Curved Beam - VTU e-Learning

CURVED BEAMS 

CONTENT: 

WHAT’S A CURVED BEAM BEAM? 

DIFFERENCE BETWEEN A STRAIGHT BEAM AND A 

CURVED BEAM 

WHY STRESS CONCENTRA 

CONCENTRATION TION OCCUR AT INNER SIDE OR 

CONCAVE SIDE OF CURV CURVED BEAM? 

DERIVATION FOR STRES STRESSES IN CURVED BEAM 

PROBLEMS.

Theory of Simple Bending 

Due to bending moment, tensile stress develops in one portion of section 

and compressive stress in the other portion across the depth. In between 

these two portions, there is a layer where stresses are zero. Such a layer 

is called neutral layer. Its trace on the cross section is called neutral axis. 

Assumption 

The material of the beam is perfectly homogeneous and isotropic. 

The cross section has an axis of symmetry in a plane along the 

length of the beam. 

The material of the beam obeys Hooke’s law. 

The transverse sections which are plane before bending remain 

plane after bending also. 

Each layer of the beam is free to expand or contract, independent of 

the layer above or below it. 

Young’s modulus is same in tension & compression. 

Consider a portion of beam between sections AB and CD as shown in 

the figure. 

Let e1f1 be the neutral axis and g1h1 an 

element at a distance y from neutral 

axis. Figure shows the same portion 

after bending. Let r be the 

radius of curvature and ѳ is the angle 

subtended by a1b1 and c1d1at centre of 

radius of curvature. Since it is a neutral 

axis, there is no change in its length (at 

neutral axis stresses are zero.) 

EF = E1F1 = RѲ

G1H1 = (R+Y) Ѳ 

Also Stress 

OR 

dF = 0 

GH = RRѲ 

∴ there is no direct force acting on the element considered.

Since Σyδa a is first moment of area about neutral axis, Σyδa/a Σ is the 

distance of centroid from neutral axis. Thus neutral axis coincides with 

centroid of the cross section. Cross sec sectional tional area coincides with neutral 

axis. 

From (1) and (2) 

CURVED BEAM 

Curved beams are the parts of machine members found in C - 

clamps, crane hooks, frames of presses, riveters, punches, shears, boring 

machines, planers etc. In straight beams the neutral axis of the section 

coincides with its centroidal axis and the stress distribution in the beam 

is linear. But in the case of curved beams the neutral axis of the section 

is shifted towards the centre of curvature of the beam causing a nonlinear 

[hyperbolic] distribution of stress. The neutral axis lies between 

the centroidal axis and the centre of curvature and will always be present 

within the curved beams.

Stresses in Curved Beam 

Consider a curved beam subjected to bending moment Mb as shown 

in the figure. The distribution of stress in curved flexural member is 

determined by using the following assumptions: 

i) The material of the beam is perfectly homogeneous [i.e., same 

material throughout] and isotropic [i.e., equal elastic properties in all 

directions] 

ii) The cross section has an axis of symmetry in a plane along the length 

of the beam. 

iii) The material of the beam obeys Hooke's law. 

iv) The transverse sections which are plane before bending remain plane 

after bending also. 

v) Each layer of the beam is free to expand or contract, independent of 

the layer above or below it. 

vi) The Young's modulus is same both in tension and compression. 

Derivation for stresses in curved beam 

Nomenclature used in curved beam 

Ci =Distance from neutral axis to inner radius of curved beam 

Co=Distance from neutral axis to outer radius of curved beam 

C1=Distance from centroidal axis to inner radius of curved beam 

C2= Distance from centroidal axis to outer radius of curved beam 

F = Applied load or Force 

A = Area of cross section 

L = Distance from force to centroidal axis at critical section 

σd= Direct stress 

σbi = Bending stress at the inner fiber 

σbo = Bending stress at the outer fiber 

σri = Combined stress at the inner fiber 

σro = Combined stress at the outer fiber

Stresses in curved beam 

Mb = Applied Bending Moment 

ri = Inner radius of curved beam 

ro = Outer radius of curved beam 

rc = Radius of centroidal axis 

rn = Radius of neutral axis 

CL = Center of curvature 

M b 

F 

F 

In the above figure the lines 'ab' and 'cd' represent two such planes 

before bending. i.e., when there are no stresses induced. When a bending 

moment 'Mb' ' is applied to the beam the plane cd rotates with respect to 

'ab' through an angle 'd 'dθ' ' to the position 'fg' and the outer fibers are 

shortened while the inner fibers s are elongated. The original length of a 

strip at a distance 'y' from the neutral axis is (y + rrn)θ. 

It is shortened by 

the amount ydθ and the stress in this fiber is, 

NA c 2 

CA 

σ = E.e 

Where σ = stress, e = strain and E = Young's Modulus 

c 1 

C L 

c i c o 

r i 

e 

r n 

r c 

r o 

F 

F 

M b

We know, stress σ = E.e 

We know, stress e 

 

 

i.e., σ = – E θ 

θ 

Ѳ 

Ѳ 

..... (i) 

Since the fiber is shortened, the stress induced in this fiber is 

compressive stress and hence negative sign. 

The load on the strip having thickness dy and cross sectional area dA is 

'dF' 

i.e., dF = σdA = – θ 

θ dA 

From the condition of equilibrium, the summation of forces over the 

whole cross-section is zero and the summation of the moments due to 

these forces is equal to the applied bending moment. 

Let 

M b = Applied Bending Moment 

r i = Inner radius of curved beam 

r o = Outer radius of curved beam

c = Radius of centroidal axis 

r n = Radius of neutral axis 

CL= Centre line of curvature 

Summation of forces over the whole cross section 

i.e. dF 0 

As θ 

θ 

∴ θ 

θ 

 

=0 

is not equal to zero, 

∴ 

 

= 0 ..... (ii) 

The neutral axis radius 'rn' can be determined from the above equation. 

If the moments are taken about the neutral axis, 

M b = – ydF 

Substituting the value of dF, we get 

M b = θ 

θ 

 

dA 

= θ 

θ y dA 

= θ 

θ 

ydA 

 

0 

Since ydA represents the statical moment of area, it may be replaced 

by A.e., the product of total area A and the distance 'e' from the 

centroidal axis to the neutral axis.

∴ M = b θ 

A.e ..... (iii) 

θ 

From equation (i) θ 

θ = – σ 

Substituting in equation (iii) 

M = – b σ . A. e. 

 

∴ σ = 

 

..... (iv) 

This is the general equation for the stress in a fiber at a distance 'y' from 

neutral axis. 

