Unit V Advanced topics in bending of beams of... · Unit V Advanced topics in bending of beams ... Dr.P.Venkateswara rao, ... load of 4 kN inclined at 200 to the vertical and passing

Unit V Advanced topics in bending of beams

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

1

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.


2

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.

Advanced topics in bending of beams


3

Unsymmetrical bending of beams:

• While using the well known simple flexure formula 𝑀

𝐼=𝑓

𝑦 , it is

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.

Unsymmetrical bending of beams


4

• 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.

Unsymmetrical bending of beams


5

• Stresses due to unsymmetrical bending:

Figure 1 shows the cross section of a beam under the action of a bending moment M acting in plane YY.

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 𝑀′ = Msin 𝜃

𝑀′′=Mcos 𝜃

𝜃 𝛼 X X

Y

Y

U

U

V

V

N

A

P (u,v)

u v dA

Figure 1

G

Stresses due to unsymmetrical bending


6

• 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,

𝜎𝑏= 𝑀′𝑢

𝐼𝑉𝑉 +

𝑀′′𝑣

𝐼𝑈𝑈

= 𝑀 sin 𝜃.𝑢

𝐼𝑉𝑉 +

𝑀 cos 𝜃.𝑣

𝐼𝑈𝑈



7

• 𝜎𝑏=M𝑣 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., M𝑣 cos 𝜃

𝐼𝑈𝑈+

𝑢 sin 𝜃

𝐼𝑉𝑉=0



8

• Or,

• 𝑣 = - 𝐼𝑈𝑈

𝐼𝑉𝑉×

sin 𝜃

cos 𝜃 𝑢

• Or, 𝑣 = - 𝐼𝑈𝑈

𝐼𝑉𝑉× tan𝜃 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,

tan𝛼 = - 𝐼𝑈𝑈

𝐼𝑉𝑉× tan𝜃

𝑣 cos 𝜃

𝐼𝑈𝑈+

𝑢 sin 𝜃

𝐼𝑉𝑉=0



9

• 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.



10

• Deflection due to unsymmetrical bending:

• Figure 2 shows the transverse section of the beam with centroid G.

• XX and YY are two rectangular co-ordinate axes and UU and VV are the principal axesinclined at an angle 𝜃 to the xy set of coordinate axes.

• W is the load acting along line YY on the section of the beam. The load W can be resolved into the following two components:

• (i) Wsin𝜃 ----along UG

• (ii) Wcos 𝜃 ------ along VG

W 𝑀′ = Wsin 𝜃

𝑀′′=Wcos 𝜃

𝜃

𝛽 X X

Y

Y

U

U

V

V

N

A

Figure 2

𝛿𝑢 𝛿𝑣

𝛿

Deflection of beams due to unsymmetrical bending


11

• 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)𝑙3

𝐸 𝐼𝑉𝑉

• and 𝛿𝑣 = 𝑘(𝑤 cos 𝜃) 𝑙3

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.



12

• The total or resultant deflection 𝛿 can then be found as follows:

• 𝛿 = 𝛿𝑢 2 + 𝛿𝑣 2

• 𝛿 = 𝑘𝑙3

𝐸

𝑤 sin 𝜃

𝐼𝑉𝑉

2+

𝑤 cos 𝜃

𝐼𝑈𝑈

2

• 𝛿 = 𝑘𝑤𝑙3

𝐸

sin 𝜃

𝐼𝑉𝑉

2+

cos 𝜃

𝐼𝑈𝑈

2

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

tan𝛽 =𝛿𝑢

𝛿𝑣=

𝐼𝑈𝑈

𝐼𝑉𝑉tan 𝜃



13

• 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.


M 𝑀′ = Msin 𝜃

𝑀′′=Mcos 𝜃

𝜃 𝛼 X X

Y

Y

U

U

V

V

N

A

P (u,v)

u v dA

Figure 1

G

W 𝑀′ = Wsin 𝜃

𝑀′′=Wcos 𝜃

𝜃

𝛽 X X

Y

Y

U

U

V

V

N

A

Figure 2

G


14

• 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.

Unsymmetrical beams - Problems


15

• Solutioin: • To find the centroid:

• 𝑦 =60×10×5 + 100×5×60

60×10 + 100×5

𝑦 = 30 mm (from top)

Now 𝐼𝑥𝑥 = 𝐼𝑈𝑈 =60×103

12+ 60 × 10 × 30 − 5 2

+5×1003

12+ 100 × 5 × 80 − 50 2

= (38 × 104) + (86.67 × 104) = 1.25 × 106 mm4 = 1.25 × 10−6 𝑚4


X, U X, U

Y, V

Y, V

10 mm

60 mm

10

0 m

m

30 mm


16

Now 𝐼𝑌𝑌 = 𝐼𝑉𝑉 =10×603

12+

100×53

12

= 181041.67 mm4

= 0.181 × 10−6 𝑚4


X, U X, U

Y, V

Y, V

10 mm

60 mm

10

0 m

m

30 mm


17

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

Bending moment:

𝑀𝑢 =𝑤𝑢𝑙

4=

1.368 × 3

4= 1.026 kNm

𝑀𝑣 =𝑤𝑣𝑙

4=

3.758 × 3

4= 7.82 kNm

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


X, U X, U

Y, V

Y, V

10

mm

60 mm

10

0 m

m

30

mm

W=4 kN

200

A B

C D


18

(i) Maximum tensile stress:

At C, 𝜎𝑐 =𝑀𝑢×2.5×10−3

𝐼𝑉𝑉+

𝑀𝑣×80×10−3

𝐼𝑈𝑈

=1.026×2.5×10−3

0.181×10−6 +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+

𝑀𝑣×30×10−3

𝐼𝑈𝑈

=1.026×30×10−3

0.181×10−6 +2.82×30×10−3

1.25×10−6

= 170055.25 + 67680

= 237735.25 𝑘𝑁/𝑚2

= 237.735 MN/𝑚2


X, U X, U

Y, V

Y, V

10

mm

60 mm

10

0 m

m

30

mm

W=4 kN

200

A B

C D


19

(iii) Deflection due to the load, 𝛿:

𝛿 =𝑘𝑤𝑙3

𝐸

𝑠𝑖𝑛2𝜃

𝐼𝑉𝑉+

𝑐𝑜𝑠2𝜃

𝐼𝑈𝑈

k=1

48 for a beam with simply supported ends

and carrying a point load at its centre.

𝛿 =𝑘𝑤𝑙3

𝐸𝐼𝑈𝑈𝑠𝑖𝑛2𝜃

𝐼𝑈𝑈

𝐼𝑣𝑣

2

+ 𝑐𝑜𝑠2𝜃

=1

48×

4 × 103 × 33

200 × 109 × 1.25 × 10−6

× 𝑠𝑖𝑛220 ×1.25 × 10−6

0.181 × 10−6

2

+ 𝑐𝑜𝑠220


X, U X, U

Y, V

Y, V

10

mm

60 mm

10

0 m

m

30

mm

W=4 kN

200

A B

C D


20

𝛿 = 9 × 10−3 × 2.542 = 0.02287 𝑚 𝛿 = 22.9 𝑚𝑚

(iv) Position of neutral axis:

tan𝛽 =𝐼𝑈𝑈

𝐼𝑉𝑉tan 𝜃

=1.25×10−6

0.181×10−6 × tan 20

∴ 𝛽 = 68.30


X, U X, U

Y, V

Y, V

10

mm

60 mm

10

0 m

m

30

mm

W=4 kN

200

A B

C D

𝛽


21

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.


2.5 mm

2.5 mm

30 mm

45 mm

600 N

200


22

Solutiion:

A Cantilever of I-section, 2.4 m long is subjected to a load of 200 N at the free end. 𝑙 = 1.8 𝑚.

Since I-section is symmetrical about XX

and YY axes, therefore XX and YY are the

principal axes UU and VV.

