Barber J.R. Intermediate Mechanics of Materials

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6.4 Closed sections 315

∗

¯y

∗

+C (6.25)

and, since B can be chosen to be any point, it completely determines the shear ﬂow

distribution except for the unknown constant C. Different values of C will give dif-

ferent but parallel lines of action for the resultant shear force V

, and these can be

determined using the method of §6.2. Conversely, if the line of action of V

is pre-

scribed, this can be used to determine C.

Thus, we see that there is a qualitative difference between the shear behaviour of

open and closed thin-walled sections. For an open section, equation (6.9) corresponds

to a unique line of action for the shear force, whereas for a closed section, equation

(6.25) with an appropriate value of the constant C describes the distribution for any

line of action of V

However, closed sections do have shear centres, in the sense of a point where

V must act in order for there to be no twist of the section. We shall return to this

question in §6.6 below.

Summary

The procedure for ﬁnding the shear ﬂow and hence the shear stress in a thin-walled

closed section loaded by a shear force with a prescribed line of action can be sum-

marized as follows:-

(i) Choose any convenient point A as a reference point and denote the shear ﬂow

there as q

=C.

(ii) Use equation (6.25) to determine general expressions for q in each segment of

the section. Remember that the area A

∗

is treated as positive if it is traversed in

passing clockwise from A to B and negative if it is traversed anticlockwise.

(iii) Equate the moment of the distribution about any convenient point to the corre-

sponding moment of the prescribed shear force and use the resulting equation to

determine the constant C.

Example 6. 5

The closed semi-circular section of Figure 6.35 is loaded by a vertical shear force V

whose line of action passes through the point O as shown. Determine the complete

distribution of shear stress in the section.

We ﬁrst determine the second moment of area I

for the section. We can use

equation (6.13) for the semi-circular part of the section and add the contribution for

the straight part, to obtain

(2a)

= 2.237a

t .

It is convenient to choose the top corner of the section to be the point A, since

segments starting at A have a simple geometric shape. The shear ﬂow in the straight

316 6 Shear and Torsion of Thin-walled Beams

portion is found from equation (6.25) by considering the shaded segment AB

Figure 6.36 (a), giving

(a −y)t(a + y)

2(2.237a

+C =

4.474a



1 −



+C .

Figure 6.35: Closed semi-circular section

(a) (b)

Figure 6.36

To ﬁnd q in the curved portion, we use the segment shaded in Figure 6.36 (b), but

since it is traversed in passing anticlockwise from A to B, the corresponding area A

∗

must be treated as negative. We obtain

= −

a sin(

)atd

2.237a

+C = −

cos

2.237a

+C .

6.4 Closed sections 317

The above equations completely deﬁne the shear ﬂow q and hence the shear

stress,

= q/t, in the section, except for the unknown constant C. To determine C,

we take moments about any convenient point, as in §6.2, to enforce the condition

that the distribution is statically equivalent to the force V . The result is most easily

obtained by taking moments about O, since the moment arm is then zero for both q

and V

. The moment arm for q

is the radius a and we obtain

×0 =

−

(

)ad

= −

2.237

−

cos

— i.e.

0 =

C −

2.237

from which

C =

2.237

= 0.285

Substituting this value back into the expressions for shear ﬂow, we obtain



0.508 −0.223



; q

(0.285 −0.447 cos

) ,

which completes the solution.

0.162

0.508

50.5

Figure 6.37: Shear stress ﬁeld due to the shear force V

passing through O

The distribution is illustrated in Figure 6.37. Notice that the shear ﬂow is negative

and hence clockwise in the segment −

, where

= cos

−1



0.285

0.447



= 50.5

The magnitude of q reaches a maximum at the two points where the section crosses

the neutral axis, being 0.508V/a at O and −0.162V/a at D.

318 6 Shear and Torsion of Thin-walled Beams

6.5 Pure torsion of cl osed thin-walled sections

For a closed section, the shear ﬂow can be non-zero even if the shear force V is zero,

in which case equation (6.25) reduces to

= C . (6.26)

In other words the shear ﬂow is constant around the section. This corresponds to the

case of a thin-walled closed section subjected to pure torsion.

dF = q dS

t dS

Figure 6.38: Calculating the torque due to a given shear ﬂow

The torque T is the same about any point if V = 0 and we can ﬁnd it by taking

moments exactly as in §§6.2, 6.4. Consider the contribution of the small element of

area tdS in Figure 6.38, where dS is an element of length around the mean line of

the section and t is the local thickness, which is not necessarily constant. The ele-

mental force associated with tdS is dF =

tdS = qdS. We assume that

is uniformly

distributed over tdS and hence the line of action of dF is locally tangent to the mean

line of the section — i.e. the line which is equidistant between the inner and outer

edges of the wall. The moment of dF about O is therefore ndF where n is the normal

distance from O to the local tangent to the mean line. Summing the contribution for

all such elements by integration, we have

T =

ndF =

qndS = C

ndS (6.27)

from equation (6.26). Notice that C can be taken out of the integral because it is

constant.