At the outer fiber, y = co 

∴ Bending stress at the outer fiber σbo i.e., σ = bo ( rn + co = ro) ..... (v) 

Where co = Distance from neutral axis to outer fiber. It is compressive 

stress and hence negative sign. At the inner fiber, y = – ci 

∴ Bending stress at the inner fiber 

σ bi = 

 

i.e., σ bi = 

 

( rn – ci = ri) ..... (vi) 

Where ci = Distance from neutral axis to inner fiber. It is tensile stress 

and hence positive sign.

Difference between a straight beam and a curved beam 

Sl.no straight beam curved beam 

1 In Straight beams the neutral 

axis of the section coincides 

with its centroidal axis and the 

stress distribution in the beam 

is linear. 

In case of curved beams the 

neutral axis of the section is 

shifted towards the center of 

curvature of the beam causing 

a non-linear stress 

distribution.

2 

3 Neutral axis and centroidal 

axis coincides 

Neutral axis is shifted 

towards the least centre of 

curvature 

Location of the neutral axis By considering a rectangular cross 

section

Centroidal and Neutral Axis of Typical Section of Curved Beams

Why stress concentration occur at inner side or concave side of 

curved beam 

Consider the elements of the curved beam lying between two axial 

planes ‘ab’ and ‘cd’ separated by angle θ. . Let fg is the final position of 

the plane cd having rotated through an angle ddθ 

about neutral axis. 

Consider two fibers symmetrically located on either side of the neutral 

axis. Deformation in both the fibers is same and equal to yd ydθ θ.

Since length of inner element is smaller than outer element, the strain 

induced and stress developed are higher for inner element than outer 

element as shown. 

Thus stress concentration occur at inner side or concave side of curved 

beam 

The actual magnitude of stress in the curved beam would be influenced 

by magnitude of curvature However, for a general comparison the stress 

distribution for the same section and same bending moment for the 

straight beam and the curved beam are shown in figure. 

It is observed that the neutral axis shifts inwards for the curved beam. 

This results in stress to be zero at this position, rather than at the centre 

of gravity. 

In cases where holes and discontinuities are provided in the beam, they 

should be preferably placed at the neutral axis, rather than that at the 

centroidal axis. This results in a better stress distribution. 

Example: 

For numerical analysis, consider the depth of the section ass twice 

the inner radius.

For a straight beam: 

Inner most fiber: 

Outer most fiber: 

For curved beam: h=2r h=2ri 

e = rc - rn = h – 0.910h = 0.0898h 

co = ro - rn= h – 0.910h = 0590h 

ci = rn - ri = 0.910h - 

= 0.410h

Comparing the stresses at the inner most fiber based on (1) and (3), we 

observe that the stress at the inner most fiber in this case is: 

σbci = 1.522σBSi 

Thus the stress at the inner most fiber for this case is 1.522 times greater 

than that for a straight beam. 

From the stress distribution it is observed that the maximum stress in a 

curved beam is higher than the straight beam. 

Comparing the stresses at the outer most fiber based on (2) and (4), we 

observe that the stress at the outer most fiber in this case is: 

σbco = 1.522σBSi 

Thus the stress at the inner most fiber for this case is 0.730 times that for 

a straight beam. 

The curvatures thus introduce a non linear stress distribution. 

This is due to the change in force flow lines, resulting in stress 

concentration on the inner side. 

To achieve a better stress distribution, section where the centroidal axis 

is the shifted towards the insides must be chosen, this tends to equalize 

the stress variation on the inside and outside fibers for a curved beam. 

Such sections are trapeziums, non symmetrical I section, and T sections. 

It should be noted that these sections should always be placed in a 

manner such that the centroidal axis is inwards. 

Problem no.1 

Plot the stress distribution about section A-B of the hook as shown in 

figure. 

Given data: 

ri = 50mm 

ro = 150mm 

F = 22X10 3 N 

b = 20mm 

h = 150-50 = 100mm 

A = bh = 20X100 = 2000mm 2

e = rc - rn = 100 - 91.024 = 8.976mm 

Section A-B B will be subjected to a combination of direct load and 

bending, due to the eccentricity of the force. 

Stress due to direct load will be, 

y = rn – r = 91.024 – r 

Mb = 22X10 3 X100 = 2.2X10 6 N-mm

Problem no.2 

Determine the value of “t” in the cross section of a curved beam as 

shown in fig such that the normal stress due to bending at the extreme 

fibers are numerically equal. 

Given data; 

Inner radius ri=150mm =150mm 

Outer radius ro=150+40+100 

=150+40+100 

=290mm 

Solution; 

From Figure Ci + CO = 40 + 100 

= 140mm…………… (1) 

Since the normal stresses due to bending at 

the extreme fiber are numerically equal we 

have, 

i.e Ci= 

Radius of neutral axis 

rn= 

= 0.51724C 0.51724Co…………… (2) 

rn =197.727 mm 

ai = 40mm; bi = 100mm; bb2 

=t; 

ao = 0; bo = 0; ri = 150mm; rro 

= 290mm;

= 

i.e., 4674.069+83.61t +83.61t = 4000+100t; 

∴ t = 41.126mm 

Problem no.3 

Determine the stresses at point A and B of the split ring shown in the 

figure. 

Solution: 

The figure shows the critical section of the split 

ring. 

Radius of centroidal axis rc = 80mm 

Inner radius of curved beam ri = 80 

Outer radius of curved beam ro = 80 + 


Applied force 

= 50mm 

= 110mm 

rn = 

= 

= 77.081mm 

F = 20kN = 20,000N (compressive)

Area of cross section A = π 

d2 = π 

x602 = 2827.433mm 2 

Distance from centroidal axis to force l = rc = 80mm 

Bending moment about centroidal axis Mb = Fl = 20,000x80 

=16x10 5 N-mm 

Distance of neutral axis to centroidal axis 

e = rc rn 

= 80 77.081=2.919mm 

Distance of neutral axis to inner radius 

ci = rn ri 

= 77.081 50=27.081mm 

Distance of neutral axis to outer radius 

Direct stress σd = 

co = ro rn 

= 110 77.081=32.919mm 

 

= 

. 

= 7.0736N/mm 2 (comp.) 

Bending stress at the inner fiber σbi = 

 

= . 

.. 

Bending stress at the outer fiber σbo = = 

Combined stress at the inner fiber 

= 105N/mm 2 (compressive) 

 

. 

.. 