Moment of inertia,

𝐼𝑈𝑈 = 𝐼𝑋𝑋 =30×503

12−

28×453

12= 99875 𝑚𝑚4

= 9.99 × 10−8 𝑚4

𝐼𝑉𝑉 = 𝐼𝑌𝑌 = 2 ×2.5 × 303

12+

45 × 23

12

= 1.128 × 10−8 𝑚4


2.5 mm

2.5 mm

30 mm

45 mm

200 N

200

2 mm

X,U X,U

Y, V

Y, V


23

Maximum bending moment, M=wl= 200× 2.4 = 480 𝑁𝑚.

Components of M, 𝑀𝑈 = M sin20 = 480 × 𝑠𝑖𝑛20 = 164.17 𝑁𝑚 𝑀𝑉 = M cos 20 = 480 × 𝑐𝑜𝑠20 = 451 𝑁𝑚.

𝑀𝑈 will cause tensile stresses at points

A and C and compressive stresses at points

B and D.

𝑀𝑉 will cause tensile stresses at points A and B

and compressive stresses at points C and D.


2.5 mm

2.5 mm

30 mm

45 mm

200 N

200

2 mm

X,U X,U

Y, V

Y, V

A B

C D


24

Now, resultant bending stresses on A and B

are as follows.

𝜎𝐴 =𝑀𝑈 × 15 × 10−3

𝐼𝑉𝑉+

𝑀𝑉 × 25 × 10−3

𝐼𝑈𝑈

=164.17×15×10−3

1.128×10−8 +451×25×10−3

9.99×10−8

= 218.31 + 112.86

= 331.17 MN/𝑚2

𝜎𝐵 = −𝑀𝑈 × 15 × 10−3

𝐼𝑈𝑈+

𝑀𝑉 × 25 × 10−3

𝐼𝑉𝑉

= −164.17×15×10−3

1.128×10−8 +451×25×10−3

9.99×10−8

= 218.31 + 112.86

= −105.4 MN/𝑚2


2.5 mm

2.5 mm

30 mm

45 mm

200 N

200

2 mm

X,U X,U

Y, V

Y, V

A B

C D


25

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:

(i) Stresses at the points A,B and C

of the mid section of the beam.

(i) Deflection of the beam at mid

section and its direction with the

load line.

(i) Position of the neutral axis.

Take E=200 GN/𝑚2

Unsymmetrical bending of beams - Problems

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

Solution:

Let (𝑥, 𝑦 ) be the co-ordinates of centroid G, with respect to the rectangular axes B𝑌1 and B𝑋1.

Now 𝑥 = 𝑦 =80×10×40 + 70×10×5

80×10 + 70×10= 23.66 mm

Moment of inertia about xx axis,

𝐼𝑋𝑋 =

80 × 103

12+ 80 × 10 × 23.66 − 5 2

+10 × 703

12+ 70 × 10 × 45 − 23.66 2

= 889833 𝑚𝑚4

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


10

mm

80 mm

80

mm

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

A

B

𝑌1

G

𝑋1

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.

Principal moment of inertia,

𝐼𝑈𝑈 =1

2𝐼𝑥𝑥 + 𝐼𝑦𝑦 +

1

2(𝐼𝑥𝑥 − 𝐼𝑦𝑦)cos 90

−𝐼𝑥𝑦 sin 90 (At 𝜃1 = 450)

=1

28.898 + 8.898 × 105 + 0

−5.2266 × 105 = 14.1246 × 105 𝑚𝑚4


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

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)]+

[70×10× (−18.66) × (21.34)] = −243923.5 − 278743 = −522666 𝑚𝑚4

= −5.2266× 105 𝑚𝑚4

(Product of inertia about the centroidal axes

is zero because portions 1 and 2 are

rectangular strips)


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

Also 𝐼𝑈𝑈 + 𝐼𝑉𝑉 = 𝐼𝑋𝑋 + 𝐼𝑌𝑌 ∴ 𝐼𝑉𝑉 = 𝐼𝑋𝑋 + 𝐼𝑌𝑌 − 𝐼𝑈𝑈

= 2 × 8.898 × 105 − 14.124 × 105

= 3.67 × 105𝑚𝑚4

Stresses at the points A, B and C:

Bending moment at the mid section,

𝑀 =𝑊𝐿

4=

400 × 2.4 × 103

4= 2.4 × 105𝑁𝑚𝑚

The components of bending moment are: 𝑀′ = Msin 𝜃 = 2.4 × 105 sin 45

= 1.697 × 105 Nmm 𝑀′′ = Mcos 𝜃 = 2.4 × 105 cos 45

= 1.697 × 105 Nmm


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

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 𝑣 = −23.66 cos 45 − (−23.66 sin 45) = 0

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


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

C

Stresses at Various points:

Point A, 𝜎𝐴 = 𝑀′𝑢

𝐼𝑉𝑉 +

𝑀′′𝑣

𝐼𝑈𝑈=

1.697×105×23.1

3.67×105 +1.697×105×56.56

14.1246×105

∴ 𝜎𝐴 = 17.47 N/𝑚𝑚2

Point B, 𝜎𝐵 =1.697×105×(−33.45)

3.67×105 +1.697×105×0

14.1246×105

∴ 𝜎𝐵 = −15.47 N/𝑚𝑚2

Point C, 𝜎𝐶 =1.697×105×(23.1)

3.67×105 +1.697×105×(−56.56)

14.1246×105

∴ 𝜎𝐶 = 3.88 N/𝑚𝑚2


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

C

Deflection of the beam 𝛿:

𝛿 =𝑘𝑤𝑙3

𝐸

𝑠𝑖𝑛2𝜃

𝐼𝑉𝑉2 +

𝑐𝑜𝑠2𝜃

𝐼𝑈𝑈2

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 × 106𝑚𝑚4, 𝐼𝑉𝑉 = 3.67 ×105 𝑚𝑚4

𝛿 =1

48×

400 × 2.4 × 103 3

200 × 103×

sin 452

3.67 × 105 2+

cos 452

14.1246 × 106 2

∴ 𝛿 = 1.146 𝑚𝑚


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

C

The deflection 𝛿 will be inclined at an angle 𝛽 clock wise with the line GV, given by

tan𝛽 =𝐼𝑈𝑈

𝐼𝑉𝑉tan 𝜃 =

14.1246 × 106

3.67 × 105tan 45 = 3.848

𝛽 = 75.430

Thus the deflection is at 75.43 - 45=30.430 clockwise with the load line

Gy’

Position of the neutral axis:

The neutral axis will be at 90- 30.43 =59.570

Anticlockwise with the load line, because the

Neutral axis is perpendicular to the line of

deflection.


𝑌1

G

A

B 𝑋1

10

mm

80 mm 8

0 m

m

U

U

V

V

Y’

Y

X X 450

10 mm

𝑥

𝑦

𝐺1

𝐺2

C

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.



35

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.

Curved beams


36

Stresses in curved bars (Winkler – Bach theory):

Curved beams

Fig. Bending of a curved bar

G

dA

y dy

𝜃

A

E B

C

D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


37


• Figure shows a bar ABCD initially in

Its unstrained state.

• Let AB’C’D be the strained position of the

bar.

• Let, R= Radius of the curvature of the

centroidal axis HG.

• y= Distance of the fibre EF from the

centroidal layer HG,

• R’=Radius of curvature of HG’

• y’=Distance between EF’ and HG’ after straining.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M

G

dA

y dy


38


• M=Uniform bending moment applied

to the beam (assumed positive when

tending to increase the curvature)

• 𝜃= Original angle subtended by the

centroidal axis HG at its centre of

curvature O.

• 𝜃′= Angle subtended by HG’ (after bending)

at the centre of curvature O’.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


39

Assumptions made in the analysis:

1. Plane sections (transverse sections)

remain plane during bending.

