1+ 2 2 2 𝜋𝐴 𝑅 + ℎ ℎ 𝑅 + 𝑦2

75

P

45

2 𝑃 0.15 180 × 106 = P 𝜋 × 0.003375 0.152 + 4.87 × 10−4 0.152 0.0375 1+ All dimensions in mm 4.87 × 10−4 0.15 + 0.0375

𝑃 = 19440 𝑁 ∴ 𝑃 = 19.44 𝑘𝑁

Curved beams - Problems Problem: A curved bar is formed of a tube of 120 mm outside diameter and 7.5 mm thickness. The centre line of this beam is a circular arc of radius 225 mm. A bending moment of 3 kNm tending to increase curvature of the bar is applied. Calculate the maximum tensile and compressive stresses setup in the bar.

Curved beams - Problems Solution: 𝑑2 = 120 mm 𝑡 = 7.5 𝑚𝑚 𝑑1 = 120 − 2 × 7.5 = 105 𝑚𝑚

105 mm 120 mm

M

𝜋 𝐴 = 0.122 − 0.1052 = 0.00265 𝑚2 4 𝑀 = 3 𝑘𝑁𝑚 𝜋 𝐴1 = × 0.1052 = 0.00866 𝑚2 4 𝜋 𝐴2 = × 0.122 = 0.01131 𝑚2 4 Bending moment, M=+3 kNm (tending to increase the curvature)

O

M

Curved beams - Problems We know, for circular section, 2 4 𝑑 1 𝑑 ℎ2 = + 16 128 𝑅2

105 mm 120 mm

M

For inner tube, 2 4 𝑑 1 𝑑 1 1 ℎ1 2 = + × 2 16 128 𝑅 =

0.1052 16

1 + 128

×

0.1054 0.2252

= 7.078 × 10−4 𝑚2

For outer circle, 2 4 𝑑 1 𝑑 2 2 2 ℎ2 = + × 2 16 128 𝑅 =

0.122 16

+

1 128

×

0.124 0.2252

= 9.32 × 10−4 𝑚2

O

M

Curved beams - Problems 𝐴ℎ2 = 𝐴2 ℎ2 2 − 𝐴1 ℎ1 2 0.00265ℎ2 = 0.01131 × 9.32 × 10−4 −0.00866 × 7.078 × 10−4 ℎ2 = 0.00166 𝑚2 Maximum tensile stress occurs at A. 𝑀 𝑅2 𝑦 ∴ 𝜎𝐴 = 1+ 2 𝐴𝑅 ℎ 𝑅+𝑦

105 mm 120 mm

M

A

3 × 103 0.2252 0.06 = 1+ 0.00265 × 0.225 0.00166 0.225 + 0.06 𝜎𝐴 = 37.32 MN/𝑚2

B O

M

× 10−6

Curved beams - Problems 105 mm 120 mm

M

Maximum compressive stress occurs at B. 𝑀 𝑅2 𝑦 ∴ 𝜎𝐵 = 1− 2 𝐴𝑅 ℎ 𝑅−𝑦

A

B O

M

3 × 103 0.2252 0.06 = 1− 0.00265 × 0.225 0.00166 0.225 − 0.06 = −50.75 𝜎𝐵 = 50.75 MN/𝑚2 (comp.)

× 10−6

Curved beams 𝒉𝟐 for T-section: Let 𝑅 + 𝑦 = 𝑎 𝑑𝑦 = 𝑑𝑎 𝑅2 𝑅3 3 𝑅 𝑑𝐴 𝑑𝐴 2 ℎ = + − 𝑅2 𝐴 𝑅1 𝑅 + 𝑦 𝑅2 𝑅 + 𝑦 𝑅3 𝑅3 𝑅2 𝑏2 𝑑𝑎 𝑡1 𝑑𝑎 = + − 𝑅2 𝐴 𝑅1 𝑎 𝑎 𝑅2 𝑅3 𝑅2 𝑅3 = 𝑏2 log 𝑒 + 𝑡1 log 𝑒 − 𝑅2 𝐴 𝑅1 𝑅2 Where 𝐴 = 𝑏1 𝑡1 + 𝑏2 𝑡2