= 58.016N/mm 2 (tensile)

σri = σd + σbi 

= 7.0736 105.00 

= - 112.0736N/mm 2 (compressive) 

Combined stress at the outer fiber 

σro = σd + σbo 

= 7.0736+58.016 

= 50.9424N/mm 2 (tensile) 

Maximum shear stress 

The figure. 

τmax = 0.5x σmax 

= 0.5x112.0736 

= 56.0368N/mm 2 , at B

Problem No. 4 

Curved bar of rectangular section 40x60mm and a mean radius of 

100mm is subjected to a bending moment of 2KN 2KN-m m tending to 

straighten the bar. Find the position of the Neutral axis and draw a 

diagram to show the variation of stress across the section. 

Solution 

Given data: 

b= 40mm h= 60mm 

rc=100mm Mb= = 2x10 

C1=C2= 30mm 

6 N-mm 

30mm 

rn= 

ro= rc+h/2=100+30=130 =130 =(r =(ri+c1+c2) 

ri= rc- h/2 = 100 - 30= 70mm (rc-c1) 

rn= 96.924mm 


e = rc - rn= 100-96.924 

=3.075mm 


ci= rn- ri = (c1-e) = 26.925mm 


co=c2+e= (ro-rn) = 33.075mm 

Area 

A= bxh = 40x60 = 2400 mm 2 

e) = 26.925mm 

) = 33.075mm 

Bending stress at the inner fiber σbi = = 

= 104.239 N/mm 2 (compressive)

Bending stress at the outer fiber σbo = = 

. . 

= -68.94 N/mm 2 (tensile) 

Bending stress at the centroidal axis = 

 

= 

 

= -8.33 N/mm 2 (Compressive) 

The stress distribution at the inner and outer fiber is as shown in the 

figure.

Problem No. 5 

The section of a crane hook is a trapezium; the inner face is b and is at a 

distance of 120mm from the centre line of curvature. The outer face is 

25mm and depth of trapezium =120mm.Find the proper value of b, if the 

extreme fiber stresses due to pure bending are numerically equal, if the 

section is subjected to a couple which develop a maximum fiber stress of 

60Mpa.Determine the magnitude of the couple. 

Solution 

ri = 120mm; bi = b; bo= 25mm; h = 120mm 

σbi = σbo = 60MPa 

Since the extreme fibers stresses due to pure bending are numerically 

equal we have, 

We have, 

But h= ci + co 

120 = ci+2ci 

 

 

= 

 

Ci/ri =co/ro =ci/co =120/240 

2ci=co 

Ci=40mm; co=80mm 

rn= ri + ci = 120+40 =160 mm

=150.34mm 

To find the centroidal axis, (C2) 

bo= 125.84mm; b=25mm; h=120mm 

= 74.313mm. 

But C1=C2 

rc= ro-c2 =240 - 74.313 =165.687mm 

e=rc- rn = 165.687 - 160 = 5.6869 mm 

Bending stress in the outer fiber, 

σ M c 

Aer 

A= . 

 

= 1050.4mm 

60 = 

 

.. 

Mb=10.8x10 6 N-mm

Problem no.6 

Determine the stresses at point A and B of the split ring shown in 

fig.1.9a 

Solution: 

Redraw the critical section as shown in the figure. 


Inner radius of curved beam ri = 80 

Outer radius of curved beam ro = 80 + 

Radius of neutral axis rn = 

 

= √√ 

 

 

 

= 50mm 

= 110mm 

=77.081mm 

Applied force F = 20kN = 20,000N (compressive) 

Area of cross section A = π 

d2 = π 

x602 = 2827.433mm 2 

Distance from centroidal axis to force l = rc = 80mm 

Bending moment about centroidal axis Mb = FI = 20,000x80 

=16x10 5 N-mm 

Distance of neutral axis to centroidal axis e = rc rn 

= 80 77.081 

=2.919mm


ci = rn ri = 77.081 50 = 27.081mm 


co = ro rn = 110 77.081 = 32.919mm 


 

= 

. 

= 7.0736N/mm 2 (comp.) 


 

Bending stress at the outer fiber σbo = 

 

Combined stress at the inner fiber 

σri = σd + σbi = 7.0736 105.00 

= . 

.. 

= 105N/mm 2 (compressive) 

= 

. 

.. 

= 58.016N/mm 2 (tensile) 

= 112.0736N/mm 2 (compressive) 

Combined stress at the outer fiber 

σro = σd + σb = 7.0736+58.016 

Maximum shear stress 

= 50.9424N/mm 2 (tensile) 

max = 0.5x σmax = 0.5x112.0736 

= 56.0368N/mm 2 , at B

The figure shows the stress distribution in the critical section.

Problem no.7 

Determine the maximum tensile, compressive and shear stress induced 

in a ‘c’ frame of a hydraulic portable riveter shown in fig.1.6a 

Solution: 

Draw the critical section as shown in 

the figure. 

80 

Inner radius of curved beam ri = 

100mm 

Outer radius of curved beam ro = 100+80 

= 180mm 

Radius of centroidal axis rc = 100+ 

= 140mm 

Radius of neutral axis rn = ln 

 

 

= ln 

 

 

50 

 

= 136.1038mm 


e = rc - rn = 140-136.1038 = 3.8962mm 

Critical 

Section 

R100 

b = 50 mm 

175 mm 

c 2 

c o 

h = 80mm 

e 

CA 

NA 

r o 

9000N 

c 1 

c i 

rn rc r = 100mm 

i 

C L 

175mm 

F 

F


ci = rn - ri = 136.1038-100 = 36.1038mm 


co = ro - rn = 180-136.1038 = 43.8962mm 

Distance from centroidal axis to force 

Applied force F = 9000N 

l = 175+ rc = 175+140 = 315mm 

Area of cross section A = 50x80 = 4000mm 2 

Bending moment about centroidal axis Mb = FI = 9000x315 


 

= 2835000 N-mm 

= 

= 2.25N/mm2 (tensile) 


 


= . 

. 


 

= . 

. 


Combined stress at the inner fiber σri = σd + σbi = 2.25+65.676 


Combined stress at the outer fiber σro = σd + σbo = 2.25 44.362 


Maximum shear stress max = 0.5x σmax = 0.5x67.926 

= 33.963 N/mm 2 , at the inner fiber 

The stress distribution on the critical section is as shown in the figure. 

