Barber J.R. Intermediate Mechanics of Materials

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Problems 505

Further reading

A.P. Boresi, R.J. Schmidt, and O.M. Sidebottom (1993), Advanced Mechanics of

Materials, John Wiley, New York, 5th edn., Chapter 9.

R.D. Cook and W.C. Young (1985) Advanced Mechanics of Materials, Macmillan,

New York, Chapter 10.

Problems

11.1. Figure P11.1 shows a curved bar of solid circular cross section, with diameter

30 mm. The axis of the bar describes an arc with a radius 40 mm and the bar is loaded

by a bending moment of 300 Nm. Find the maximum tensile stress. What would be

the percentage error involved in using the elementary bending equation

=My/I?

300 Nm

40 mm

30 mm

Figure P11.1

11.2. Find the maximum bending moment that can be transmitted by the curved beam

of Figure P11.1 if the allowable tensile stress in the material is

all

=120 MPa.

11.3. The elementary bending theory predicts that the ends of an initially straight

beam of length L will experience a relative rotation

∆φ

when subjected to a bending moment M [see equation (3.9)].

Use this result to estimate the relative rotation of the ends of the curved beam

of Figure 11.5, assuming that the ‘length’ of the beam is the mean length, L = ¯r

∆θ

where

∆θ

is the subtended angle (in radians). Repeat the estimate using L = ˆr

∆θ

Comment on your results.

506 11 Curved Beams

11.4. Figure P11.4 shows a U-shaped curved beam with a T-shaped cross section,

loaded by two opposed forces of magnitude 500 lb. Find the maximum circumferen-

tial stress

max

θθ

0.25 in

500 lb

5 in

1 in

3.5 in

0.25 in

1.9 in

1 in

Figure P11.4

11.5. The crane hook in Figure P11.5 is loaded by a force of 8000 lb. Find the cir-

cumferential stress at the point A, where the hook has the trapezoidal section shown

with a= 1 in., b = 4 in., c =2 in., d = 1 in.

8000 lb

4 in

1 in

Figure P11.5

Problems 507

You may use without proof the following results for the trapezoidal section:-

A =

(c + d)(b −a)

; ¯r =

(2a + b)c + (a + 2b)d

3(c + d)

;

ˆr

= (d −c) +

(bc −ad)

(b −c)

ln(b/a) .

11.6. Use integration to derive the expressions for A, ¯r,A/ˆr for the trapezoidal section

given in Problem 11.5.

11.7.* In a curved beam, the stress concentrating effect of the inner radius can be

reduced by making the thickness there larger than that at the outer radius. This is

the rationale for using the trapezoidal section in Figure P11.5. If the ratio between

outer and inner radii is b/a = 2, ﬁnd the ratio c/d for the trapezoidal section, if the

circumferential stresses at the inner and outer radii are to be equal and opposite. The

results given in Problem 11.5 may be used without proof.

11.8. Figure P11.8 shows part of the frame of a press. Find the maximum tensile

stress at A, when a vertical load of 20 tons is applied as shown.

2.25

8.5

20 tons

22.5

Section A −B

all dimensions in inches

Figure P11.8

508 11 Curved Beams

11.9. The C-clamp of Figure P11.9 is manufactured from a steel with a tensile yield

stress S

= 380 MPa. The frame has the T-section shown. Find the maximum clamp-

ing force P, if there is to be a factor of safety of 2.5 against yielding at the point

7 R

all dimensions in mm

Figure P11.9

11.10. Figure P11.10 shows a circular ring of mean diameter 350 mm and circular

cross section of diameter 100 mm, loaded by equal and opposite 50 kN forces. Find

the maximum stress in the ring.

Method: This indeterminate problem is similar to that of Example 3.16, except

that we need to use equation (11.16) to replace the integrand in (3.110).

Use Castigliano’s second theorem to resolve the indeterminacy and hence ﬁnd the

maximum bending moment. Then use the results of §11.1 to determine the maximum

tensile stress.

Problems 509

100 mm

50 kN

350 mm

Figure P11.10

11.11. The steel U-beam of Figure P11.11 has uniform 2 inch square cross section

and is loaded by two forces P, as shown.

Use equations (11.16, 3.24) to determine the total strain energy in the beam as a

function of P and hence use the method of §3.3 to determine the relative displacement

of the points A,B, when P =4000 lbs (E

steel

=30×10

psi).

0.5 R

2.5 R

all dimensions in inches

Figure P11.11

510 11 Curved Beams

11.12. A curved beam of square cross section a×a and inner radius na is loaded by

a bending moment M.

Find the maximum circumferential stress

max

θθ

and the maximum radial stress

max

Sketch the variation of these quantities as functions of n. Sketch also the ratio

max

θθ

and comment on the results.

11.13. The curved section of the beam in Figure P11.4 is loaded by a bending mo-

ment of 200 lb ft. Find the location and magnitude of the maximum radial stress

max

11.14. Find the maximum radial stress

max

for the U-beam of Figure P11.11 loaded

by two forces P= 4000 lbs.