2. The material obeys Hooke’s law (limit of

proportinality is not exceeded)

3. Radial strain is negligible.

4. The fibres are free to expand or contract

without any constraining effect from the

adjacent fibres.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


40

• For finding the strain and stress normal

to the section, consider the fibre EF at a

distance y from the centroidal axis.

• Let 𝜎 be the stress in the strained layer EF’

under the bending moment M and e is the

strain in the same layer.

Strain, 𝑒 =𝐸𝐹′−𝐸𝐹

𝐸𝐹=

(𝑅′+𝑦′)𝜃′−(𝑅+𝑦)𝜃

(𝑅+𝑦)𝜃,

or 𝑒 =(𝑅′+𝑦′)𝜃′

(𝑅+𝑦)𝜃− 1

1 + 𝑒 =(𝑅′+𝑦′)𝜃′

(𝑅+𝑦)𝜃 --------------(1)

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


41

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

𝑒0 =𝑅′𝜃′

𝑅𝜃− 1

1 + 𝑒0 =𝑅′𝜃′

𝑅𝜃 -------(2)

1 + 𝑒

1 + 𝑒0=

(𝑅′+𝑦′)𝜃′

(𝑅 + 𝑦)𝜃𝑅′𝜃′

𝑅𝜃

=𝑅′ + 𝑦′

𝑅 + 𝑦×

𝑅

𝑅′=

1 +𝑦′𝑅′

1 +𝑦𝑅

• 𝑒 = 1 + 𝑒0

1+𝑦′

𝑅′

1+𝑦

𝑅

− 1

• or, 𝑒 =𝑒0

𝑦′

𝑅′ +

𝑦′

𝑅′ + 𝑒0 −

𝑦

𝑅

1 + 𝑦

𝑅

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


42

• According to assumption (3) i.e., radial strain is zero.

• ∴ 𝑦 = 𝑦′

• 𝑒 =𝑒0

𝑦′

𝑅′ +

𝑦

𝑅′ + 𝑒0 −

𝑦

𝑅

1 + 𝑦

𝑅

• Adding and subtracting the term 𝑒0𝑦

𝑅, we get

• 𝑒 =𝑒0

𝑦′

𝑅′ +𝑒0

𝑦′

𝑅′

𝑦

𝑅′ + 𝑒0 −

𝑦

𝑅 +𝑒0

𝑦

𝑅 −𝑒0

𝑦

𝑅

1 + 𝑦

𝑅

= 𝑒0 1+

𝑦

𝑅+𝑦

1

𝑅′−

1

𝑅 + 𝑒0 𝑦

1

𝑅′−

1

𝑅

1 + 𝑦

𝑅

𝑒 = 𝑒0 + 1+𝑒0

1

𝑅′−

1

𝑅𝑦

1+𝑦

𝑅

-------------(3)

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


43

• From Figure, it is obvious that for the given bending moment the layers above the centroidal layer are

in tension and layers below the centroidal

layer are in compression.

𝜎 = 𝐸𝑒 = 𝐸 𝑒0 +1 + 𝑒0

1𝑅′

−1𝑅

𝑦

1 +𝑦𝑅

− −(4)

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

Total force on the section, 𝐹 = 𝜎 𝑑𝐴.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


44

Consider a small strip of elementary area dA, at a distance y from the centroidal layer HG, we have

𝐹 = 𝐸 𝑒0 dA + E 1 + 𝑒0

1𝑅′

−1𝑅

𝑦

1 +𝑦𝑅

dA

= 𝐸 𝑒0 dA + E 1 + 𝑒0

1

𝑅′ −1

𝑅

𝑦

1 +𝑦𝑅

dA

= 𝐸𝑒0 A + E 1 + 𝑒01

𝑅′ −1

𝑅

𝑦

1+𝑦

𝑅

dA ----(5)

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

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


45

The total resisting moment is given by,

𝑀 = 𝜎𝑦 𝑑𝐴

= 𝐸 𝑒0 𝑦 𝑑𝐴 +𝐸 1 + 𝑒0

1𝑅′

−1𝑅

𝑦2𝑑𝐴

1 +𝑦𝑅

= 𝐸𝑒0 × 0 + 𝐸 1 + 𝑒0

1

𝑅′−

1

𝑅

𝑦2𝑑𝐴

1 +𝑦𝑅

(since 𝑦 𝑑𝐴 = 0)

𝑀 = 𝐸 1 + 𝑒01

𝑅′−

1

𝑅

𝑦2𝑑𝐴

1+𝑦

𝑅

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


46

𝐿𝑒𝑡, 𝑦2𝑑𝐴

1+𝑦

𝑅

= Aℎ2 -------------(6)

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

𝑀 = 𝐸 1 + 𝑒01

𝑅′−

1

𝑅𝐴ℎ2 ----------(7)

𝑁𝑜𝑤, 𝑦

1 +𝑦𝑅

𝑑𝐴 = 𝑅𝑦

𝑅 + 𝑦dA

= 𝑦 −𝑦2

𝑅 + 𝑦dA = 𝑦𝑑𝐴 −

𝑦2𝑑𝐴

𝑅 + 𝑦

= 0 −1

𝑅

𝑦2

1 +𝑦𝑅

dA

∴ 𝑦

1 +𝑦𝑅

𝑑𝐴 = −1

𝑅

𝑦2

1 +𝑦𝑅

dA = −1

𝑅Aℎ2 −−−−− −(8)

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


47

𝑦

1 +𝑦𝑅

𝑑𝐴 = −1

𝑅Aℎ2 −−−−− −(8)

𝐹 = 𝐸𝑒0 A + E 1 + 𝑒01

𝑅′ −1

𝑅

𝑦

1+𝑦

𝑅

dA ---(5)

Hence equation (5) becomes

𝐹 = 𝐸𝑒0 A − E 1 + 𝑒01

𝑅′ −1

𝑅

𝐴ℎ2

𝑅 −− −-(9)

Since transverse plane section remain plane

during bending , ∴ 𝐹 = 0.

or 0 = 𝐸𝑒0 A − E 1 + 𝑒01

𝑅′ −1

𝑅

𝐴ℎ2

𝑅

or, 𝐸𝑒0 A = E 1 + 𝑒01

𝑅′ −1

𝑅

𝐴ℎ2

𝑅

or, 𝑒0 = 1 + 𝑒01

𝑅′ −1

𝑅

ℎ2

𝑅

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


48

or, 𝑒0𝑅

ℎ2 = 1 + 𝑒01

𝑅′ −1

𝑅---------(10).

Substituting the value of 1 + 𝑒01

𝑅′ −1

𝑅 in

Equation (7), we get

𝑀 = 𝐸 𝑒0𝑅

ℎ2 × 𝐴ℎ2 = 𝑒0EAR

or, 𝑒0 =𝑀

𝐸𝐴𝑅 −−− − 11

𝜎 = 𝐸 𝑒0 +1 + 𝑒0

1𝑅′

−1𝑅

𝑦

1 +𝑦𝑅

− −(4)

Substituting the value of 𝑒0 in equation (4), we get

𝜎 = 𝐸𝑀

𝐸𝐴𝑅+

𝑦

1 +𝑦𝑅

𝑒0

𝑅

ℎ2

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


49

𝜎 =𝑀

𝐴𝑅+ 𝐸

𝑦

1 +𝑦𝑅

𝑒0

𝑅

ℎ2

= 𝑀

𝐴𝑅+ 𝐸

𝑦

1 +𝑦𝑅

𝑒0

𝑅

ℎ2

=𝑀

𝐴𝑅+ 𝐸

𝑦

1 +𝑦𝑅

×𝑀

𝐸𝐴𝑅×

𝑅

ℎ2

=𝑀

𝐴𝑅+

𝑀

𝐴𝑅

𝑅𝑦

1 +𝑦𝑅

` ×1

ℎ2

𝜎 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2×

𝑦

𝑅 + 𝑦−−−− − 12 Tensile

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


50

On the other side of HG, y will be negative,

and stress will be compressive

𝜎 =𝑀

𝐴𝑅1 −

𝑅2

ℎ2×

𝑦

𝑅 − 𝑦−−− − 13

When the bending moment is applied in such

a manner that it tends to decrease the curvature,

then the equation (13), tensile.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


51

Position of neutral axis:

At the neutral axis, 𝜎 = 0 𝑀

𝐴𝑅1 +

𝑅2

ℎ2×

𝑦

𝑅 + 𝑦= 0

𝑅2

ℎ2×

𝑦

𝑅 + 𝑦= −1

𝑅2y = −ℎ2 𝑅 + 𝑦 = −Rℎ2 − ℎ2y 𝑦 𝑅2 + ℎ2 = −Rℎ2

𝒚 = −𝑹𝒉𝟐

𝑹𝟐 + 𝒉𝟐

Hence neutral axis is located below the centroidal axis.