Combined stress = -42.112 N/mm 

σ ro 

2 

Bending stress σbo=-44.362 N/mm 

Direct stress ( σd) 2 

CA 

NA 

h =80 mm 

σ ri=67.926 N/mm 2 

σ bi=65.676 N/mm 2 

σ d=2.25 N/mm 2 

b = 50 mm

Problem no.8 

The frame punch press is shown in fig. 1.7s. Find the stress in inner and 

outer surface at section A-B the frame if F = 5000N 

Solution: 

Draw the critical section as shown in the 

figure. 

Inner radius of curved beam ri = 25mm 


= 65mm 

Distance of centroidal axis from inner fiber c1 = 

= 

 

 

b = 6 mm 

o 

c 2 

c o 

 

 

= 16.667mm 

h = 40mm 

e 

CA 

NA 

 

r o 

c 1 

c i 

r c 

r n 

b = 18 mm 

i 

r = 25mm 

i 

C L 

100mm 

F 

F

∴ Radius of centroidal axis rc = ri c1 

Radius of neutral axis rn = 

= 25+16.667 = 41.667 mm 

= 

 

 

 

 

 

 

 

 

 

 


Distance of neutral axis to inner radius ci = rn ri 

Distance of neutral axis to outer radius co = ro rn 

=38.8175mm 

 

 

= 1.66738.8175 

=2.8495mm 

= 38.817525=13.8175mm 

= 65-38.8175=26.1825mm 

Distance from centroidal axis to force l = 100+ rc = 100+41.667 

Applied force F = 5000N 

Area of cross section A = 

b b = 

 

= 141.667mm 

x4018 6 = 480mm2 

Bending moment about centroidal axis Mb = FI = 5000x141.667 

= 708335 N-mm


 

= 



 


 

= . 

. 

= 286.232N/mm 2 (tensile) 

= . 

. 



= 296.649N/mm 2 (tensile) 

Combined stress at the outer fiber σro = σd + σbo = 10.417286.232 





Combined stress =-198.189 N/mm 

σ ro 

2 


Direct stress (σ d) 

b = 6 mm 

o 

2 

CA 

NA 

h =40 mm 

σ ri=296.649 N/mm 2 

σ bi=286.232 N/mm 2 

σ d=10.417 N/mm 2 

b = 18 mm 

i

Problem no.9 

Figure shows a frame of a punching machine and its various dimensions. 

Determine the maximum stress in the frame, if it has to resist a force of 

85kN 

Solution: 


figure. 


Outer radius of curved beam ro = 550mm 


rn = 

 

 

 

 

 

ai = 75mm; bi = 300mm; b2 = 75mm; ao = 0; bo = 0 

A=a1+a2=75x300+75x225 =39375mm 2 

∴ rn = 

75 

Let AB be the ref. line 

550 

75 

300 

250 

750 mm 

85 kN 

b =75mm 

2 

 

 

= 333.217mm 

 

 

c o 

225 mm 

CA 

a 2 

e 

NA 

X 

c i 

a = 75mm 

a 1 

r =550 mm 

o 

i 

B 

b=300mm 

i 

A 

r n 

r c 

r = 250 mm 

i 

C L 

750 

F 

F

x 

 

= 

 

 

= 101.785mm 

 

Radius of centroidal axis rc = ri +x 

= 250+101.785=351.785 mm 


Distance of neutral axis to inner radius ci = rn ri 

Distance of neutral axis to outer radius co = ro rn 

Distance from centroidal axis to force l = 750+ rc 

Applied force F = 85kN 

Bending moment about centroidal axis Mb = FI 


 

= 351.785-333.217=18.568mm 

= 333.217 250=83.217mm 

= 550 333.217=216.783mm 

= 750+351.785 = 1101.785mm 

= 85000x1101.785 

= 93651725N-mm 

= 



 

= . 

. 


Bending stress at the outer fiber σbo = = 

. 

. 







The below figure shows the stress distribution. 

Combined stress =-48.33 N/mm 

σ ro 

2 


Direct stress ( σd) 2 

b = 75 mm 

2 

= 24.165N/mm 2 , at the outer fiber 

a 2 

CA 

225 

NA 

a =75mm 

i 

a 1 

σ ri=44.8 N/mm 2 

σ bi=42.64 N/mm 2 

b = 300 mm 

i 

σ d=2.16 N/mm 2

Problem no.10 

Compute the combined stress at the inner and outer fibers in the critical 

cross section of a crane hook is required to lift loads up to 25kN. The 

hook has trapezoidal cross section with parallel sides 60mm and 30mm, 

the distance between them being 90mm .The inner radius of the hook is 

100mm. The load line is nearer to the surface of the hook by 25 mm the 

centre of curvature at the critical 

section. What will be the stress at inner 

and outer fiber, if the beam is treated as 

straight beam for the given load? 

Solution: 

Draw the critical section as shown in the figure. 


Outer radius of curved beam ro = 100+90 = 

190mm 

Distance of centroidal axis from inner fiber 

c1 = 

= 40mm 

= 

 

x 

 

100 mm 

30mm 

90mm 

60mm 25mm 

F = 25 kN 

NA 

CA 

h = 90 mm 

c 2 

c o 

e 

c 1 

c i 

l 

ri rn rc ro F C L

Radius of centroidal axis rc = ri + c1 = 100+40 


= 

= 140 mm 

 

 

 

 

 

 

 

 

 

 

 

 

 


= 135.42mm 

= 140 135.42=4.58mm 

Distance of neutral axis to inner radius ci = rn ri = 135.42 100 

=35.42mm 

Distance of neutral axis to outer radius co = ro rn = 190 135.42 

= 54.58mm 

Distance from centroidal axis to force l = rc 25= 140 25 

Applied force F = 25,000N = 25kN 

= 115mm 


b b = 

x90x60 30 = 4050mm2 

 

Bending moment about centroidal axis Mb = FI = 25,000x115 


 

= 2875000 N-mm 

= 

= 6.173N/mm2 (tensile)


 


 

= . 

. 

= 54.9 N/mm 2 (tensile) 

= . 

. 






Maximum shear stress τmax = 0.5x σmax = 0.5x61.072 



) Beam is treated as straight beam 

From DDHB refer table, 

b = 30mm 

bo = 60-30 = 30mm 

h = 90 

c1 = 40mm 

c2 = 90-50 = 40mm 

A = 4050 mm 2 

Mb = 28750000 N/mm 2 

Also 

C2 = 

 

---------------------- From DDHB

C2 = 

= 50mm 

 

c1 = 90-50= 40mm 

Moment of inertia I = 

Direct stress σb = 

 

= 

 

= 2632500mm 4 

 

= 



 

Bending stress at the outer fiber σbo = - 

 

= 

2632500 

= 43.685 N/mm 2 (tensile) 

= 2875000x50 

 

= -54.606N/mm 2 (compressive) 



Combined stress at the outer fiber σro = σd + σbo = 6.173-54.606 

= -48.433N/mm 2 (compressive) 

The stress distribution on the straight beam is as shown in the figure.