𝑡1

𝑏1

dy y G

𝑡2

𝑏2 𝑅2

𝑅3

𝑅1

O

Curved beams 𝑏1

𝒉𝟐 for I-section: Let 𝑅 + 𝑦 = 𝑎 𝑑𝑦 = 𝑑𝑎 𝑅2

3

𝑅 ℎ = 𝐴 2

3

𝑅 = 𝐴

𝑅1 𝑅2

𝑅1

𝑡1 dy

𝑑𝐴 + 𝑅+𝑦

𝑏2 𝑑𝑎 + 𝑎

𝑅3 𝑅2

𝑅3 𝑅2

𝑑𝐴 + 𝑅+𝑦

𝑡3 𝑑𝑎 + 𝑎

𝑅4

𝑅3

𝑅4 𝑅3

𝑑𝐴 − 𝑅2 𝑅+𝑦

𝑏3

𝑅2 𝑅3 𝑅4 = 𝑏2 log 𝑒 + 𝑡3 log 𝑒 +𝑏1 log 𝑒 − 𝑅2 𝐴 𝑅1 𝑅2 𝑅3

𝑡3 G

𝑏1 𝑑𝑎 − 𝑅2 𝑡2 𝑎

𝑅3

y

𝑏2

𝑅2

𝑅4 𝑅3

𝑅1

Where 𝐴 = 𝑏1 𝑡1 + 𝑏2 𝑡2 + 𝑏3 𝑡3 O

Curved beams -Problems 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: A curved beam has a T-section (shown in Figure). The inner radius is 300 mm. What is the 60 𝑚𝑚 Eccentricity of the section?

20 𝑚𝑚

dy y G

20 𝑚𝑚

80 𝑚𝑚 320 𝑚𝑚

380 𝑚𝑚

300 𝑚𝑚

O

Curved beams -Problems 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Area, 𝐴 = 𝑏1 𝑡1 + 𝑏2 𝑡2 = (60 × 20) + (80 × 20) ↑∴ 𝐴 = 2800 𝑚2 dy Centre of gravity of T-section, taking from O, 60 𝑚𝑚 y 80 × 20 × 310 + 60 × 20 × 350 𝑅= 2800 ∴ 𝑅 = 327.14 𝑚𝑚

20 𝑚𝑚

G

20 𝑚𝑚

3 𝑅 𝑅2 𝑅3 2 ℎ = 𝑏2 log 𝑒 + 𝑡1 log 𝑒 − 𝑅2 𝐴 𝑅1 𝑅2 320 𝑚𝑚 320 327.143 80 log 𝑒 300 + = − 327.142 380 2800 20 log 𝑒 320 2 2 ∴ ℎ = 512.08 𝑚𝑚

80 𝑚𝑚 380 𝑚𝑚

300 𝑚𝑚

O

Curved beams -Problems Eccentricty, e=distance of the neutral axis from the centroidal axis. 𝑅ℎ2 ∴𝑒=− 2 𝑅 + ℎ2 327.14 × 512.08 𝑒=− = −1.56 mm 2 327.14 + 512.08 -ve sign indicates that neutral axis is located below the centroidal axis.

20 𝑚𝑚

60 𝑚𝑚

dy y G

20 𝑚𝑚

80 𝑚𝑚 320 𝑚𝑚

380 𝑚𝑚

300 𝑚𝑚

O

Curved beams 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: A central horizontal section of a hook is an I-section with dimensions shown in Figure. The hook carries a load P, the load line passing 60 mm from the inside edge of the section, and the centre of curvature being in the load line. Determine the magnitude of the load P if the maximum stress in the hook is not to exceed the permissible 𝑅2 stress of 108 N/𝑚𝑚2 . What will be the maximum compressive stress in hook for that value of load?

40 𝑚𝑚 20 𝑚𝑚 dy y

30 mm

20 𝑚𝑚

40 𝑚𝑚

G

60 𝑚𝑚 𝑅4 𝑅3

60 𝑚𝑚

O

Curved beams 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Permissible stress, 𝜎 = 105 𝑁/𝑚𝑚2 Load P calculation: Maximum compressive stress:

40 𝑚𝑚 20 𝑚𝑚 dy y

Area of cross section, A=𝑏1 𝑡1 + 𝑏2 𝑡2 + 𝑏3 𝑡3 = 40 × 20 + 60 × 30 + 40 × 20 = 3400 𝑚𝑚2 .