The ﬁnal integral in (6.27) can be given a physical interpretation. From Figure

6.39 (a), we see that ndS is the product of the base and the height of the triangle OPQ

and hence is twice the area of that triangle. Summing ndS over all elements dS, we

therefore obtain twice the area A enclosed by the mean line of the section. In other

words

6.5 Pure torsion of closed thin-walled sections 319

ndS = 2A

and hence

T = C

ndS = 2AC . (6.28)

The mean line and the corresponding enclosed area A are illustrated in Figure

6.39 (b). Notice however that since t ≪a, only a small error would be involved in

using either the inner or the outer wall of the section to deﬁne A.

mean

line

(a) (b)

Figure 6.39: (a) Contribution to the integral in equation (6.27) from the element dS,

(b) Interpretation of the enclosed area A

The shear ﬂow and shear stress can be determined from (6.26, 6.28) as

q = C =

;

2At

. (6.29)

Notice that the shear ﬂow is constant around the section, but the shear stress will vary

inversely with thickness t if the latter is not constant. In particular, the maximum

shear stress will occur where the thickness t is a minimum.

6.5.1 Torsional stif fness

It is also useful to know the angle of twist

, and hence the torsional stiffness of the

section. This is most conveniently found using an energy method. As in §3.3, we can

equate the work done by the torque T during loading to the stored strain energy. The

work done is

W =

, (6.30)

since the system is elastic and the torque will be everywhere proportional to the

instantaneous angle of twist [see equation (3.30)].

320 6 Shear and Torsion of Thin-walled Beams

The stored strain energy can be obtained from equation (3.33) as

U =

ZZZ

dV , (6.31)

where V is the volume of material in the beam and we have used the relation G =

E/2(1+

). The shear stress is assumed to be uniform across the thickness t and

hence is constant in the volume element dV =LtdS shown in Figure 6.40. The integral

for the strain energy can therefore be written

U =

tdS , (6.32)

where S is the perimeter of the closed section.

L t dS

Figure 6.40: Volume element for the integration of equation (6.31)

Equating the work done and the strain energy U and using (6.29) for the shear

stress

, we then obtain



2At



tdS =

, (6.33)

from which

T L

, (6.34)

where

K = 4A



(6.35)

is a measure of the torsional stiffness of the cross section. Notice that K has the

dimensions of length to power 4. It reduces to the polar moment of area J in the

special case where the section is axisymmetric.

In many problems, the thickness t is constant and the integral in equations (6.34,

6.35) reduces to

. (6.36)

Alternatively, if the perimeter can be divided into segments S

,... with thickness

,... respectively, we have

+ ... . (6.37)

6.5 Pure torsion of closed thin-walled sections 321

Example 6. 6

The symmetrical closed thin-walled section of Figure 6.41(a) is twisted by a torque

of 20 Nm. Find the maximum shear stress and the twist of the beam per unit length if

the material is steel (G= 80 GPa).

6.5

16.5

all dimensions in mm

(a) (b)

Figure 6.41: (a) A closed thin-walled section, (b) the mean line

We ﬁrst determine the area enclosed by the mean line, which is

A = 5 ×15 + 5 ×15 + 5 ×15 = 225 mm

from Figure 6.41 (b). The maximum shear stress will occur where the thickness t is

a minimum (t

min

=1 mm) and hence is

20,000

2 ×225 ×1

= 44.5 MPa.

The perimeter S of the section comprises a segment of length S

=5 +10+15 +10+

5=45 mm of thickness 1 mm and a segment S

=15+25+15 =55 mm of thickness

2 mm. We therefore have

= 72.5

and

K =

4 ×225

72.5

= 2973 mm

from equation (6.35). The twist per unit length is

80 ×10

×2973 ×10

−12

= 0.084 rad/m .

322 6 Shear and Torsion of Thin-walled Beams

6.5.2 Design considerations in torsion

Equations (6.29, 6.34) are extremely simple and efﬁcient to use for determining the

shear stress and twist in a thin-walled closed section loaded in torsion. In the sprit of

§1.2.2 they are therefore very useful as a design tool and there is also considerable

justiﬁcation for using them ‘out of context’ to get a rough estimate of the torsional

behaviour of closed sections whose walls are not strictly thin.

Another interesting consideration here is that the torsional shear stresses are re-

duced and the stiffness increased by increasing the enclosed area A, which however

does not contribute to either the material cost or the weight of the beam. If reducing

weight and cost are primary concerns in the design, the optimal choice for torsion

is a thin-walled circular section, since this has the greatest enclosed area for a given

perimeter. For a cylinder of mean radius a and wall thickness t, the area enclosed

is A =

, the perimeter is S = 2

a and the volume of material per unit length is

V =St =2

at. It follows that

t =

(6.38)

and the shear stress and stiffness are respectively

(6.39)

K =

)

= 2a

V . (6.40)

Thus, for a given material volume V , the torsional shear stress decreases inversely

with the radius a of the tube and the stiffness increases with a

Of course, there are practical reasons limiting the values of a that can be used in

particular applications. The beam has to ﬁt into a given space and connect to other

components at the ends. Also, the thickness for given V is inverse with a from (6.38)

and for very thin sections, local buckling of the wall is a limiting failure mode. The

reader can verify this by rolling up a piece of paper, taping it to make a closed section

and then loading it in torsion. We shall discuss failure modes involving buckling in

Chapter 12.