σro N/mm 2 

=-48.433 

σ bo=-54.606 N/mm 2 

σ d 

b = 30 mm 

c =50mm 

2 

NA, CA 

h =90 mm 

c =40mm 

1 

b 

σ ri= 49.858 N/mm 2 

σ bi= 43.685 N/mm 2 

σ d= 6.173 N/mm 2 

b /2 = 15 

o 

b /2 = 15 

o 

60 mm

Problem no.11 

The section of a crane hook is rectangular in shape whose width is 

30mm and depth is 60mm. The centre of curvature of the section is at 

distance of 125mm from the inside section and the load line is 100mm 

from the same point. Find the capacity of hook if the allowable stress in 

tension is 75N/mm 2 

Solution: 


figure. 



= 185mm 

Radius of centroidal axis rc =100+ 

 

= 130mm 

Radius of neutral axis rn = ln 

 

 

= 153.045mm 

h=60mm 

b=30mm 

 

= ln 

 

 

125 mm 

Distance of neutral axis to centroidal axis e = rc - rn 

b = 30 mm 

100 

F = ? 

c 2 

c o 

h = 60mm 

e 

CA 

NA 

= 155-153.045 = 1.955mm 

r o 

c 1 

c i 

l 

rn rc Load line 

100 

r = 125mm 

i 

F 

C L

Distance of neutral axis to inner radius ci = rn - ri 

Distance of neutral axis to outer radius co = ro - rn 

= 153.045-125 = 28.045mm 

= 185-153.045 = 31.955mm 

Distance from centroidal axis to force l = rc -25 = 155-25 = 130mm 

Area of cross section A = bh = 30x60 = 1800mm 2 

Bending moment about centroidal axis Mb = Fl = Fx130 


 

= 

 


 

= 130F 

+ 

 

Combined stress at the inner fiber σri = σd + σbi 

i.e., 75 = . 

+ 

. 

F = 8480.4N =Capacity of the hook.

Problem no.12 

Design of steel crane hook to have a capacity of 100kN. Assume factor 

of safety (FS) = 2 and trapezoidal section. 

Data: Load capacity F = 100kN = 10 5 N; 

Trapezoidal section; FS = 2 

Solution: Approximately 1kgf = 10N 

∴ 10 5 = 10,000 kgf =10t 

Selection the standard crane hook dimensions from table 25.3 when safe 

load =10t and steel (MS) 

∴ c =11933; Z = 14mm; M = 71mm and 

h = 111mm 

bi= M = 7133 

bo = 2xZ = 2x14 = 28 mm 

r1 = 

= = 59.5mm 

 

h = 111mm 

b o 

Z 

H 

M = b i 

CA 

NA 

h = 111 mm 

c 2 

c o 

e 

c 1 

b =28 

o b =71 

i 

c i 

r =59.5 mm 

i 

r n 

r = 

c 

r o 

l 

C L 

F

Assume the load line passes through the centre of hook. Draw the 

critical section as shown in the figure. 

Inner radius of curved beam ri = 59.5mm 

Outer radius of curved beam ro = 59.5+111 = 170.5mm 


= 

 

 

 

 

.. 

 

= 98.095mm 

 

 

 

 

. 

 

. 

Distance of centroidal axis from inner fiber c1 = 

Radius of centroidal axis rc = ri + c1 

= 

 

 

 

= 47.465+59.5= 106.965 mm 


=106.965-98.095 =8.87mm 

Distance of neutral axis to inner radius ci = rn - ri 

= 98.095-59.5=38.595mm 


= 170.5-98.095=72.0405mm 

Distance from centroidal axis to force l = rc -106.965 

= 47.465mm

Applied force F = 10 5 N 


b b 

= 

x111x71 28 = 5494.5mm2 

 

Bending moment about centroidal axis Mb = Fl = 10 5 x141.667 


 

= 

. 



 


= 106.965x10 5 N-mm 

= . . 

... 

= 142.365/mm 2 (tensile) 

 

= . . 

.5x8.. 

= -93.2 N/mm 2 (compressive) 


= 160.565N/mm 2 (tensile) 

Combined stress at the outer fiber σro = σd + σbo = 18.2-93.2 

= -75N/mm 2 (compressive) 

Maximum shear stress τmax = 0.5x σmax = 0. 160.565 



σro N/mm 2 

=-75 

σ bo=-93.2 N/mm 2 

σ d 

b = 28 mm 

o 

h = 111 mm 

CA 

NA 

σ ri=160.565 N/mm 2 

σ bi=142,365 N/mm 2 

σ d=18.2 N/mm 2 

b = 71 mm 

i

Problem no.13 

The figure shows a loaded offset bar. What is the maximum offset 

distance ’x’ if the allowable stress in tension is limited to 50N/mm 2 

Solution: 

Draw the critical section as shown in the figure. 


Inner radius ri = 100 – 100/2 = 50mm 

Outer radius ro = 100 + 100/2 = 150mm 

Radius of neutral axis rn = r 2 

o√ri 

4 

93.3mm 

mm 

= √150√50 

2 

= 

4 

e = rc - rn = 100 - 93.3 = 6.7mm 

ci = rn – ri = 93.3 – 50 = 43.3 

co = ro - rn = 150 - 93.3 = 

56.7mm 

A = 

x d2 = 

x 1002 = 7853.98mm 2 

Mb = Fx = 5000 x 

Combined maximum stress at the inner fiber 

(i.e., at B)

σri= Direct stress + bending stress 

= 

 

 

50 . 

 

. .. 

∴ 

x= 599.9 = Maximum offset distance. 

Problem no.14 

An Open ‘S’ made from 25mm diameter 

rod as shown in the figure determine the 

maximum tensile, compressive and shear 

stress 

Solution: 

(I) Consider the section P-Q

Draw the critical section at PP-Q 

as shown in the figure. 

