Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 1

Example: 1

A circular ring of rectangular section with a slit is

loaded as shown in Fig. Determine the magnitude of the

force P if the maximum resultant stress along the

section ab does not exceed 100 N/mm2. Draw the

stress distribution diagram along ab.

Solution:

Mean radius of curvature, R = 11 cm

Radius of curvature of inner surface, R1 = 8 cm

Radius of curvature of outer surface, R2 = 14 cm

Breadth, B = 4 cm

Depth, D = 6 cm

Maximum resultant stress will occur at the inner radius

along section ab, that is, at the point b.

Bending moment, M = P × R = 11P N-cm

Resultant stress at the point

Therefore,

Stress distribution (along Gb)

where y varies from 0 to 3, compressive stresses at x =

0 , y = 0 cm

Stresses distribution Along Ga where y varies from 0 to 3

(tensile stress)

Since, the direct stress is compressive

1. T-section: Figure 17.5 shows a T section with

following dimensions:

Breadth of the flange = B

Breadth of the web = b

Radius of curvature up to centroid G of the section

= R

Radius up to extreme outer edge of web = R1

Radius up to inner edge of flange = R2

Radius up to outer edge of flange = R3

Area of cross-section of T section = A

2. I-section: Figure 17.6 shows an I section with flange

and web of breadths B and b, respectively.

R = Radius of curvature up to centroid G of the

section

R1 = Radius up to outer edge of inner flange

R2 = Radius up to inner edge of inner flange

R3 = Radius up to inner edge of outer flange

R4 = Radius up to outer edge of outer flange

A = area of cross-section

= B(R4 − R3) + b(R3 − R2) + B(R2 − R1)

Figure 17.5

Figure 17.6

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 2

3. Channel Section: Figure 17.7 shows a channel

section with

B = breadth of web

b = breadth of flanges

R = Radius of curvature up to centroid G of the

section

R1 = Radius up to inner surface

R2 = Radius up to outer edge of web

R3 = Radius up to the outer edge of flange

A = Area of cross-section = B(R2 − R1) +

2b(R3 − R2)

Example: 2

A curved beam whose centroidal line is a circular arc of

12 cm radius. The cross-section of the beam is

of T shape with dimensions as shown in Fig. Determine

the maximum tensile and compressive stresses set up

by a bending moment of 70,000 N cm; tending to

decrease the curvature.

Solution:

Figure show the curved bar with T section subjected to

a bending moment M tending to decrease the

curvature. Therefore, there will be tensile stresses

between A to G, and compressive stresses

between G to B.

Let us first calculate the distance of centroid from the

outer edge of web.

Radius of curvature, R = 12 cm (given).

Radius up to inner surface, R1 = 12 −1.864 = 10.136

cm

Radius up to outer edge of flange, R2 = 11.136 cm.

Radius up to outer edge of web, R3 = R1 + 6

= 10.136 + 6 = 16.136 cm

B = 6 cm, b = 1 cm, Area A = 6 × 1 + 1 × 5 = 11 cm2

Maximum compressive stress at point B

Maximum tensile stress at point A

Example: 3

Figure shows a press applying a 150-kN force on a job.

Determine the stresses at the points a and b. The

section is hollow as shown.

Solution:

Figure 17.7

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 3

Let us first determine the position of the centroid

Radius of curvature, R = 24 + 13 = 37 cm

Area of cross-section,

A = 24 × 6 + 2 × 4 × 20 + 4 × 16 = 398 cm2

Bending moment, M = Force × (60 + R)

= 150 × 97 kN cm, where R = 37 cm

Bending stress due to M at

(tensile)

Bending stress due to M at

(compressive)

Resultant stress at the point a = 6.869 + 0.408

= 7.277 kN/cm2 (tensile)

Resultant stress at the point b = 5.682 − 0.408

= 5.274 kN/cm2 = 52.74 N/mm2 (compressive)

Ah2 for a Trapezoidal Section:

Figure 17.12 shows a trapezoidal section of a curved

bar with breadths B1 and B2, depth D and radius of

curvature R. Say, C is the centre of curvature and G is

the centroid of the section. Then,

Consider a strip of depth dy at a distance of y from

the centroidal layer.

If b is the breadth of the strip

Area of the strip,

Figure 17.12

Now

Example: 4

Determine the maximum compressive and tensile

stresses in the critical section of a crane hook lifting a

load of 40 kN. The dimensions of the hook are shown

in Fig. The line of application of the load is at a distance

of 8 cm from the inner fibre (rounding off the corners of

the cross-section are not taken into account).

Solution:

Figure shows a crane hook and the trapezoidal section.

The load line KK′ is away from the centre of the

curvature C.

B1 = 4 cm; B2 = 8cm;D = 12 cm

Position of CG of the section:

So, y2 = 16/3 cm

Area of cross-section,

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 4

Now,

Substituting the values,

The bending moment tends to reduce the curvature, so

the portion GA will be in compression and

portion GB will be in tension.

Maximum compressive stress at A,

as R + y1 = 18 cm

Maximum tensile stress at B

Ah2 for a Circular Section:

Figure 17.14 shows the circular section of diameter d of

a curved bar of radius of curvature R, from the centre of

curvature C up to the centroid G of the section.