Elastic Stability

In many design contexts, it is desirable to use components that have the minimum

weight and cost compatible with acceptable values of strength and stiffness. The

methods of the preceding chapters will often predict that the optimum design ac-

cording to this criterion is a thin-walled structure. For example, we saw in §6.5 that

closed sections are more effective at transmitting torques and that the strength and

stiffness of such sections increases with the area enclosed by the section. It follows

that the theoretical optimum torsion member for a given mass of material is a thin-

walled cylindrical tube and the strength and stiffness theoretically increase without

limit as the radius of the tube increases and the wall thickness correspondingly de-

creases [see equations (6.39, 6.40)].

Thin-walled structures can indeed be quite surprisingly strong and lightweight

— a result that is best appreciated by constructing simple structures out of paper and

testing them to destruction.

Experiment

Take a single sheet of 8

×11 (or A4) paper, roll it up into a cylindrical tube with a

small overlap on the longer edge and tape up the join with Scotch tape. Now stand

the tube vertically on a ﬂat table and carefully rest this book on the top of the tube.

Then add another book of similar weight.

Probably you will be able to support at least two books, with a total weight ex-

ceeding 5 lbs, on this single sheet of paper.

However, add a bit more weight and the complete structure will collapse. The

paper will not tear and the joint will stay intact, but the walls will fold in a complex

way and no longer support the load. This collapse mode is known as elastic instability

or buckling and it places limits on the thinness of components.

Somehow, it seems even more surprising that with (for example) 10 sheets of paper suitably

placed, you can support 50 lbs! Try asking your non-engineering friends to guess how many

sheets of paper (and if you want to be exact, inches of tape) are needed to support a 5 lb

weight, 11 inches above the table and then do a demonstration.

J.R. Barber, Intermediate Mechanics of Materials, Solid Mechanics and Its Applications 175,

2nd ed., DOI 10.1007/978-94-007-0295-0_12, © Springer Science+Business Media B.V. 2011

512 12 Elastic Stability

Buckling generally only occurs when the component is loaded in compression.

This again is easily demonstrated with a sheet of paper. If you grasp two opposite

edges and apply a tensile force as shown in Figure 12.1 (a), there will be no instability

and in fact the paper is quite strong. It requires considerable strength to fracture it

in tension. However, if you try to load it in compression as in Figure 12.1 (b), it will

immediately bend out of plane and support hardly any load.

(a)

(b)

Figure 12.1: A piece of paper loaded (a) in tension and (b) in compression

12.1 Uniform beam in compression

A simple system that exhibits this behaviour is the uniform beam of Figure 12.2 (a)

loaded by a compressive axial force. A straight beam loaded purely in compression

is sometimes referred to as a column. We suppose that the beam is simply-supported

at its ends, z = 0,L, and that the force P is sufﬁcient to cause the beam to deform

into the buckled shape shown in Figure 12.2 (b), in which the lateral deﬂection is

described by an as yet unknown function u(z).

12.1 Uniform beam in compression 513

(a)

(b)

u(z)

Figure 12.2: (a) Simply-supported beam loaded in compression; (b) deformed shape

In general, we anticipate reaction forces R

at the simple supports, as shown

in the free-body diagram of Figure 12.3 (a), but by taking moments about the ends,

A,B, respectively, it can be shown that these reactions must both be zero.

(a)

(b)

u(z)

M(z)

Figure 12.3: Free-body diagram of (a) the whole beam and (b) a segment of length z

at one end

Figure 12.3 (b) shows a free-body diagram for a segment of the beam of length z

and by taking moments about one end of this segment, we conclude that the bending

moment M is given by

M = Pu . (12.1)

514 12 Elastic Stability

It follows that the displacement u must satisfy the equation

= −M = −Pu , (12.2)

from equation (1.17), and hence

= 0 . (12.3)

This is a homogeneous ordinary differential equation whose solution can be

shown by substitution to be

u = Acos

z + Bsin

z , (12.4)

where

(12.5)

and A,B are arbitrary constants to be determined from the boundary conditions

u = 0 ; z = 0 (12.6)

u = 0 ; z = L . (12.7)

Substituting (12.4) into (12.6, 12.7), we obtain

A = 0 (12.8)

Bsin

L = 0 . (12.9)

Clearly this problem has the trivial solution A =B =0, corresponding to the case

where u(z) = 0 for all z. In other words, the beam transmits the compressive force

whilst remaining undeformed.

However, if

sin

L = 0 , (12.10)

i.e. if

L =

, 2

, 3

, ..., (12.11)

the boundary conditions are satisﬁed for non-zero B. Indeed, in this case, B can take

any value. This deﬁnes a neutrally stable solution. The beam can adopt any one of a

continuous series of equilibrium conﬁgurations corresponding to different values of

the constant, B.

A simple system that exhibits neutral stability is the ball resting on a horizontal

surface, as shown in Figure 12.4 (a). The ball can be placed at any point on the

surface and it will remain there in equilibrium. By contrast, if the surface is concave,

as in Figure 12.4 (b), there is only one position of equilibrium (at the lowest point)

and it is stable, because the ball will roll back if it is displaced from the equilibrium

position. If instead the surface is convex, as in Figure 12.4 (c), there is again a single

equilibrium position, but it is unstable because an arbitrarily small displacement will

generate forces tending to accelerate the ball away from the equilibrium position.