30 mm

20 𝑚𝑚 G

60 𝑚𝑚 𝑅4

𝑅2

𝑅3

60 𝑚𝑚

To find c.g. of the section from ‘O’,

40 𝑚𝑚

O

40 × 20 × 140 + 40 × 20 × 110 + 60 × 30 × 75 𝑅= = 98.5 mm. 3400

Curved beams 40 𝑚𝑚 20 𝑚𝑚 3 𝑅 𝑅 𝑅 𝑅 2 3 4 ℎ2 = 𝑏2 log 𝑒 + 𝑡3 log 𝑒 +𝑏1 log 𝑒 − 𝑅2 𝐴 𝑅1 𝑅2 𝑅3 dy 98.53 90 130 150 20 𝑚𝑚 2 ℎ = 60 log 𝑒 + 20 log 𝑒 + 40 log 𝑒 − 98.52 y 40 𝑚𝑚 3400 60 90 130

ℎ2 = 810.14 𝑚𝑚2 .

30 mm

G

60 𝑚𝑚 𝑃 𝐴

Direct stress, =

𝑃 3400

N/𝑚𝑚2

𝑅4 𝑅2

𝑅3

60 𝑚𝑚

Bending moment, 𝑀 = −𝑃 × 𝑅 = −98.5 𝑃 𝑁𝑚𝑚. O

Curved beams 40 𝑚𝑚

The bending stress at any point is given by 20 𝑚𝑚 𝑀 𝑅2 𝑦 𝜎𝑏 = 1+ 2 dy 𝐴𝑅 ℎ 𝑅+𝑦 20 𝑚𝑚 −98.5𝑃 98.52 𝑦 y = 1+ 3400 × 98.5 810.14 98.5 + 𝑦 G −𝑃 𝑦 30 mm = 1 + 11.97 3400 98.5 + 𝑦 60 𝑚𝑚 Maximum bending stress (tensile) occurs at Y=-38.5 mm 𝑅2 −𝑃 −38.5 60 𝑚𝑚 𝜎𝑏 𝑚𝑎𝑥 = 1 + 11.97 3400 98.5 − 38.5 O 𝑃 2 = N/𝑚𝑚 508.9

40 𝑚𝑚

𝑅4 𝑅3

Curved beams ∴ 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑡 𝑚𝑎𝑥 = 𝜎𝑑 + 𝜎𝑏 𝑚𝑎𝑥 𝑃 𝑃 108 = + 3400 508.9 𝑃 108 = 442.65

40 𝑚𝑚 20 𝑚𝑚 dy y

30 mm

𝑃 = 108 × 442.65 = 47806.2 𝑁 = 47.8 𝑘𝑁

20 𝑚𝑚

40 𝑚𝑚

G

60 𝑚𝑚 𝑅4 𝑅2

𝑅3

60 𝑚𝑚

O

Curved beams ∴ Maximum compressive stress Occurs at y = 𝑦2 = 51.5 𝑚𝑚, 𝜎𝑐 𝑚𝑎𝑥 = 𝜎𝑑 + 𝜎𝑐 𝑚𝑎𝑥 𝜎𝑐 𝑚𝑎𝑥

𝑃 = 3400

𝑃 51.5 − 1 + 11.97 3400 98.5 + 51.5 47806.2 3400 47806.2 51.5 − 1 + 11.97 3400 98.5 + 51.5

40 𝑚𝑚 20 𝑚𝑚 51.5 mm

dy y

38.5 mm

30 mm

20 𝑚𝑚

40 𝑚𝑚

G

60 𝑚𝑚 𝑅4

𝜎𝑐 𝑚𝑎𝑥 =

𝜎𝑐 𝑚𝑎𝑥 = 57.79 N/𝑚𝑚2

𝑅2

𝑅3

60 𝑚𝑚

O

Shear centre Shear Centre definition:  To avoid twisting and cause only bending, the transverse forces must act through a point which may not necessarily coincide with the centroid, but will depend upon the shape of the section. Such a point is known as the ‘Shear Centre’.

