Economical sections of beams

The efficient use of material in a force-resisting member involves (a) the selection of available material to be well suited to the use of the member, and (b) the distribution of the material in the member so that it can best resist the forces acting on the member. The use of material, however, always, must be considered in relation to the cost involved.

The main point to be considered here is the effect of the distri­bution of the material in a beam on the resistance of the beam to ex­ternal loads; this involves (a) the effect of the shape of cross-section and (b) the effect of a variation of size of cross section along the beam.

(a) Effect of Shape of Cross Section. — If a beam having a constant cross section safely resists static loads, the maximum fiber unit stress at the dangerous section must not exceed the allowable or working unit stress that experiments and experience have shown to be permis­sible. That is, s in the flexure formula, s = Mc/I, must not be greater than the working unit stress when M is the maximum bending moment. The flexure formula shows that, when a given unit stress s is developed in the beam, the bending moment M required to develop this unit stress is large when l\c is large. Furthermore, I\c is made large by forming the cross section so that the greater part of the area is as far from the neutral axis as is practicable, for, although both I and с are increased by this procedure; I is increased much more than is c. Thus; metal beams made of steel, aluminum, etc, are rolled in the form of I sections channel sections, etc. Built-up metal beams of various shapes are made to conform to this principle. Steel beams are made with the two flanges equal in area, the proportional limits in compression and tension being approximately equal. Cast-iron beams, however, frequently are cast with the tensile flanges larger in area than the compressive flange, the compressive strength of cast iron being much (about four times) larger than the tensile strength. Although cast-iron beams rarely are used in buildings or bridges, they frequently are used in machine frames that are subjected to bending.

(b) Effect of a Varying Section along the Beam. —The bend­ing moment, in general, varies along a beam and is maximum at one section of the beam. If, then, a beam has a constant cross section (I/c = a constant), the maximum fiber unit stress will occur on the outer fiber of the section at which the bending moment is maximum, and the unit stress in the outer fibers at all other sections will be less than that at the dangerous section. Therefore, when the beam is car­rying the load that causes the allowable fiber unit stress in the beam, there is much material in the beam on either side of the dangerous section that is under-stressed, and hence this under-stressed mate­rial could be saved by varying the cross section of the beam so that the I/c would vary as the bending moment M varies.

Torsion torsion of a circular shaft

Consider a circular shaft built in at the upper end and twisted by a couple applied to the lower end (Fig. 17a). It can be shown by measurements at the surface that circular sections of the shaft remain circular during twist, and that their diameters and the distances be­tween them do not change provided the angle of twist is small.

A disk-like element of the shaft, such as that adjacent to the section mn and shown isolated as in Fig. 17b, will be in the follow­ing state of strain. There will be a rotation of its bottom cross section with reference to its top through an angle d. An element

abcd of the surface of the disc whose sides were vertical before strain takes the form shown in Fig 17b. The lengths of the sides remain essentially the same and only the angles at the corners Change. The element is in a state of pure shear and the magnitude of the shearing strain, measured by the angle cac’, is given very closely by

=c’c/ac’

Since c'c is the small arc of radius r or d/2 subtended by the angle d, c'c = rd. Then

=c’c/ac’=rd/dx

For a shaft twisted by a torque at the end, the angle of twist is pro­portional to the length and the quotient d/dx is constant. It repre­sents the angle of twist per unit length of the shaft and will be called . Then, from (13),

 = r (14)

The shearing stresses, which act on the sides of the element and pro­duce the above shear, have the directions shown. The magnitude of each

S=Gr=1/2Gd

So much for the state of stress of an element at the surface of the shaft.

As for the state of stress within the shaft, the assumption will now be made that not only the circular boundaries of the cross sec­tions of the shaft remain undistorted but also that the cross sections themselves remain plane and rotate as if absolutely rigid; that is, every diameter of the cross section remains straight and rotates through the same angle. The tests of circular shafts show that the theory developed on this assumption is in very good agreement with the experimental results. Such being the case, the discussion for the element; abcd at the surface of the shaft (Fig. 17b) will hold also for a similar element of the surface of an inner cylin­der, whose radius  replaces r (Fig. 17c). The thickness of the element in the radial direction is considered as very small. Such elements are then also in pure shear, and the shearing stress on their sides is

s=G (16)

This states that the shearing stress varies directly as the distance p from the axis of the shaft. Figure 18 pictures this stress distribution in the plane of the cross section and also the accompanying shearing stresses in an axial plane. The maximum stress occurs in the outer surface of the shaft.