The contribution of the enclosed area A to torsional stiffness can have some un-

expected consequences when design changes are made. In my undergraduate days,

a popular student pastime was to buy an old family saloon car and make something

more sporty by removing the roof and parts of the door surrounds. However, many of

these cars were built using ‘monocoque’ construction, in which there is no chassis.

Instead, the car body is an integral part of the structure and front and rear axle assem-

blies are attached only through local sub-frames. In this case, the torsional stiffness

between the front and rear of the car relies on the roof (and to some extent the doors)

which render the car body a closed section. Removing the roof therefore reduced

this torsional stiffness dramatically, resulting in a vehicle with quite bizarre handling

characteristics, particularly on rough roads.

It is in fact quite a challenge to obtain sufﬁcient torsional stiffness in an open car,

even with a chassis. Closed sections can be used for the chassis members, but this

6.6 Finding the shear centre for a closed section 323

is somewhat limited by space considerations. The Lotus Seven — an inexpensive

kit car manufactured by Lotus in the 60s — solved this problem by boxing in the

transmission tunnel.

6.6 Finding the shear centre for a closed section

The energy argument of §6.5.1 can be generalized to determine the location of the

shear centre for a closed section. Remember that the shear centre is deﬁned as the

point through which V

must act if there is to be no twist.

We ﬁrst apply the reciprocal theorem of §3.8 to the bar subjected to the two

separate loads comprising (i) a torque T and (ii) a force V

acting through the shear

centre C. By the deﬁnition of C, V

produces no twist, so T does no work against

the deﬂections due to V

. It follows by the reciprocal theorem that V

does no work

against the deﬂections due to T and hence that during the application of T, the shear

centre does not move. Thus, the shear centre is also the centre of rotation of the

section when the beam is loaded in pure torsion.

Now a shear force with a more general line of action can be decomposed into

an equal force V

through the shear centre and a torque T = V

d, where d is the

perpendicular distance from C to the line of action of V

. With this decomposition,

the shear force acting through C will produce a displacement u

of C but no twist,

whereas the torque T will produce a twist

of the section, but no displacement of C.

Thus, the work done when the two components are applied simultaneously is

W =

u +

. (6.41)

In other words, there will be no coupling terms between V

,u and T,

The twist

due to T must be in the same direction as T and hence the second

term in equation (6.41) must be positive for all T and hence for all non-zero values

of d. But the ﬁrst term is independent of d and hence W will be a minimum for a

given magnitude and direction of V

, if we choose the line of action of V

such that

d = 0. In other words, the work done by V

is a minimum when its line of action

passes through the shear centre. Thus, we can ﬁnd the shear centre by writing the

strain energy for a shear force of general line of action, using the expression for

shear stress from equation (6.25), and requiring it to be a minimum.

From equation (6.32), we have

U =

tdS =

(6.42)

and hence

∂

. (6.43)

This must be zero if V

acts through the shear centre C.

Notice how these energy theorems are really quite useful!

324 6 Shear and Torsion of Thin-walled Beams

Now changing the line of action of V

(without changing its direction) is equiv-

alent to superposing a torque and we know from §6.5 that this merely changes q

everywhere by a constant. It follows that

∂

= constant (6.44)

and this constant is non-zero (as is G), so the condition

∂

d =0 simpliﬁes to

qdS

= 0 , (6.45)

which must be satisﬁed if V

is to act through the shear centre.

Summary

The procedure for ﬁnding the shear centre of a closed thin-walled section can there-

fore be summarized as follows:-

(i) Choose any convenient point A as a reference point and denote the shear ﬂow

there as q

=C.

(ii) Use equation (6.25) to determine general expressions for q in each segment of

the section. Remember that the area A

∗

is treated as positive if it is traversed in

passing clockwise from A to B and negative if it is traversed anticlockwise.

(iii) Substitute the result into condition (6.45), which then serves as an equation for

the unknown constant C.

(iv) Substitute for C into the original expressions for q and take moments about any

point as in §6.2 for open sections, to ﬁnd the line of action of V

, using equation

(6.18).

6.6.1 Twist due to a shear force

If the shear force V

does not pass act through the shear center, it will cause the

section to twist. If the location of the shear centre is known, this twist is easily found

by decomposing the shear force into an equal force through the shear centre and

an equivalent torque T =V

d, where d is the perpendicular distance from the shear

centre to the line of action of V

. In this representation, only the torque produces

twist, which is therefore given by

T L

, (6.46)

from equation (6.34).

If the location of the shear centre is not known, a similar decomposition together

with the relation (6.45) yields a general expression for the twist of the section. We

ﬁrst write the shear ﬂow

q = q

+ q

, (6.47)