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


52

Values of ℎ2for various sections

We know, ℎ2 =1

𝐴

𝑦2

1+𝑦

𝑅

dA

=𝑅

𝐴

𝑦2

𝑅 + 𝑦dA

=𝑅

𝐴 𝑦𝑑𝐴 − 𝑅𝑑𝐴 +

𝑅2

𝑅 + 𝑦𝑑𝐴

=𝑅

𝐴0 − 𝑅𝐴 +

𝑅2

𝑅 + 𝑦𝑑𝐴

∴ 𝒉𝟐 =𝑹𝟑

𝑨

𝒅𝑨

𝑹 + 𝒚− 𝑹𝟐

Curved beams

𝜃

A

E B

C D

G’

G

H

B’ F’

F

C’

O

O’ 𝜃′

R

R’

M

M


53

Value of ℎ2for Rectangular section:

Figure shows the rectangular section with centre

of curvature O lying on YY-axis and XX-axis is

the centroidal bending axis.

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

𝑊𝑒 𝑘𝑛𝑜𝑤, 𝒉𝟐 =𝑹𝟑

𝑨

𝒅𝑨

𝑹 + 𝒚− 𝑹𝟐

∴ ℎ2 =𝑅3

𝐵𝐷

𝐵𝑑𝑦

𝑅 + 𝑦

𝐷/2

−𝐷/2

− 𝑅2

Curved beams

G

dA

y dy

D

Y

Y

R

O

B

X X


54

∴ ℎ2 =𝑅3

𝐵𝐷

𝐵𝑑𝑦

𝑅 + 𝑦

𝐷/2

−𝐷/2

− 𝑅2

=𝑅3

𝐷log𝑒 𝑅 + 𝑦

𝐷/2−𝐷/2

− 𝑅2

∴ 𝒉𝟐 =𝑹𝟑

𝑫𝐥𝐨𝐠𝒆

𝟐𝑹 + 𝑫

𝟐𝑹 − 𝑫− 𝑹𝟐

Curved beams

G

dA

y dy

D

Y

Y

R

O

B

X X


55

Problem:

Figure shows a frame subjected to a load of 2.4 kN.

Find (i) The resultant stresses at points 1 and 2.

(ii) Position of the neutral axis.

Curved beams - Problems

48

18 2.4 kN

2.4 kN Dimensions in mm

120

1 2 48


56

Solution:

Area of section at 1-2,

𝐴 = 48 × 18 × 10−6 = 8.64 × 10−4 𝑚2

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

= −403.2 𝑁𝑚.

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


48 18 2.4 kN


120

1 2 48


57

(i) Resultant stresses at points 1 and 2:

Direct stress, 𝜎𝑑 =2.4×103

8.64×10−4 × 10−6

= 2.77 MN/𝑚2 (tensile)

ℎ2 =𝑅3

𝐷log𝑒

2𝑅 + 𝐷

2𝑅 − 𝐷− 𝑅2

Here, 𝑅 = 48 𝑚𝑚 = 0.048 𝑚,

𝐷 = 48 𝑚𝑚 = 0.048 𝑚

ℎ2 =0.0483

0.048log𝑒

2 × 0.048 + 0.048

2 × 0.048 − 0.048− 0.0482

= 0.0482 log𝑒 3 − 1 = 2.27 × 10−4 𝑚2


48 18 2.4 kN


120

1 2 48


58

Bending stress due to M at point 2,

𝜎𝑏2 =𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅 − 𝑦

=−403.2

8.64 × 10−4 × 0.0481 −

0.0482

2.27 × 10−4

0.024

0.048 − 0.024× 10−6 MN/𝑚2

∴ 𝜎𝑏2 = −9.722 1 − 10.149

= +88.95 MN/𝑚2( tensile)


48 18 2.4 kN


120

1 2 48


59

Bending stress due to M at point 1,

𝜎𝑏1 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦

=−403.2

8.64 × 10−4 × 0.0481 +

0.0482

2.27 × 10−4

0.024

0.048 + 0.024× 10−6 MN/𝑚2

= −42.61 𝑀𝑁/𝑚2= 42.61 MN/𝑚2(Comp.)


48 18 2.4 kN


120

1 2 48


60

Resultant stress at point 2, 𝜎2 = 𝜎𝑑 + 𝜎𝑏2 = 2.77 + 88.95

= 91.72 MN/𝑚2(tensile).

Resultant stress at point 1, 𝜎1 = 𝜎𝑑 + 𝜎𝑏1 = 2.77 − 42.61

= 39.84 MN/𝑚2(Comp.)


48 18 2.4 kN


120

1 2 48


61

Position of neutral axis:

We know, 𝑦 = −𝑅ℎ2

𝑅2+ℎ2

= −0.048×2.27×10−4

0.0482+2.27×10−4

= −0.00435 𝑚 = −4.35 𝑚𝑚

Hence, neutral axis is at a radius of 4.35 mm below the centroidal axis.


48 18 2.4 kN


120

1 2 48


62

𝒉𝟐for 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, A=𝜋

4𝑑2

Consider a strip of width b and a depth dy

at a distance y from the centroidal layer as shown.

𝑏 = 2𝑑

2

2

− 𝑦2

Curved beams

G

Y

dy

b

R

O

𝒉𝟐for circular section (contd…):

Area of strip, 𝑑𝐴 = 𝑏. 𝑑𝑦 = 2𝑑

2

2− 𝑦2 . dy

ℎ2 =𝑅3

𝐴

𝑑𝐴

𝑅 + 𝑦− 𝑅2

=𝑅3

𝐴

2𝑑2

2

− 𝑦2

𝑅 + 𝑦

+𝑑/2

−𝑑/2

dy − 𝑅2

=8𝑅3

𝜋𝑑2

2𝑑2

4− 𝑦2

𝑅 + 𝑦

+𝑑/2

−𝑑/2

dy − 𝑅2

Curved beams

G

Y

dy

b

R

O

𝒉𝟐for circular section (contd…):

=8𝑅3

𝜋𝑑2

2𝑑2

4− 𝑦2

𝑅 + 𝑦

+𝑑/2

−𝑑/2

dy − 𝑅2

Equating the integral by binomial expression

and then integrating, we get

∴ 𝒉𝟐 =𝒅𝟐

𝟏𝟔+

𝟏

𝟏𝟐𝟖

𝒅𝟒

𝑹𝟐+ ⋯

Curved beams

G

Y

dy

b

R

O

𝒉𝟐 for triangular section:

Let 𝑅 + 𝑦 = 𝑎 𝑑𝑦 = 𝑑𝑎

Width of elementary strip,

𝑏′ =𝑅2 − 𝑎 𝑏

𝑑

Area of elementary strip, 𝑑𝐴 = 𝑏′𝑑𝑦 = b′. da

Now, ℎ2 =𝑅3

𝐴

𝑑𝐴

𝑅 + 𝑦

𝑅2

𝑅1

− 𝑅2

=𝑅3

𝐴

𝑏′𝑑𝑎

𝑎− 𝑅2

𝑅2

𝑅1

Curved beams

G

Y

dy

b

R

O

d

𝑅1

𝑅2

𝑏′

d b 𝑅2- a ?