 The point of intersection of the bending axis with the cross section of the beam is defined as ‘shear centre’.

Shear centre Shear Centre definition:  The shear center always falls on a cross-sectional axis of symmetry.  If the cross section contains two axes of symmetry, then the shear center is located at their intersection. Notice that this is the only case where shear center and centroid coincide.  Shear centre is the point through which the resultant shear forces acts.  Shear flows from tension side to compression side when looking on section in the direction of increasing bending moment.

Shear centre Shear Centre of thin walled open sections: Examples Channel section: b ′ Shear stress at a distance ‘𝑥 in the bottom horizontal member AB as shown in C 𝐹1 Figure is given by , h 𝑡𝑤 𝐹𝐴𝑦 𝑞1 = 𝑧 G 𝐼𝑡 𝑥 S 𝐹 ℎ e = × 𝑥𝑡 × 𝐼𝑡 2 𝑏 𝐹2 a b𝛿𝑥 𝐹1 = 𝑞1 𝑡 𝑑𝑥 0

𝐹1

𝑏

∴ 𝐹1 = 0

𝐹 ℎ × 𝑥𝑡 × × 𝑡 𝑑𝑥 𝐼𝑡 2

B

T+ 𝛿𝑇

𝑦

𝑡𝑓

T d

𝑥 A u

c

𝑡

Shear centre 𝑏

∴ 𝐹1 = 0

𝐹 ℎ × 𝑥𝑡 × × 𝑡 𝑑𝑥 𝐼𝑡 2

b

𝐹𝑡ℎ 𝑏2 𝐹𝑡ℎ𝑏 2 = × = 2𝐼 2 4𝐼 The shear force 𝐹1 is same in the top 𝑥 S member also.

𝑡𝑓 C

𝐹1 𝑡𝑤

h

𝑧

G O

e

Shearing stress in the vertical member BC at a distance ′𝑦′ from B (as shown in Fig.) 𝐹𝐴𝑦 𝑞2 = = 0 (since 𝑦 = 0) 𝐼𝑡 ∴ 𝐹2 = 0

T

𝐹2

d a b𝛿𝑥

𝐹1

B

T+ 𝛿𝑇

𝑦

𝑥 A u

c

𝑡

Shear centre Taking moments about ‘O’, 𝐹1 × ℎ = 𝐹 × 𝑒 b

𝐹𝑡ℎ𝑏 2 ×ℎ =𝐹×𝑒 4𝐼 𝑏 2 ℎ2 𝑡 ∴𝑒= 4𝐼

𝑡𝑓 C

𝐹1 𝑡𝑤

h

G

𝑥 S

𝑧

O

e

T

𝐹

𝐹2

d a b𝛿𝑥

𝐹1

B

T+ 𝛿𝑇 u

𝑦

𝑥 A

c

𝑡

Shear centre

S

G

X

Y

Fig. Shear centre of an equal –leg angle section

Shear centre S

S

S

S S

Fig. Shear centres S of sections consisting of two intersecting narrow rectangles

Shear centre of a semi circular section: 𝐹𝐴𝑦 𝑞= 𝐼𝑡

𝑑𝛼

𝜃

𝐴𝑦 =

𝑅 𝑑𝛼 . 𝑡 × 𝑅 cos 𝛼 =

0 𝜋

𝑦 2 dA

𝐼= 0

𝑅2 t sin 𝜃

𝑞

𝜋

=

𝑅 cos 𝜃

𝑑𝜃

2 𝑅𝑑𝜃. 𝑡

𝛼

O

e

0

𝜋𝑅3 𝑡 = 2 2𝐹 ∴𝑞= sin 𝜃 𝜋𝑅𝑡

Force acting on the area 𝑅𝑑𝜃 × 𝑡 = 𝑞 × 𝑅𝑑𝜃 × 𝑡

𝜃

𝑅 cos 𝛼

Shear centre

𝑡

R

Shear centre of a semi circular section: Moment of this force about O =𝑞 × 𝑅𝑑𝜃 × 𝑡 × 𝑅 Total moment of the force due to shearing stress on the cross section 𝑞 𝜋 𝜋 𝑞𝑅2 𝑡 𝑑𝜃 =