Radius of centroidal axis rc =100mm 

Inner radius ri =100 

Outer radius ro = 100+ 


rn = = 

= 99.6mm 

= 87.5mm 

+ = 112.5mm 

Distance of neutral axis from centroidal axis e =r =rc - rn 

Distance of neutral axis to inner fiber cci 

= rn – ri 

=100 - 99.6 = 0.4mm 

= 99.6 – 87.5 =12.1 mm

Distance of neutral axis to outer fiber co = ro -rn 

=112.5 – 99.6 = 12.9 mm 

Area of cross-section A = π 

4 d2 = π 

4 x252 = 490.87 mm 2 

Distance from centroidal axis I = rc = 100mm 

Bending moment about centroidal axis Mb = F.l = 100 x 100 

= 100000Nmm 

Combined stress at the outer fiber (i.e., at Q) =Direct stress +bending 

stress 

σro= F MbCo 

- = 

A Aeo 

1000 

490.87 – 

100000 X12.9 

490.87 X 0.4 X 112.5 

= - 56.36 N/mm2 (compressive) 

Combined stress at inner fibre (i.e., at p) 

σri= Direct stress + bending stress 

= F 

A +M bc i 

Aer i 

= 1000 

490.87 

= 72.466 N/MM2 (tensile) 

(ii) Consider the section R -S 

+ 100000 X 12.1 

490.87 X 0.4 X 87.5 

Redraw the critical section at R –S as shown in fig. 

rc = 75mm 

ri = 75 - 25 

2 

=62.5 mm

25 

ro = 75 + 

2 

= 87.5 mm 

A = π 

4 d2 = π 

4 X 252 = 490.87 mm 2 

rn = r 

2 

o √ri 

4 

= 

√87.5 √62.5 

e = rc - rn = 75 -74.4755 =0.5254 mm 

4 

 

2 

=74.4755 mm 

ci = rn - ri =74.4755 – 62.5 =11.9755 mm 

co = ro - rn = 87.5 – 74.4755 = 13.0245 mm 

l = rc = 75 mm 

Mb = Fl = 1000 X 75 = 75000 Nmm 

Combined stress at the outer fibre (at R) = Direct stress + Bending stress 

σro = F 

A – M b c o 

Aero 

= 1000 

490.87 - 

= - 41.324 N/mm 2 (compressive) 

75000 X13.0245 

490.87 X 0.5245 X 62.5

Combined stress at the inner fiber (at S) = Direct stress + Bending stress 

σri = F 

A 

+ Mbco 

Aero 

= 1000 

490.87 + 

= 55.816 N/mm 2 (tensile) 

75000 X 11.9755 

490.87 X 0.5245 X 62.5 

∴ Maximum tensile stress = 72.466 N/mm 2 at P 

Maximum compressive stress = 56.36 N/mm 2 at Q 

Maximum shear stress τmax=0.5 σmax= 0.5 X 72.466 

Stresses in Closed Ring 

= 36.233 N/mm 2 at P 

Consider a thin circular ring subjected to symmetrical load F as shown 

in the figure.

The ring is symmetrical and is loaded symmetrically in both the 

horizontal and vertical directions. 

Consider the horizontal section as shown in the figure. At the two ends 

A and B, the vertical forces would be F/2. 

No horizontal forces would be there at A and B. this argument can be 

proved by understanding that since the ring and the external forces are 

symmetrical, the reactions too must be symmetri symmetrical. 

Assume that two horizontal inward forces H, act at A and B in the upper 

half, as shown in the figure. In this case, the lower 

half must have forces H acting outwards as shown. 

This however, results in violation of symmetry and 

hence H must be zero. BBesides 

the forces, moments 

of equal magnitude M0 act at A and B. It should be 

noted that these moments do not violate the 

condition of symmetry. Thus loads on the section can 

be treated as that shown in the figure. The unknown 

quantity is M0. . Again Consi Considering symmetry, We 

conclude that the tangents at A and B must be 

vertical and must remain so after deflection or MM0 

does not rotate. By Castigliano’s theorem theorem, the partial 

derivative of the strain energy with respect to the load gives the 

displacement of the he load. In this case, this would be zero. 

……………………….(1)

The bending moment at any point C, located at angle θ, , as shown in the 

figure. 

Will be 

As per Castigliano’s theorem, 

From equation (2) 

And, ds = Rdθ 

………..(2)

As this quantity is positive the direction assumed for MMo 

is correct and it 

produces tension in the inner fibers and compression on the outer. 

It should be noted that these equations are valid in the region, 

θ = 0 to θ = 90 0 . 

The bending moment Mbb 

at any angle θ from equation (2) 2) will be:

It is seen that numerically, MMb-max 

is greater than Mo. 

The stress at any angle Ѳ can be found by considering the 

forces as shown in the figure. 

Put θ = 0 in Bending moment equation (4) then we will 

get, 

At A-A Mbi = 0.181FR 

Mbo = - 0.181FR 

And θ = 90, 

At B-B Mbi = - 0.318FR 

Mbo = 0.318FR 

The vertical force F/2 2 can be resolved in two 

components (creates normal direct stresses) and S 

(creates shear stresses). 

The combined normal stress across any section will be:

The stress at inner (σ1Ai) ) and outer points ( (σ1Ao) at A-A A will be (at Ѳ = 0) 

On similar lines, the stress at the point of application of load at i 

outer points will be (at Ѳ = 90 0 lines, the stress at the point of application of load at inner and 

) 

It should be noted that in calculating the bending stresses, it is assumed 

that the radius is large compared to the depth, or the beam is almost a 

straight beam.

A Thin Extended Closed Link 

Consider a thin closed ring subjected to symmetrical load F as shown in 

the figure. At the two ends C and D, the vertical forces would be F/2. 

No horizontal forces would be there at C and D, as discussed earlier 

ring. 

The unknown quantity is MM0. 

Again considering symmetry, we 

conclude that the tangents at C and D must be vertical and must remain 

so after deflection or M0 does not rotate. 

There are two regions to be considered in this case: 

The straight portion, (0 < y < L) where 

Mb b = MO 

The curved portion, where

As per Castigliano’s theorem 

\

It can be observed that at L = 0 equation reduces to the same expression 

as obtained for a circular ring. Mo produces tension in the inner fibers 

and Compression on the outer. 

The bending moment Mbb 

at any angle Ѳ will be 

Noting that the equation are valid in the region, Ѳ = 0 to Ѳ = p/2, p/2 

At Ѳ = 0 

At section B-B B bending moment at inner and outer side of the fiber is 

At section A-A A the load point, i.e., at Ѳ = p/2, the maximum value of 

bending moment occurs (numerically), as it is observed that the second 

part of the equation is much greater than the first part. 

It can be observed that at L = 0, equation (v) reduces to the same 

expression as obtained for a circular ring. 

It is seen that numerically, MMb-max 

is greater than Mo.

The stress at any angle Ѳ can be found by considering the force as 

shown in the figure. 

