Barber J.R. Intermediate Mechanics of Materials

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Problems 595

Figure PA.12

A.13. The beam of Figure PA.13 is simply supported at the points z=0, 2L and loaded

by equal forces F at z = L,3L. The ﬂexural rigidity of the beam is EI. Estimate

the displacement under each of the loads, using a discretization with three equal

elements, each of length L. What is the location and magnitude of the maximum

displacement?

Figure PA.13

A.14. Figure PA.14 (a) shows one element of length

∆

of a beam of ﬂexural rigidity

EI, loaded by a concentrated force a distance a from one end.

∆

i+1

Figure PA.14(a)

Use equations (A.82, A.83) to ﬁnd the equivalent nodal forces F

i+1

and nodal

moments M

i+1

Use your results and the stiffness matrix in (A.86) to approximate the displace-

ment u(z) of the beam of Figure PA.14 (b), based on a representation of the beam as

a single element of length L.

596 A The Finite Element Method

Figure PA.14(b)

A.15. Figure PA.15 (a) shows one element of length

∆

in a beam discretized by the

ﬁnite element method. The only loading is the linearly varying load

w(z) =

(z −z

)

∆

; z

< z < z

i+1

per unit length, where

∆

i+1

−z

Use equations (A.82, A.83) to ﬁnd the equivalent nodal forces F

i+1

and nodal

moments M

i+1

Use your results and the stiffness matrix in (A.86) to approximate the displace-

ment u(z) of the beam of Figure PA.15 (b), based on a representation of the beam as

a single element of length L. The ﬂexural rigidity of the beam is EI.

(a)

∆

i+1

(b)

w z

Figure PA.15

Problems 597

A.16. Solve the problem of Figure A.13, using two elements with two terminal nodes

and one internal node at the mid-point.

Find the displacement predicted at the mid-point and compare it with the exact

value 5w

/384EI.

If you are familiar with matrix operations and have access to Maple, MathCad,

Mathematica or Matlab etc., re-solve the problem with 4,6,8 elements, etc and plot

the percentage error in the predicted central displacement as a function of the number

of elements used.

How many elements are needed to achieve 0.1% accuracy?

A.17. The beam of Figure PA.17 has ﬂexural rigidity EI, is built in at both ends

z = 0, L, and is subjected to a uniformly distributed load w

per unit length.

Estimate the central displacement, using two elements of length L/2.

w per unit length

Figure PA.17

A.18*. A beam of length 2L is simply supported at the ends and at the mid-point. It

is subjected to a uniformly distributed load w

per unit length. Estimate the reaction

at the central support, using a discretization with two equal elements, each of length

A.19. Use the Rayleigh-Ritz method to obtain an approximate solution for the de-

ﬂection of a beam of ﬂexural rigidity EI and length L, simply supported at its ends

and subjected to a uniformly distributed load w

per unit length. Use a one degree of

freedom quadratic approximation to the deformed shape.

Verify that the predicted displacement is identical to that obtained in equation

(A.87).

Properties of Areas

In §4.3 we found that the coordinates ¯x, ¯y of the centroid of a plane area A can be

determined from equations (4.28, 4.29) — i.e.

A ¯x =

′

dA ; A ¯y =

′

dA , (B.1)

whilst the second moments of area about the general set of axes O

′

are deﬁned

by the integrals

′

′2

dA ; I

′

′2

dA ; I

′

dA . (B.2)

In many practical cases, the area can be decomposed into simpler shapes for

which the appropriate quantities are tabulated and the required results can then be

obtained by superposition, as demonstrated in §4.3. Otherwise, the ﬁrst and second

moments must be obtained directly from the deﬁnitions by performing the integra-

tions.

The biggest challenge in this procedure is to establish the appropriate limits on

the double integral. Once this has been done, the double integral operator can be

applied to each of the quantities in equations (B.1,B.2) and the resulting integrals are

generally straightforward.

One of the two integrations can often be performed by inspection. For example, if

we wish to determine I

′

for the section of Figure B.1(a), we can reduce it immediately

to a single integral by dividing the section into strips parallel to the x

′

-axis as shown.