Consider a strip of depth dy at a distance of y from the

centroidal layer as shown.

Figure 17.14

Example: 5

A curved bar is formed of a tube of an outside diameter

of 8 cm and a thickness of 0.5 cm. The centre line of this

beam is a circular arc of radius 15 cm. Determine the

greatest tensile and compressive stresses set up by a

bending moment of 1.2 kN m tending to increase its

curvature.

Solution:

Figure shows the cross-section of a curved bar of

radius of curvature R = 15 cm.

Area of cross-section,

Area of inner circle,

Area of outer circle,

Bending moment, M = 1.2kNm = 1.2 × 105 N cm

Figure 17.15

For a circular section

For inner circle,

For outer circle,

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 5

Maximum tensile stress at b,

Maximum compressive stress at a,

Example: 6

A ring is made of round steel bar, 2 cm diameter and

the mean diameter of the ring is 12 cm. Determine the

greatest intensities of tensile and compressive stresses

along a diameter XX if the ring is subjected to a pull of 5

kN along diameter YY.

Solution:

Figure shows a ring of a mean diameter of 12 cm, a bar

diameter of 2 cm, subjected to a diametral pull P.

Radius of curvature, R = 6 cm

Bar diameter, d = 2 cm

Pull, P = 10 kN

Area of cross-section,

Stresses:

(tensile)

= -4641.75 N/cm2 = -46.42 (compressive)

Example: 7

A chain link is made of round steel rod of 1 cm

diameter. If R = 3 cm and l = 5 cm, determine the

maximum stress along the section where tensile load is

applied. If P = 0.5 kN.

Solution:

R = 3 cm, d = 1 cm, l = 5 cm, and P = 1 kN

Now,

Then,

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 6

θ = 00 , Therefore

Maximum stress at intrados,

Maximum stress at extrados,

Maximum stress occurs at the intrados, i.e., where

the load is applied.

Example: 8

A ring with a mean diameter of 120 mm and a circular

cross-section of 40 mm diameter is subjected to a

diametral compressive load of 20 kN. Calculate the

deflection of the ring along the load line. E = 200

GN/m2.

Solution:

Since the diametral load is compressive, there will

be reduction in diameter along the load line and

increase in diameter perpendicular to the load line.

R = 60 mm = 6 cm; d = 40 mm = 4 cm; P = 20 × 103 N

E = 2,000 GN/m2 = 200 × 105 N/cm2

Deflection along the load line

Example: 9

A chain link is made of a steel rod of 12 mm diameter.

The straight portion is 60 mm in length and the ends

are 60 mm in radius. Determine the deflection of the

link along the load line when subjected to a load of 1

kN. Given E = 200 × 103 N/mm2.

Solution:

Rod diameter, d = 1.2 cm

Area of cross-section, A = π2/4 = 1.31 cm2

Length of straight portion, l = 6 cm

Radius of curvature, R = 6 cm

Load, P = 1.0 kN, E = 200 × 105 N/cm2

Radius of gyration, k = d/4 = 0.3 cm

Deflection along the load line

= 0.213 mm

Example: 10

For the frame of a punching machine shown in Fig.

17.23. Determine the circumferential stresses at A and

B on a section inclined at an angle θ = 45° to the

vertical.

Solution:

Force, P = 200 kN.

Perpendicular force on the section AB

Tangential force on the section AB

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 7

Figure 17.23

Area of cross-section, A = 30 × 10 + 5 × 20 + 15 × 10 =

550 cm2

Location of G

y1 = 40-15.9 =24.1 cm

Radius of curvature, R = R1 + y2 = 20 + 15.9 = 35.9 cm

Bending moment on the section,

Direct force on the section,

Tensile stress at point A

= 6293.2 N/cm2 = 62.93 N/mm2

Compressive stress at point B

= 1270.23 [3.906]-257.8 = 4704.5 N/cm2 = 47.04 N/mm2

Example: 11

The radius of the inner fibres of a curved bar of

trapezoidal section is equal to the depth of the cross-

section. The base of the trapezium on the concave side

is four times the base on the convex side. Determine the

ratio of the stresses in the extreme fibres of the curved

bar to the stresses in the same fibres of a straight bar

subjected to the same bending moment.

Solution:

Now, R2 = D

Therefore, R1 = 2R2 = 2D

B2 = 4B1

Therefore, y2 = 0.4D

R = R2 + y2 = 1.4D

Figure 17.24

Let us consider that this curved bar is subjected to a

bending moment, M tending to reduce the

curvature. The maximum tensile stress, (when y =

−y2)

The maximum compressive stress, (when y = +y1)

Stresses in the straight bar

Let us divide the section in two triangles as shown, so

as to calculate the moment of inertia IYY.

Area of triangle I,

Mechanics of Solids (NME-302) Bending of curved bars

Yatin Kumar Singh Page 8

Area of triangle II,

IYY of triangle I, about its CG = B1D3/36

Distance,

IYY of triangle II, about its CG

IYY of the whole section

The maximum tensile stress,

The maximum compressive stress, (y = + y1)

Therefore, the ratios