𝑏′ =𝑅2 − 𝑎 𝑏

𝑑

𝒉𝟐 for triangular section (contd…):

ℎ2 =𝑅3

𝐴

𝑅2 − 𝑎 𝑏

𝑑

𝑅2

𝑅1

𝑑𝑎

𝑎− 𝑅2

=𝑅3

𝐴

𝑅2𝑏

𝑑log𝑒

𝑅2

𝑅1−

𝑏

𝑑𝑅2 − 𝑅1 − 𝑅2

Since, 𝑅2 = R +2

3d ; 𝑅1 = R −

𝑑

3

and 𝑅2 − 𝑅1 = d

ℎ2 =𝑅3

𝐴

R +23d 𝑏

𝑑log𝑒

R +23d

R −𝑑3

− 𝑏 − 𝑅2

Curved beams

G

Y

dy

b

R

O

d

𝑅1

𝑅2

𝑏′

𝒉𝟐 for triangular section (contd…):

ℎ2 =𝑅3

𝐴

R +23d 𝑏

𝑑log𝑒

R +23d

R −𝑑3

− 𝑏 − 𝑅2

=2𝑅3

𝑏𝑑

3R + 2𝑑 𝑏

3𝑑log𝑒

3R + 2𝑑

3R − 𝑑− 𝑏 − 𝑅2

∴ 𝒉𝟐=𝟐𝑹𝟑

𝒅

𝟑𝐑 + 𝟐𝒅

𝟑𝒅𝒍𝒐𝒈𝒆

𝟑𝐑 + 𝟐𝒅

𝟑𝐑 − 𝒅− 𝟏 − 𝑹𝟐

Curved beams

G

Y

dy

b

R

O

d

𝑅1

𝑅2

𝑏′

𝒉𝟐 for trapezoidal section :

Consider an elementary strip of

width b’ and depth dy at a distance y

from the centroidal axis.


𝑏′ = b +𝐵 − 𝑏

𝑑1 + 𝑑2𝑅2 − 𝑎

Area of the strip,

𝑑𝐴 = 𝑏′dy = b +𝐵−𝑏

𝑑1+𝑑2𝑅2 − 𝑎 da

Curved beams

G

Y

dy

B

R

O

d

𝑅1

𝑅2

𝑏′

b

𝑑1

𝑑2

𝐵 − 𝑏

2

𝑅2 − 𝑎

𝑑

𝑅2 − 𝑎

𝒉𝟐 for trapezoidal section (contd…):

ℎ2 =𝑅3

𝐴

𝑑𝐴

𝑅 + 𝑦

𝑅2

𝑅1

− 𝑅2

=𝑅3

𝐴 𝑏 +

𝐵 − 𝑏

𝑑1 + 𝑑2𝑅2 − 𝑎

𝑑𝑎

𝑎

𝑅1

𝑅1

− 𝑅2

=𝑅3

𝐴 𝑏

𝑑𝑎

𝑎

𝑅1

𝑅1

+𝐵 − 𝑏

𝑑1 + 𝑑2

𝑅2 − 𝑎

𝑎

𝑅2

𝑅1

𝑑𝑎 − 𝑅2

=𝑅3

𝐴𝑏 log𝑒

𝑅2

𝑅1+

𝐵 − 𝑏

𝑑1 + 𝑑2𝑅2 log𝑒

𝑅2

𝑅1− 𝑅2 − 𝑅1 − 𝑅2

Curved beams

G

Y

dy

B

R

O

d

𝑅1

𝑅2

𝑏′

b

𝑑1

𝑑2

𝒉𝟐 for trapezoidal section (contd…):

ℎ2 =𝑅3

𝐴

𝑏 log𝑒

𝑅 + 𝑑2

𝑅 − 𝑑1+

𝐵 − 𝑏

𝑑𝑅 + 𝑑2 log𝑒

𝑅 + 𝑑2

𝑅 − 𝑑1−

𝐵 − 𝑏

− 𝑅2

Where,

𝐴 =𝐵 + 𝑏

2d

𝑑1 =𝑑

3

𝐵 + 2𝑏

𝐵 + 𝑏

𝑑2 = d − 𝑑1

Curved beams

G

Y

dy

B

R

O

d

𝑅1

𝑅2

𝑏′

b

𝑑1

𝑑2

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.


P

P

0.05 m

0.15 m

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,

𝐴 =𝜋

4𝑑2 =

𝜋

4× 0.102= 7.854 × 10−3 𝑚2

Bending moment, M=P(0.15+0.10)=0.25P

ℎ2 =𝑑2

16+

1

128

𝑑4

𝑅2

=0.102

16+

1

128

0.104

0.102 = 7.031 × 10−4𝑚2


P

P

0.05 m

0.15 m

Solution:

Direct stress, 𝜎𝑑 =𝑃

𝐴 (comp.)

Bending stress at point1 due to M,

𝜎𝑏1 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦tensile

Total stress at point 1, 𝜎1 = 𝜎𝑑 + 𝜎𝑏1

150 = −𝑃

𝐴+

𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦tensile

150 = −𝑃

7.854 × 10−3

+0.25 𝑃

7.854 × 10−3 × 0.101 +

0.102

7.031 × 10−4

0.05

0.10 + 0.05

150 = −127.32𝑃 + 318.31𝑃 × 5.74 = 1699.78𝑃


P

P

0.05 m

0.15 m

1 2 0.10 m

∴ 𝑃 =150

1699.78𝑀𝑁 =

150 × 103

1699.78 kN = 88.25 kN − (i)

Bending stress at point 2 due to M,

𝜎𝑏2 =𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅 − 𝑦

Total stress at point 2, 𝜎2 = 𝜎𝑑 + 𝜎𝑏2

200 = −𝑃

𝐴+

𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅 − 𝑦

200 = −𝑃

7.854 × 10−3

+0.25 𝑃

7.854 × 10−3 × 0.101 −

0.102

7.031 × 10−4

0.05

0.10 − 0.05

200 = −127.32𝑃 − 318.31𝑃 × 13.22 = −4335.38𝑃


P

P

0.05 m

0.15 m

1 2 0.10 m

𝑃 = −200

4335.38𝑀𝑁 = −

200 × 103

4335.38 kN = −46.13 kN

∴ 𝑃 = 46.13 𝑘𝑁 𝐶𝑜𝑚𝑝. − 𝑖𝑖

Comparing (i) and (ii) values, the safe load

P will be lessor of these.

Hence, P= 46.13 kN.


P

P

0.05 m

0.15 m

1 2 0.10 m

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

exceed 225 MN/𝑚2.

(ii) Draw the stress distribution

along 1-2.


P

1 2 G

6 cm

9 cm

G

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

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.)


P

1 2 G

6 cm 9 cm

G

ℎ2 =𝑅3

𝐷log𝑒

2𝑅 + 𝐷

2𝑅 − 𝐷− 𝑅2

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

ℎ2 =0.1653

0.09log𝑒

2 × 0.165 + 0.09

2 × 0.165 − 0.09− 0.1652

= 0.0499 × 0.5596 − 0.1652 = 6.99 × 10−4

Resultant stress at point 2,

𝜎𝑚𝑎𝑥 = −𝑃

𝐴+

𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅−𝑦

225× 106 = −185.18P +0.165𝑃

0.0054×0.1651 −

0.1652

6.99×10−4

0.045

0.165−0.045


P

1 2 G

6 cm 9 cm

G

𝑃 = −83189 𝑁 𝑜𝑟 𝑃 = 83.189 𝑘𝑁(𝑐𝑜𝑚𝑝. )

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

Stress distribution along G1:

Resultant stress at point 1,

𝜎𝑚𝑎𝑥 = −𝑃

𝐴+

𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅+𝑦

𝜎1 = −83.189

0.0054

+0.165 × 83.189

0.0054 × 0.1651 +

0.1652

6.99 × 10−4

0.045

0.165 + 0.045

= 128.6 𝑀𝑁/𝑚2

Curved beams - Problems P

1 2 G

6 cm 9 cm

G

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

Curved beams - Problems P

1 2 G

6 cm 9 cm

G 128.6 MN/𝑚2

225 MN/𝑚2

0

Hyperbolic

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.