The vertical force F/2 2 can be resolved in two components (creates 

normal direct stresses) and S (creates shear stresses). 

The combined normal stress across any section will be 

The stress at inner fiber 

be (at Ѳ = 0): 

The stress at inner fiber 

will be (at the loading point Ѳ = 90 0 ): 

fiber σ1Bi and outer fiber σ1Bo and at section B 

and at section B-B will 

fiber σ1Ai and outer fiber σ1Ao and at section A 

and at section A-A

Problem 15 

Determine the stress induced in a circular ring of circular cross section 

of 25 mm diameter subjected to a tensile load 6500N. The inner 

diameter of the ring is 60 mm. 

Solution: the circular ring and its critical section are as shown in fig. 

1.29a and 1.29b respectively. 

Inner radius ri = = 30mm 

Outer radius = 30+25 = 55mm 

Radius of centroidal axis rrc 

= 30 + = 42.5mm 

Radius of neutral axis rrn 

= 

= =42.5mm 

Distance of neutral axis to centroidal axis e = rrc 

- rn 

= 42.5 – 41.56 = 0.94mm 

Distance of neutral axis to inner radius cci 

= rn - ri 

= 41.56 – 30 = 11.56mm


= 55 - 41.56 = 13.44mm 

Direct stress at any cross section at an angle θ with horizontal 

σd = 

θ 

 

Consider the cross section A – A 

At section A – A, θ = 90 0 with respect to horizontal 

Direct stress σd = 

 

 

Bending moment Mb = - 0.318Fr 

= 0 

Where r = rc, negative sign refers to tensile load 

M = - 0.318x6500x42.5 = -87847.5 N-mm 

This couple produces compressive stress at the inner fiber and tensile 

stress at the at outer fiber 

Maximum stress at the inner fiber σ =Direct stress + Bending stress 

= 0 - 

 

= .. 

.. 

= - 73.36N/mm 2 (compressive) 

Maximum stress at outer fiber σ = Direct stress + Bending stress 

=0+ 

 

Consider the cross section B – B 

= .. 

.. 


At section B – B, θ = 0 0 with respect to horizontal 


 

2A 

Bending moment Mb = 0.182Fr 

= cos 

. 

Where r = rc, positive sign refers to tensile load 

M = 0.182x6500x42.5 = 50277.5 N-mm 

= 6.621 N/mm2 




= σd - 

 

= 6.621 + .. 

.874x0. 



= σd + 

 

=6.621+ .. 

.. 

= -20 N/mm 2 (compressive)

Problem 16 

Determine the stress induced in a circular ring of circular cross section 

of 50 mm diameter rod subjected to a compressive load of 20kNN. The 

mean diameter of the ring is 100 mm. 

Solution: the circular ring and its critical section are as shown in fig. 

1.30a and 1.30b respectively. 

Inner radius ri = - = 25mm 

Outer radius = + = 75mm 


= = 50mm 


= 

Distance of neutral axis to centroidal axis e = rrc 

- rn 

= 50 - 46.65 = 3.35mm 


= rn - ri 

= 46.65 46.65-25 = 21.65 mm 


= 75 - 46.65 = 28.35mm 

= = 46.65mm 

Area of cross section A = x55 2 = 1963.5mm 2


σd = 

 

 

Consider the cross section A – A 



 

 

Bending moment Mb = + 0.318Fr 

= 0 

Where r = rc, positive sign refers to tensile load 

= + 0.318x20000x50 = 318000 N-mm 



Maximum stress at the inner fiber =Direct stress + Bending stress 

= 0 + 

 

= 31800021. 

.. 


Maximum stress at outer fiber = Direct stress + Bending stress 

=0 - 

 

= - .35 

.. 

= - 18.27 N/mm 2 (compressive)

Consider the cross section B – B 



 

 

Bending moment Mb = -0.1828Fr 

= 

. 

= 5.093 N/mm 2 (compressive) 

Where r = rc, negative sign refers to tensile load 

M = - 0.182x20000x50 = -182000 N-mm 




= σd - 

Aer 

= -5.093 + . 

.5x3. 



= σd + 

 

= -5.093 + . 

.. 

= 5.366 N/mm 2 (tensile)

Problem 17 

A chain link is made of 40 mm diameter rod is circular at each end the 

mean diameter of which is 80mm. The straight sides of the link are also 

80mm. The straight sides of the link are also 80mm.If the link carries a 

load of 90kN; estimate the tensile and compress compressive ive stress in the link 

along the section of load line. Also find the stress at a section 90 from the load line 

0 away 

Solution: refer figure 

= 80mm; dc = 80mm; 

F = 90kN = 90000N 

Draw the critical cross section as shown in fig.1.32 

Inner radius ri = 40 - = 20mm 

Outer radius = + = 60mm 


= 40mm 


= 

rc = 40mm; 

= = 37.32mm 


=40-37.32 = 2.68mm 


= rn - ri 

= 37.32-20 = 17.32 mm


= 60 – 37.32 = 22.68mm 


σd = 

θ 

 

Consider the cross section A – A [i.e., Along the load line] 



 

 

= 0 

Bending moment M = - 

π 

M = 

π 

where r = rc, 

= 1.4x10 6 N-mm 




= 0 + 

 

= . . 

π 

. 

= - 360 N/mm 2 (tensile) 


= 0 - 

 

= - . . 

π 

. 


Consider the cross section B – B [i.e., 90 0 away from the load line] 



(compressive) 

Bending moment M = - π 

π 

M = π 

π 

θ 

= . 

π 

= 35.81 N/mm 2 

where r = rc, 

= - 399655.7N-mm 




= σd - 

 

= 35.81 + .. 

π 

. 

= 138.578 N/mm 2 (tensile) 


= σd + 

 

= 35.81 - .. 

π 

. 


Maximum tensile stress occurs at outer fiber of section A –A and 

maximum compressive stress occurs at the inner fiber of section A –A.

Using usual notations prove that the moment of resistance M of a 

curved beam of initial radius RR1 

when bent to a radius R2 R by 

uniform bending moment is 

M = EAeR1 

Consider a curved beam of uniform cross section as shown in Figure 

below. Its transverse section is symmetric with respect to the y axis and 

in its unstressed state; its upper and lower surfaces intersect the vertical 

xy plane along the arcs of circle AAB 

B and EF centered at O [Fig. 1(a)]. 1( 

Now apply two equal and opposite couples M and d M' as shown in Fig. 1. 

