LECTURE XXIII.

CONTENTS.-Stresses on Chains-Shearing Stress and Strain-Example 1. -Torque or Twisting Movement-Strength of Solid Round ShaftsExample II.-Pressures on and Reactions from the Supports of Beams -Examples III. IV.—Transverse Stress or Bending Moment of Beams -(1) Load at Middle; (2) Load Distributed-Example V.-Questions. In this Lecture we will continue the subject of "strength of materials,” and finish the course with reasons for the shapes generally given to sections of cast iron, wrought iron, and steel girders.

Stresses on Chains.-The only stress to which the sides of the links of chains are subjected under ordinary circumstances, is that of tension. This stress tends to bring the sides of the links closer together, and consequently we find that large chain cables for mooring ships (where very sudden and severe stresses are encountered) have a cast-iron stud or wedge fitted between the inner sides of the links. These studs most effectually keep the sides of the links apart, and prevent any link jamming a neighbouring one. They add materially to the strength of the chain, for they are in compression whilst the sides of the links are in tension. Being composed of cast iron, which offers the immense resistance to compression of fully 45 tons per square inch,* there is not much fear of their giving way before the sides of the links.

The strength of a stud-link may be taken as equal to double the strength of a rod of wrought iron, of the same diameter and quality of material as that of which the chain is composed, whereas the strength of an open-link chain is only about 70 per cent. of this amount, even with perfect welding.†

In Molesworth's "Pocket-Book of Engineering Formulæ," the student will find at page 54 a formula for the safe load on chains, viz.—

W = 7.1d2

* See Table of the Ultimate Strengths and Safe Working Loads given in Lecture XXII.

† Some well-known authorities give less than 70 per cent.

Now, such a formula is very easy of application, but the student should never rest content until he finds out how the constants have been arrived at, and what relation the various symbols have towards each other. If he refers back to the short table of "Ultimate Strengths and Working Loads" given in the previous Lecture, he will find opposite wrought-iron bars and under tension, the value 5 tons per square inch as the safe working load. Consequently, applying what was said above about perfect stud-link chains, he will see that—

W = {

twice the load of a rod of the same diameter and quality as that of which the chain is composed.

= 2 × 5 × cross area of the chain iron,

.. W

This is near enough to the constant given by the above empirical formula to enable him to see how it has been obtained.

Chains which are subjected to many sudden jerks (such as lifting chains for cranes and slings) become in time crystalline, or short in the grain, and consequently brittle and unsafe. The best precaution to adopt in order to periodically remove this enforced internal condition, is to draw them once a year very slowly through a fire, thus allowing them to become heated to a dull red, and then to cool them slowly in a heap of ashes. This method is followed at Woolwich Arsenal and some other Government works.

Shearing Stress and Strain.-The action which is produced by shearing and punching machines on iron, steel, or copper plates, &c., is to force one portion of the metal across an adjacent portion. The shearing stress is the reaction per square inch opposing the load or pressure applied to the shears or punch, and the shearing strain is the deformation per unit length or volume. Rivets holding boiler plates together, fulcra of levers, the pins of the links of the chain of a suspension bridge, the cotter keys of a pump rod, are all subjected to shearing stresses and strains. The ultimate and the working shearing stresses for a few engineering materials were given in a table in Lecture XXII.

In the case of loaded beams (which we will consider shortly in connection with bending moments) the shearing force at any point or any transverse section thereof is equal to the algebraical sum of all the forces on either side of the point or section.

EXAMPLE I.-A steel punch 1" diameter is used in a large shipyard punching machine to make holes in steel plates 1" thick. What will be the total shearing stress or least pressure required?

ANSWER.-Referring to the table in last Lecture, we see that the ultimate shearing strength or shearing stress for steel bars (which we will assume to be the same as for plates) is 30 tons per square inch. Now a hole 1" diameter has a circumference

=

= πα

314", and since the plate is " thick, the area of the resisting section must be the circumference of the hole x its depth, or = 3·14′′ × 1′′ = 3·14 square inches.

.. The total pressure required = 30 tons x 3.14 = 94.2 tons.

Torque, or Twisting Moment.*-In the case of a shaft having a lever, pulley, or wheel fixed to it with a force P lbs., applied at radius R feet from the centre of the shaft, then

The twisting moment is Px R foot-lbs.

Or if R be in inches,

The torque

=

P x R inch-lbs.

Strength of Solid Round Shafts.-It is evident from the above, that a shaft subjected to a twisting moment must offer a sufficient resistance thereto, otherwise it would be twisted, or sheared, or ruptured through by the torque. It may be proved that in the case of solid round shafts their resistance to torsion is directly proportional to the cubes of their diameters when made of the same materal and quality.t

The term torque was devised by the late Professor James Thomson, of Glasgow University, to signify twisting or tortional moment. The footlbs. of torque must not be confused with ft.-lbs. of work or with resilience, which is the work done in straining a body as measured by the elongation or compression in feet x the mean load causing the strain. It will there fore, perhaps, save confusion, to calculate torques in inch-lbs.—i.e., to take the leverage or arm of the moment in inches, and the force applied in lbs.

†This is evident from the fact that the shaft must offer a moment of resistance, or shearing moment, equal to the twisting moment at the instant of rupture. Now, the area to be sheared is the cross area of the shaft =D, where D is the diameter of the shaft. The mean arm or leverage at which this resistance acts is equal to half the radius of the shaft, for at the centre the arm iso, and at the circumference it is=r, the radius of the shaft.

square inch of cross section of the material be=S, the product of these three quantities will be the total shearing moment, and must equal the twisting moment—viz. = P × R, where P is the force applied at the end of the

Diameters of three shafts, 1", 2′′, and 3′′ diameter respectively.

Torques which they will respectively resist when stressed to the same extent.

T1:T,T,:: D: D,3 : D,"

2

2

T1: T,:T,:: 13 : 2o : 33

8

:: I : 8:27.

In other words, the strengths of the three solid shafts will be as I: 8:27.

A good wrought-iron shaft of 1" diameter has been found to withstand a torque of 800 ft.-lbs., or 9600 inch-lbs., which means that they will resist 800 lbs. force at I foot, or 12′′ leverage, or 400 lbs. at 2 feet, or 24", and so on.

800 feet-lbs. of torque 9600 inch-lbs. torque.

EXAMPLE II.-On the above basis, what force acting at the circumference of a pulley 20" diameter will break a wroughtiron shaft 2" diameter?

lever or circumference of the pulley, and R the length of the arm or radius of the wheel or pulley.

Consequently,

PXR=SD2x

D) =

=S_D3
16

4

But S is a constant quantity for any particular material. Also, and 16 are constants. .. Px R vary as D3.

At the instant of rupture the strength of the shaft just balances or is equal to the twisting moment P× R.

. The strength of shaft varies as D3.

This is the same as the general statement in the text above. Without some such algebraical explanation, students are sorely puzzled how the cube of the diameter crops up; or still inore so when they see the following which appears in some text-books.

The moment of resistance of) 31416.

a round shaft to torsion

}

=

16

x diameter x shearing stress.

Such a statement is, however, quite evident after the above analysis. (We must leave the consideration of hollow shafts, tubes, &c., to our Advanced Course.)

For power of wrought-iron shafts take 70 per cent. of steel shafts of the same size.