150 kN

A B

Dimensions in mm

d=180

B=135 b=45

𝑑2 𝑑1

G A B

Solution:

B=135 mm=0.135 m;

b=45 mm=0.045 m;

d=180 mm;

P=150 kN

Area of section, 𝐴 =0.135+0.45

2× 0.18 = 0.0162 𝑚2

𝑑1 =𝑑

3

𝐵 + 2𝑏

𝐵 + 𝑏=

0.18

3

0.135 + 2 × 0.045

0.135 + 0.045= 0.075 m

𝑑2 = 0.18 − 0.075 = 0.105 m

R=0.105+0.075=0.18 m


150 kN

A B

Dimensions in mm

d=180

B=135 b=45

𝑑2 𝑑1

G A B

ℎ2 =𝑅3

𝐴

𝑏 log𝑒

𝑅 + 𝑑2

𝑅 − 𝑑1+

𝐵 − 𝑏

𝑑𝑅 + 𝑑2 log𝑒

𝑅 + 𝑑2

𝑅 − 𝑑1−

𝐵 − 𝑏

− 𝑅2

By substituting the values , ℎ2 = 0.00256 𝑚2

Bending moment , 𝑀 = −150 × 103 × 0.18

∴ 𝑀 = −27000Nm

Direct stress, 𝜎𝑑 =𝑃

𝐴=

150×103

0.0162× 10−6

= −9.26 MN/𝑚2(tensile)


150 kN

A B

Dimensions in mm

d=180

B=135 b=45

𝑑2 𝑑1

G A B

Bending stress calculations,

Bending stress at A,

𝜎𝑏𝐴 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦= −52.43 MN/𝑚2

𝜎𝑏𝐴 = 52.43 MN/𝑚2 (comp.)

Bending stress at B,

𝜎𝑏𝐵 =𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅 − 𝑦= 74.45 MN/𝑚2

𝜎𝑏𝐵 = 74.45 MN/𝑚2 (Tensile.)


150 kN

A B

Dimensions in mm

d=180

B=135 b=45

𝑑2 𝑑1

G A B

Stress at A, 𝜎𝐴 = 𝜎𝑑 + 𝜎𝑏𝐴 = 9.26 − 52.43 = −43.17 MN/𝑚2

∴ 𝜎𝐴= 43.17 MN/𝑚2 (Comp.)

Stress at B, 𝜎𝐵 = 𝜎𝑑 + 𝜎𝑏𝐵 = 9.26 + 74.45 = 83.71 MN/𝑚2

∴ 𝜎𝐵= 83.71 MN/𝑚2 (Tensile)


150 kN

A B

Dimensions in mm

d=180

B=135 b=45

𝑑2 𝑑1

G A B

Stresses in a ring:

Figure shows a circular ring under the action of an axial pull P.

(A) Stresses on a section taken along the line of

action of P,

(i) At outside ring, i.e., at point A,

𝜎𝐴 =𝑃

𝜋𝐴

𝑅2

𝑅2 + ℎ21 +

𝑅2

ℎ2

𝑦2

𝑅 + 𝑦2 (tensile)

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

𝜎𝐵 =𝑃

𝜋𝐴

𝑅2

𝑅2 + ℎ2 1 −𝑅2

ℎ2

𝑦1

𝑅 − 𝑦1 (Comp. )

Curved beams

P

P

A

B

C D

Stresses in a ring (contd…):

(B) Stresses on a section perpendicular to the line of

action of P,

(i) At outside ring, i.e., at point C,

𝜎𝐶 =𝑃

𝐴

𝑅2

𝜋 𝑅2 + ℎ2 +𝑅2

2ℎ2

2𝑅2

𝜋 𝑅2 + ℎ2 − 1𝑦2

𝑅 + 𝑦2

+𝑃

2𝐴 (Comp. )

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

𝜎𝐷 =𝑃

𝐴

𝑅2

𝜋 𝑅2 + ℎ2 +𝑅2

2ℎ2

2𝑅2

𝜋 𝑅2 + ℎ2 − 1−𝑦1

𝑅 − 𝑦1+

𝑃

2𝐴(Ten. )

Curved beams

P

P

A

B

C D

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.


Solution:

d=30 mm=0.03 m

𝐴 =𝜋

4× 0.032= 7.068 × 10−4 𝑚2

P=12 kN

R=180 mm=0.18 m

ℎ2 =𝑑2

16+

1

128

𝑑4

𝑅2

=0.032

16+

1

128×

0.034

0.182 = 5.64 × 10−5 𝑚2


12 kN

12 kN

A

B

C D

30 mm

Solution:

Stresses:

𝜎𝐴 =𝑃

𝜋𝐴

𝑅2

𝑅2 + ℎ21 +

𝑅2

ℎ2

𝑦2

𝑅 + 𝑦2 (tensile)

𝜎𝐴 =12 × 103

𝜋 × 7.068 × 10−4

0.182

0.182 + 5.64 × 10−5

1 +0.182

5.64 × 10−5

0.015

0.18 + 0.015× 10−6

∴ 𝜎𝐴 = 243.7 MN/𝑚2(tensile)


12 kN

12 kN

A

B

C D

30 mm

Stresses:

𝜎𝐵 =𝑃

𝜋𝐴

𝑅2

𝑅2 + ℎ21 −

𝑅2

ℎ2

𝑦1

𝑅 − 𝑦1

𝜎𝐵 =12 × 103

𝜋 × 7.068 × 10−4

0.182

0.182 + 5.64 × 10−5

1 −0.182

5.64 × 10−5

0.015

0.18 − 0.015× 10−6

∴ 𝜎𝐵 = −276.3 = 276.3 MN/𝑚2(Comp. )


12 kN

12 kN

A

B

C D

30 mm

Stresses:

𝜎𝐶 =𝑃

𝐴

𝑅2

𝜋 𝑅2 + ℎ2 +𝑅2

2ℎ2

2𝑅2

𝜋 𝑅2 + ℎ2 − 1𝑦2

𝑅 + 𝑦2+

𝑃

2𝐴 (Comp. )

𝜎𝐶 =12 × 103 × 10−6

7.068 × 10−4

0.182

𝜋 0.182 + 5.64 × 10−5

+0.182

2 × 5.64 × 10−5 2 × 0.182

𝜋 0.182 + 5.64 × 10−5 − 10.015

0.18 + 0.015

+12 × 103 × 10−6

2 × 7.068 × 10−4 (Comp. )

𝜎𝑐 = −131.3 + 8.49 = −122.8 MN/𝑚2

∴ 𝜎𝑐 = 122.8 MN/𝑚2 (Comp.)


12 kN

12 kN

A

B

C D

30 mm

Stresses:

𝜎𝐷 =𝑃

𝐴

𝑅2

𝜋 𝑅2 + ℎ2 −𝑅2

2ℎ2

2𝑅2

𝜋 𝑅2 + ℎ2 − 1𝑦1

𝑅 − 𝑦1+

𝑃

2𝐴 (Tensile)

𝜎𝐷 =12 × 103 × 10−6

7.068 × 10−4

0.182

𝜋 0.182 + 5.64 × 10−5

−0.182

2 × 5.64 × 10−5 2 × 0.182

𝜋 0.182 + 5.64 × 10−5 − 10.015

0.18 − 0.015

+12 × 103 × 10−6

2 × 7.068 × 10−4

𝜎𝐷 = 166.99 + 8.49 = 175.48 MN/𝑚2 (tensile)

∴Maximum tensile stress = 243.7 MN/𝑚2

Maximum compressive stress=276.3 MN/𝑚2


12 kN

12 kN

A

B

C D

30 mm

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.