(c). The length of neutral surface remains the same. θ and θ' are the 

central angles before and after applying the moment M. Since the length 

of neutral surface remains the same 

R1θ =R2θ'' 

..... (i) 

Figure 

Consider the arc of circle JK located at a distance y above the neutral 

surface. Let r1 and r2 be the radius of this arc before and after bending 

couples have been applied. Now, the deformation of JK, 

.... (ii) 

From Fig. 1.2 a and c, rr1 

= R1 – y; r2 = R2 – y ..... (iii)

∴ δ = (R2 – y) θ' – (R1 – y) θ 

=R2θ'' 

– θ'y – R1θ + θy 

= – y ( (θ' – θ) [ R1θ = R2θ' ' from equ (i)] 

∴ δ =– y∆θ ∆θ 

[ θ' − θ = θ + ∆θ − θ = ∆θ] ..... (iv) 

The normal strain ∈∈x 

in the element of JK is obtained by dividing 

the deformation δ by the original length rr1θ 

of arc JK. 

∴ ∈x = 

The normal stress σxx 

may be obtained from Hooke's law σx = E∈x 

∴ σx =–E 

i.e. σx = – E 

( r1 = R1 – y) .(vii) 

Equation (vi) shows that the normal stress σx does not vary linearly 

with the distance y from the neutral surface. Plotting σx versus y, we 

obtain an arc of hyperbola as shown in Fig. 1.3. 

From the condition of equilibrium the summation of forces over the 

entire area is zero and the summation of the moments due to these forces 

is equal to the applied bending moment. 

∴ ∫δF =0 

i.e., ∫σxdA =0 

..... (viii) 

and ∫(– yσxdA)=M 

..... (v) 

..... (vi) 

..... (ix)

Substituting the value of the σx from equation (vii) into equation 

(viii) 

– ∆ 

dA=0 

Since ∆ 

is not equal to zero dA = 0 

i.e. R1 

=0 

1 i.e., . R1 

=0 

1 ∴ R1 = 

 

1 ∴ It follows the distance R1 from the centre of curvature O to the 

neutral surface is obtained by the relation R1 = 

..... (x) 

The value of R1 is not equal to the 1 distance from O to the centroid 

of the cross-section, since 1 is obtained by the relation, 

1 = 

1 ..... (xi) 

Hence it is proved that in a curved member the neutral axis of a 

transverse section does not pass through the centroid of that section. 

Now substitute the value of σx from equation (vii) into equations (ix) 

∆ 

 

y dA =M 

i.e., ∆ 

 

i.e., ∆ 

 

 

 

1 

dA = M ( r1 = R1 - y from iii) 

 

dA = M 

i.e., ∆ 

 

2 

 

1= M 

 

i.e., ∆ 

 

2 

= M [using equations (x) and (xi)]

i.e., ∆ 

2 

i.e., ∆ 

=M 

i.e., ∆ 

 

i.e., ∆ 

= 

Substituting ∆ 

 

σx = 

1 

∴ σx= M 1 1 

1 

 

1 1 

= 

(e = 1– R1 from Fig. 1.2a) 

………xiii) 

into equation (VI) 

..... (xii) 

..... (xiv) 

( r1 = R1 – y ..... (xv) 

An equation (xiv) is the general expression for the normal stress σx in a 

curved beam. 

To determine the change in curvature of the neutral surface caused by 

the bending moment M 

From equation (i), 

 

 

 

 

 

 

 

 

∆

1 ∆ 

= 1 

 

 

{From equation (xiii) ∆ 

 

i.e. 

 

 

Hence proved 

References: 

= M 

EAeR 1 

= 

} 

1 

 

 

∴ M =EAe R1

ASSIGNMENT 

1. What are the assumptions made in finding stress distribution for a curved flexural member? Also 

give two differences between a straight and curved beam 

2. Discuss the stress distribution pattern in curved beams when compared to straight beam with 

sketches 

3. Derive an expression for stress distribution due to bending moment in a curved beam 

EXERCISES 

1. Determine the force F that will produce a maximum tensile stress of 60N/mm 2 in section 

A - B and the corresponding stress at the section C - D 

2. A crane hook has a section of trapezoidal. The area at the critical section is 115 mm2 . The hook 

carries a load of 10kN and the inner radius of curvature is 60 mm. calculate the maximum tensile, 

compressive and shear stress. 

Hint: b i = 75 mm; b o = 25 mm; h = 115 mm 

3. A closed ring is made of 40 mm diameter rod bent to a mean radius of 85 mm. If the pull along 

the diameter is 10,000 N, determine the stresses induced in the section of the ring along which it is 

divided into two parts by the direction of pull. 

4. Determine of value of t in the cross section of a curved beam shown in Figure such that the normal 

stresses due to bending at the extreme fibers are numerically equal.

VTU,Jan/Feb.2005 

Fig.1.35 

5. Determine a safe value for load P for a machine element loaded as shown in Figure limiting the 

maximum normal stress induced on the cross section XX to 120 MPa. 

6. The section of a crane hook is trapezoidal, whose inner and outer sides are 90 mm and 

25 mm respectively and has a depth of 116 mm. The center of curvature of the section is at a 

distance of 65 mm from the inner side of the section and load line passes through the center of 

curvature. Find the maximum load the hook can carry, if the maximum stress is not to exceed 70 

MPa. 

7. a) Differentiate between a straight beam and a curved beam with stress distribution in each of the 

beam. 

b) Figure shows a 100 kN crane hook with a trapezoidal section. Determine stress in the outer, 

inner, Cg and also at the neutral fiber and draw the stress distribution across the section AB.

A B 

62.5 mm 

25 

112.5 

F = 100kN 

8. A closed ring is made up of 50 mm diameter steel bar having allowable tensile stress of 200 

MPa. The inner diameter of the ring is 100 mm. For load of 30 kN find the maximum stress in 

the bar and specify the location. If the ring is cut as shown in part -B of 

Fig. 1.40, check whether it is safe to support the applied load. 

87.5

REFERENCE BOOKS 

• “MECHANICAL ENGINEERING DESIGN” by Farazdak Haideri 

• “MACHINE DESIGN” by Maleev and Hartman 

• “MACHINE DESIGN” by Schaum’s out lines 

• “DESIGN OF MACHINE ELEMENTS” by V.B.Bhandari 

• “DESING OF MACHINE ELEMENTS-2” by J.B.K Das and P.L. Srinivasa murthy 

• “DESIGN OF SPRINGS” Version 2 ME, IIT Kharagpur, IITM

Curved Beam - VTU e-Learning

Create successful ePaper yourself

Delete template?

Save as template?