For a ductile material, plastic flow begins first in this outer surface. For a material, which is weaker in shear longitudinally than transversely, for instance, a wooden shaft with the fibers parallel to the axis, the first cracks will be pro­duced by shearing stresses acting in the axial sections and they will appear on the surface of the shaft in the longitudinal direction. In the case of a material, which is weaker in tension than in shear, for instance, a circular shaft of cast iron or a cylindrical piece of chalk, a crack along a helix inclined at 45° to the axis of the shaft often occurs. The explanation is simple. The state of pure shear is equivalent to one of tension in one direction and equal compression in the perpendicular direction. A rectangular element cut from the outer layer of a twisted shaft with sides at 45° to the axis will be subjected to such stresses. The tensile stresses shown produce the helical crack mentioned.

The relationship between the applied twisting couple M_t and the stresses, which it produces, will now be found. From the equilibrium of that portion of the shaft between the bottom and the imaginary section mn, it can be concluded that the shearing stresses distributed over the cross section mn are statically equivalent to a couple equal and opposite to the torque M_t. For each element of area dA (Fig. 17c), the shearing force = sdA. The moment of this force about the axis of the shaft is (sdA)  = G²dA, from eq. (15). —The total moment M_t about the axis of the shaft is the summation, taken over the entire cross-sectional area, of these moments on the individual elements; that is,

M_t = GI_p (17)

where I_p is the polar moment of inertia of the circular cross section. For a circle of diameter d, I_p=d⁴/32;

.M_t = Gd⁴/32 (18)

and

 = M_t/G I_p = M_t32/Gd⁴

Thus , the angle of twist per unit of length of the shaft, varies directly as the applied torque and inversely as the modulus of shear G and the fourth power of the diameter. If the shaft is of length l, the total angle of twist will be

=l (20)

This equation is useful in the experimental verification of the theory, and is checked by numerous experiments, which prove the assump­tions made in deriving the theory. It should be noted that experiments in twist are commonly used for determining the modulus of mate-rials in shear. If the angle of twist produced in a given shaft by a given torque be measured, the magnitude of G can be easily calcu­lated from eq.(20).

Substituting  from eq. (15), we obtain an equation for calculating the maximum shearing stress in twist of a solid circu­lar shaft

(s)max= M_t r/ I_p =16 M_t /d³ (22)

That is, this stress is proportional to the torque M_t and inversely proportional to the cube of the diameter of the shaft.

Eq. (22) holds only within the elastic limit but sometimes it is used up to rupture. In this case the fictitious shearing stress, found by substituting in eq. (22) the maximum torque carried by a torsion spe­cimen when tested^41/1 to rupture, is called the torsional modulus of rupture. This modulus is not a true stress but it provides a convenient measure of the relative strengths of specimens of various materials and diameters.

In practical applications the diameter of the shaft must usually be calculated from the horsepower, which it transmits. Given the horsepow­er H. P., the speed n in r. p. m. and the torque M_t in inch pounds, a formula connecting these quantities is derived as follows. Imagine a hoisting drum winding up a rope which is subjected to a constant pull P. Let the radius in inches from the axis of the drum to the center line of the rope be r and let the drum turn n r. p. m. In one minute 2rn /12 feet of rope will be wound onto the drum and, since work = force times distance, the total work per minute done against the force P is 2rnP/12. Since one horsepower is defined as the rate of doing 33,000 ft. lbs. of work per minute, the horsepower demanded of the rotating drum is

H. P.= 2rnP/1233,000. (23)

But the product Pr is the torque M_t exerted by the force P about the axis of the drum. Rearranging eq. (23) and substituting M_t = P_r we have

M_t = 63,000(H.P.)/n

This expression, derived for the particular case of a hoisting drum, can also be used in all cases where a torque or couple M_t is acting on or being exerted by a drum, a shaft, a pulley or any rotating body. To get the maximum shearing stress in a shaft, this value of M_t may be substituted in eq. (22) or, to get the angle of twist, in eq. (20).