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

and maximum tensile

stress is limited to

180 MN/m2 .

Calculate the tensile load

the ring can carry.


75

45

P

P

All dimensions in mm

A

Area of cross section, 𝐴 = 0.075 × 0.045 = 0.003375 𝑚2

ℎ2 =𝑅3

𝐷log𝑒

2𝑅 + 𝐷

2𝑅 − 𝐷− 𝑅2

ℎ2 =0.153

0.075log𝑒

2 × 0.15 + 0.075

2 × 0.15 − 0.075− 0.152

∴ ℎ2 = 4.87 × 10−4𝑚2



75

45

P

P

Load P calculation:

The maximum tensile stress will occur at A.

∴ 𝜎𝐴 = 180 × 106N/𝑚2

𝑁𝑜𝑤, 𝜎𝐴 =𝑃

𝜋𝐴

𝑅2

𝑅2 + ℎ21 +

𝑅2

ℎ2

𝑦2

𝑅 + 𝑦2

180 × 106 =𝑃

𝜋 × 0.003375

0.152

0.152 + 4.87 × 10−4

1 +0.152

4.87 × 10−4

0.0375

0.15 + 0.0375

𝑃 = 19440 𝑁 ∴ 𝑃 = 19.44 𝑘𝑁



75

45

P

P

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.


Solution: 𝑑2 = 120 mm 𝑡 = 7.5 𝑚𝑚 𝑑1 = 120 − 2 × 7.5 = 105 𝑚𝑚

𝐴 =𝜋

40.122 − 0.1052 = 0.00265 𝑚2

𝑀 = 3 𝑘𝑁𝑚

𝐴1 =𝜋

4× 0.1052 = 0.00866 𝑚2

𝐴2 =𝜋

4× 0.122 = 0.01131 𝑚2

Bending moment, M=+3 kNm (tending to increase the curvature)


O

M

M

120 mm 105 mm

We know, for circular section,

ℎ2 =𝑑2

16+

1

128

𝑑4

𝑅2

For inner tube,

ℎ12 =

𝑑12

16+

1

128×

𝑑14

𝑅2

=0.1052

16+

1

128×

0.1054

0.2252 = 7.078 × 10−4 𝑚2

For outer circle,

ℎ22 =

𝑑22

16+

1

128×

𝑑24

𝑅2

=0.122

16+

1

128×

0.124

0.2252 = 9.32 × 10−4 𝑚2


O

M

M

120 mm 105 mm

𝐴ℎ2 = 𝐴2ℎ22 − 𝐴1ℎ1

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.

∴ 𝜎𝐴 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦

=3 × 103

0.00265 × 0.2251 +

0.2252

0.00166

0.06

0.225 + 0.06× 10−6

𝜎𝐴 = 37.32 MN/𝑚2


O

M

M

120 mm 105 mm

A B

Maximum compressive stress occurs at B.

∴ 𝜎𝐵 =𝑀

𝐴𝑅1 −

𝑅2

ℎ2

𝑦

𝑅 − 𝑦

=3 × 103

0.00265 × 0.2251 −

0.2252

0.00166

0.06

0.225 − 0.06× 10−6

= −50.75

𝜎𝐵 = 50.75 MN/𝑚2(comp.)


O

M

M

120 mm 105 mm

A B

𝒉𝟐for T-section:


ℎ2 =𝑅3

𝐴

𝑑𝐴

𝑅 + 𝑦

𝑅2

𝑅1

+ 𝑑𝐴

𝑅 + 𝑦

𝑅3

𝑅2

− 𝑅2

=𝑅3

𝐴

𝑏2𝑑𝑎

𝑎

𝑅2

𝑅1

+ 𝑡1𝑑𝑎

𝑎

𝑅3

𝑅2

− 𝑅2

=𝑅3

𝐴𝑏2 log𝑒

𝑅2

𝑅1+ 𝑡1 log𝑒

𝑅3

𝑅2− 𝑅2

Where 𝐴 = 𝑏1𝑡1 + 𝑏2𝑡2

Curved beams

𝑏2

𝑏1

𝑡1

G

dy

y

𝑡2

O

𝑅2 𝑅1

𝑅3

𝒉𝟐for I-section:


ℎ2 =𝑅3

𝐴

𝑑𝐴

𝑅 + 𝑦

𝑅2

𝑅1

+ 𝑑𝐴

𝑅 + 𝑦+

𝑅3

𝑅2

𝑑𝐴

𝑅 + 𝑦

𝑅4

𝑅3

− 𝑅2

=𝑅3

𝐴

𝑏2𝑑𝑎

𝑎

𝑅2

𝑅1

+ 𝑡3𝑑𝑎

𝑎

𝑅3

𝑅2

+ 𝑏1𝑑𝑎

𝑎

𝑅4

𝑅3

− 𝑅2

=𝑅3

𝐴𝑏2 log𝑒

𝑅2


𝑅3

𝑅2+𝑏1 log𝑒

𝑅4

𝑅3− 𝑅2

Where 𝐴 = 𝑏1𝑡1 + 𝑏2𝑡2 + 𝑏3𝑡3

Curved beams

𝑏2

𝑏3

𝑏1

G

dy

y

𝑡2

O

𝑅2

𝑅1 𝑅3

𝑡3

𝑡1

𝑅4

𝐏𝐫𝐨𝐛𝐥𝐞𝐦:

A curved beam has a T-section (shown in Figure).

The inner radius is 300 mm. What is the

Eccentricity of the section?

Curved beams -Problems

80 𝑚𝑚

60 𝑚𝑚

20 𝑚𝑚

G

dy

y

20 𝑚𝑚

O

320 𝑚𝑚 300 𝑚𝑚

380 𝑚𝑚

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

Area, 𝐴 = 𝑏1𝑡1 + 𝑏2𝑡2 = (60 × 20) + (80 × 20) ↑∴ 𝐴 = 2800 𝑚2

Centre of gravity of T-section, taking from O,

𝑅 =80 × 20 × 310 + 60 × 20 × 350

2800

∴ 𝑅 = 327.14 𝑚𝑚

ℎ2 =𝑅3

𝐴𝑏2 log𝑒

𝑅2


𝑅3

𝑅2− 𝑅2

=327.143

2800

80 log𝑒

320

300+

20 log𝑒

380

320

− 327.142

∴ ℎ2 = 512.08 𝑚𝑚2


80 𝑚𝑚

60 𝑚𝑚

20 𝑚𝑚

G

dy

y

20 𝑚𝑚

O

320 𝑚𝑚 300 𝑚𝑚

380 𝑚𝑚

Eccentricty, e=distance of the neutral axis

from the centroidal axis.

∴ 𝑒 = −𝑅ℎ2

𝑅2 + ℎ2

𝑒 = −327.14 × 512.08

327.142 + 512.08= −1.56 mm

-ve sign indicates that neutral axis is

located below the centroidal axis.


80 𝑚𝑚

60 𝑚𝑚

20 𝑚𝑚

G

dy

y

20 𝑚𝑚

O

320 𝑚𝑚 300 𝑚𝑚

380 𝑚𝑚

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

stress of 108 N/𝑚𝑚2. What will be the

maximum compressive stress in hook

for that value of load?

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

Permissible stress, 𝜎 = 105 𝑁/𝑚𝑚2

Load P calculation:

Maximum compressive stress:

Area of cross section, A=𝑏1𝑡1 + 𝑏2𝑡2 + 𝑏3𝑡3 = 40 × 20 + 60 × 30 + 40 × 20

= 3400 𝑚𝑚2.