Thus,

′

′2

dA =

w(y

′

′2

′

, (B.3)

where w(y

′

) is the width of the section (the length of the strip) as a function of y

′

. In

effect, all we have done here is to perform the integration with respect to x

′

in our

heads, since the integrand (y

′2

) does not depend on x

′

600 B Properties of Areas

(a) (b)

Figure B.1: Determining the second moment of area by integration

However, if we have to calculate all the ﬁrst and second moments, it is safer to

make both stages of integration explicit. For example, in rectangular coordinates the

element of area dA is dx

′

and if we perform the x

′

integration ﬁrst, we have

dA =

d(y

′

)

c(y

′

)

′

. (B.4)

In this expression, notice how the limits c,d on the inner integral are functions of

the outer variable y

′

, whereas the limits a, b on the outer integral are constants. This

is illustrated in Figure B.1(b). The inner integral is performed over the strip CD and

its limits depend on which strip (which value of y

′

) is being considered, whereas the

outer integral sums the contributions of the separate strips.

Usually we require the centroidal second moments, but the coordinates of the

centroid may not be known initially. We therefore choose any convenient origin for

the integration and then use the parallel axis theorem of §4.3.2 to deduce the cen-

troidal values.

Example B.1

Determine the location of the centroid and the centroidal second moments I

for the quarter circle of radius a shown in Figure B.2.

The equation of the circular boundary is

′2

+ y

′2

= a

c(y )

δy

w(y )

δy

d(y )

B Properties of Areas 601

and hence the right-hand boundary of the area at height y

′

−y

′2

A suitable realization of the double integral is therefore

dA =

√

−y

′2

′

Figure B.2

Using this expression, we have

A =

√

−y

′2

′

−y

′2

′

which of course we could have written down as the area of a quarter circle of radius

Applying the same operator to the deﬁnition of A ¯x, we have

A ¯x =

′

dA =

√

−y

′2

′



−y

′2



′

and hence

¯x =

A ¯x



A similar procedure could be used to determine the coordinate ¯y, but the symmetry

of the section shows that this must also be

¯y =

Since we now know the location of the centroid, it would be possible to transfer

the origin there and calculate the centroidal second moments of area directly. How-

ever, It is more efﬁcient to calculate the second moments about the axes through O

′

and then use the parallel axis theorem to transfer the values to the centroidal axes.

602 B Properties of Areas

We have

′

′2

dA =

√

−y

′2

′

′2

−y

′2

′

The centroidal value I

can then be deduced from the parallel axis theorem (4.33) as

= I

′

−A ¯y

−







−



= 0.05488a

The same results are obtained for I

′

and I

, by symmetry.

For the product inertia, we have

′

′2

dA =

√

−y

′2

′



−y

′2



′

and the parallel axis theorem (4.35) then gives

= I

′

−A ¯x¯y =

−







−



= −0.01647a

Stress Concentration Factors

These ﬁgures are plotted from the approximate expressions given by W.D. Pilkey

(1994), Formulas for Stress, Strain and Structural Matrices, John Wiley, New York

and R.J. Roark and W.C. Young (1975), Formulas for Stress and Strain, McGraw-

Hill, New York, 5th edn. A wider range of charts for stress intensity factors can be

found in R.E. Peterson (1974), Stress Concentration Factors, John Wiley, New York

and in numerous books on mechanical design. The reader should be warned that

these various sources often disagree signﬁcantly as to numerical values, particularly

at small values of the ﬁllet radius in Figures C.3 – C.7.

Figure C.1: Rectangular bar with a transverse hole in tension or compression;

nom

= F/A, where A = (w−d)h is the reduced cross-sectional area at the hole and

h is the plate thickness.

604 C Stress Concentration Factors

Figure C.2: Rectangular bar with a hole in transverse bending;

nom

=Mh/2I, where

I = (w−d)h

/12 is based on the reduced cross section at the hole and h is the plate

thickness.

Figure C.3: Rectangular bar with a change in section through a ﬁllet radius r, loaded

in tension or compression;

nom

=F/dh, where h is the plate thickness.

radius r