= 0 𝜋

= 0

𝑞𝑅2 𝑡 𝑑𝜃 0

2𝐹 2 4𝑅 𝑅 𝑡sin 𝜃 𝑑𝜃 = 𝐹 × 𝜋𝑅𝑡 𝜋

This must be equal to 𝐹 × 𝑒 4𝑅 𝐹×𝑒 =𝐹× 𝜋

4𝑅 ∴𝑒= ≅ 1.27𝑅 𝜋

𝑑𝛼 𝑑𝜃

𝜃 𝛼

𝑅 cos 𝛼

Shear centre

O

e

𝑡

R

Shear centre - Problem Problem: A channel section has flanges 12 cm x 2 cm and web 16 cm x 1 cm. determine the shear centre of the channel.

Shear centre - Problem Solution: Flanges dimensions=12 cm x 2 cm web dimension=16 cm x 1 cm.

b 𝑡𝑓

Shear centre, 𝑒 =

𝑏2 ℎ2 𝑡 4𝐼

C 𝑡𝑤

h b= 12 cm, h=16 cm, t= 2 cm 𝑥 S 1 × 163 12 × 23 𝐼= +2× + e 12 12 2 × 12 × 2 × 92 = 4245.33 𝑐𝑚2 𝐹

122 × 162 × 2 ∴𝑒= = 4.3413 cm 4 × 4245.33

𝐹1

G

𝑧

O T

𝐹2

d a b𝛿𝑥

𝐹1

B

T+ 𝛿𝑇 u

𝑦

𝑥 A

c

𝑡

A.U. Question Paper Problems • A beam of rectangular section 20 mm x 40 mm has its centre line curved to a radius of 50 mm. The beam is subjected to a bending moment of 4x105 Nmm. Determine the intensity of maximum stresses in the beam. Also plot the bending stress across the section. (Nov/Dec 2014)

A.U. Question Paper Problems • Problem: • A ring as shown in the following Figure is carrying a load of 30 kN. Calculate the stresses at 1 and 2. (May/June 2013) d=12 cm

G

1 G

2

30 kN

A.U. Question Paper Problems • Problem: • A chain link made of steel is of 10 mm diameter. Its straight portion is 300 mm in length and its ends are 50 mm in radius. Determine the deflection of the link along the load line when subjected to a load of 10 kN. E=200 Gpa.(Nov/Dec 2012)

A.U. Question Paper Problems • Problem: • A curved beam of rectangular cross section is subjected to pure bending with a moment of 400 Nm. The beam has width of 20 mm, depth of 40 mm and is curved in plane parallel to the depth. The mean radius of curvature is 50 mm. Determine the position of neutral axis and the ratio of maximum to the minimum stress. May/June 2012)

A.U. Question Paper Problems • Problem: • Figure shows a frame subjected to a load of 3.4 kN, find the resultant stress at A and B. (Nov/Dec 2011) 18

120

3.4 kN

48 1

2

48

Dimensions in mm

3.4 kN

A.U. Question Paper Problems • Problem: • A central horizontal section of hook is a symmetrical trapezium 60 mm deep, the inner width being 60 mm and the outer being 30 mm. Estimate the extreme intensities of stress when the hook carries load of 30 kN, the load line passing 40 mm from the inside edge of the section and the centre of curvature being in the load line. (Apr/May 2011)

A.U. Question Paper Problems • Problem: • Write a brief technical note on: (a) Unsymmetrical bending of beams (b) Curved beams (c) Stress concentration (d) Significance of shear centre. (Nov/Dec 2010)

A.U. Question Paper Problems • Problem: • A curved bar of rectangular section, initially unstressed is subjected to bending moment of 2000 Nm tends to straighten the bar. The section is 5 cm wide and 6 cm deep in the plane of bending and the mean radius of curvature is 10 m. Find the position of neutral axis and the stress at the inner and outer face. (Apr/May 2010)

2 marks Questions and Answers 1. Give the reasons for unsymmetrical bending. 2. Write Winkler Bach formula.