To find c.g. of the section from ‘O’,

𝑅 =40 × 20 × 140 + 40 × 20 × 110 + 60 × 30 × 75

3400= 98.5 mm.

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

ℎ2 =𝑅3

𝐴𝑏2 log𝑒

𝑅2


𝑅3

𝑅2+𝑏1 log𝑒

𝑅4

𝑅3− 𝑅2

ℎ2 =98.53

340060 log𝑒

90

60+ 20 log𝑒

130

90+ 40 log𝑒

150

130− 98.52

ℎ2 = 810.14 𝑚𝑚2.

Direct stress, 𝑃

𝐴=

𝑃

3400 N/𝑚𝑚2

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

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

The bending stress at any point is given by

𝜎𝑏 =𝑀

𝐴𝑅1 +

𝑅2

ℎ2

𝑦

𝑅 + 𝑦

=−98.5𝑃

3400 × 98.51 +

98.52

810.14

𝑦

98.5 + 𝑦

=−𝑃

34001 + 11.97

𝑦

98.5 + 𝑦

Maximum bending stress (tensile) occurs at

Y=-38.5 mm

𝜎𝑏 𝑚𝑎𝑥 =−𝑃

34001 + 11.97

−38.5

98.5 − 38.5

=𝑃

508.9 N/𝑚𝑚2

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

∴ 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎𝑡 𝑚𝑎𝑥 = 𝜎𝑑 + 𝜎𝑏 𝑚𝑎𝑥

108 =𝑃

3400+

𝑃

508.9

108 =𝑃

442.65

𝑃 = 108 × 442.65 = 47806.2 𝑁 = 47.8 𝑘𝑁

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

∴ Maximum compressive stress

Occurs at y = 𝑦2 = 51.5 𝑚𝑚, 𝜎𝑐 𝑚𝑎𝑥 = 𝜎𝑑 + 𝜎𝑐 𝑚𝑎𝑥

𝜎𝑐 𝑚𝑎𝑥 =𝑃

3400

−𝑃

34001 + 11.97

51.5

98.5 + 51.5

𝜎𝑐 𝑚𝑎𝑥 =47806.2

3400

−47806.2

34001 + 11.97

51.5

98.5 + 51.5

𝜎𝑐 𝑚𝑎𝑥 = 57.79 N/𝑚𝑚2

Curved beams

G

60 𝑚𝑚

40 𝑚𝑚

40 𝑚𝑚

dy

y

30 mm

O

𝑅2

60 𝑚𝑚 𝑅3

20 𝑚𝑚

20 𝑚𝑚

𝑅4

38.5 mm

51.5 mm

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’.

Shear centre

The point of intersection of the bending axis with the cross section of the beam is defined as ‘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:

Shear stress at a distance ‘𝑥′ in the bottom

horizontal member AB as shown in

Figure is given by ,

𝑞1 =𝐹𝐴𝑦

𝐼 𝑡

=𝐹

𝐼 𝑡× 𝑥𝑡 ×

ℎ

2

𝐹1 = 𝑞1𝑡 𝑑𝑥

𝑏

0

∴ 𝐹1 = 𝐹

𝐼 𝑡

𝑏

0

× 𝑥𝑡 ×ℎ

2× 𝑡 𝑑𝑥

Shear centre

b

𝑥

𝐹1

A B

C

𝛿𝑥

b

𝑡𝑓

𝐹1

𝑡𝑤

𝑥

𝑦

𝑧 G

𝐹2

h

a

c d

S

e

𝑡

T

T+ 𝛿𝑇 u

∴ 𝐹1 = 𝐹

𝐼 𝑡

𝑏

0

× 𝑥𝑡 ×ℎ

2× 𝑡 𝑑𝑥

=𝐹𝑡ℎ

2𝐼×

𝑏2

2=

𝐹𝑡ℎ𝑏2

4𝐼

The shear force 𝐹1is same in the top

member also.

Shearing stress in the vertical member BC

at a distance ′𝑦′ from B (as shown in Fig.)

𝑞2 =𝐹𝐴𝑦

𝐼 𝑡= 0 (since 𝑦 = 0)

∴ 𝐹2 = 0

Shear centre

b

𝑥

𝐹1

A B

C

𝛿𝑥

b

𝑡𝑓

𝐹1

𝑡𝑤

𝑥

𝑦

𝑧 G

𝐹2

h

a

c d

S

e

𝑡

T

T+ 𝛿𝑇 u

O

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

𝐹𝑡ℎ𝑏2

4𝐼× ℎ = 𝐹 × 𝑒

∴ 𝑒 =𝑏2ℎ2𝑡

4𝐼

Shear centre

b

𝑥

𝐹1

A B

C

𝛿𝑥

b

𝑡𝑓

𝐹1

𝑡𝑤

𝑥

𝑦

𝑧 G

𝐹2

h

a

c d

S

e

𝑡

T

T+ 𝛿𝑇 u

O

𝐹

Fig. Shear centre of an equal –leg angle section

Shear centre

G X

Y

S

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

Shear centre

S S

S

S S

Shear centre of a semi circular section:

𝑞 =𝐹𝐴𝑦

𝐼 𝑡

𝐴𝑦 = 𝑅 𝑑𝛼

𝜃

0

. 𝑡 × 𝑅 cos 𝛼 = 𝑅2t sin 𝜃

𝐼 = 𝑦2

𝜋

0

dA = 𝑅 cos 𝜃 2

𝜋

0

𝑅𝑑𝜃. 𝑡

=𝜋𝑅3𝑡

2

∴ 𝑞 =2𝐹

𝜋𝑅𝑡sin 𝜃

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

Shear centre

e O

R 𝜃

𝑑𝛼

𝛼 𝑞 𝑅co

s𝛼

𝑡

𝑑𝜃

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

= 𝑞𝑅2𝑡 𝑑𝜃 = 𝑞𝑅2𝑡 𝑑𝜃

𝜋

0

𝜋

0

= 2𝐹

𝜋𝑅𝑡𝑅2𝑡sin 𝜃 𝑑𝜃 = 𝐹

𝜋

0

×4𝑅

𝜋

This must be equal to 𝐹 × 𝑒

𝐹 × 𝑒 = 𝐹 ×4𝑅

𝜋

∴ 𝑒 =4𝑅

𝜋≅ 1.27𝑅

Shear centre

e O

R 𝜃

𝑑𝛼

𝛼 𝑞 𝑅co

s𝛼

𝑡

𝑑𝜃

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.

Shear centre, 𝑒 =𝑏2ℎ2𝑡

4𝐼

b= 12 cm, h=16 cm, t= 2 cm

𝐼 =1 × 163

12+ 2 ×

12 × 23

12+

2 × 12 × 2 × 92 = 4245.33 𝑐𝑚2

∴ 𝑒 =122 × 162× 2

4 × 4245.33= 4.3413 cm

Shear centre - Problem

b

𝑥

𝐹1

A B

C

𝛿𝑥

b

𝑡𝑓

𝐹1

𝑡𝑤

𝑥

𝑦

𝑧 G

𝐹2

h

a

c d

S

e

𝑡

T

T+ 𝛿𝑇 u

O

𝐹

• 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)


30 kN

1 2 G

d=12 cm

G

• 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)


• 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)


• Problem:

• Figure shows a frame subjected to a load of 3.4 kN, find the resultant stress at A and B. (Nov/Dec 2011)


48

18 3.4 kN


120

1 2 48

• 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)


• 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)


• 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)


1. Give the reasons for unsymmetrical bending.

2. Write Winkler Bach formula.

2 marks Questions and Answers

Documents

Unit V Advanced topics in bending of beams of... · Unit V Advanced topics in bending of beams ... Dr.P.Venkateswara rao, ... load of 4 kN inclined at 200 to the vertical and passing