4. An I beam is 14 ins. deep, areas of bottom flange and web equal, each being four times the top flange, the area of which is 3 sq. ins. It is 20 ft. long, and supported at the ends. Find the greatest tensile and compressive stresses produced by a load of 6 tons placed 8 ft. from one end.

Ans. 2.06; 4.12 tons per sq. inch.

5. An I beam has flanges 4 ins. by 1 inch, and web 9 ins. by inch; span 10 ft. Find the greatest central load if the stress is limited to 5 tons per sq. inch. If the material be steel, how would the weight of the beam affect the result?

Ans. 7 tons; the carrying power would be less by 208 lbs.

6. Find the limiting span of a cast-iron pipe 94 ins. internal diameter, inch thick, the weight being 100 lbs. per foot length, and the stress not to exceed 2 tons per sq. inch.

Ans. 41 ft.

7. Compare the resistances to bending of a wrought-iron I beam, flanges 6 inches by 1 inch, web 8 ins. by inch, when upright and when laid on its side thus

.

Ans. 4.6: 1.

8. Determine the weight which may be carried at the middle of a wooden spar 6 inches diameter, 15 feet long, supported at the ends; allowing a stress of 1500 lbs. per sq. inch.

tooth per H. P.

1000 lbs. per sq. inch.

Ans. 707 lbs. 9. In question 4, page 291, find the necessary width of Thickness at root 1 inch. Stress allowed Ans. .2 ins. nearly. 10. In questions I and 2, page 308, find the value of the greatest stress produced. Ans. 78.5; 850 lbs. per sq. inch. II. In question 3, page 308, the section of the beams is I shaped, equal flanges each four times as wide as it is thick, thickness of web one half that of a flange, and its area equal that of one flange. Find the necessary dimensions, stress not to exceed 3 tons per square inch. Neglecting the resistance of the web to bending, what shape should the elevation of the beam take, its width being uniform?

Ans. Thickness of flange in., and the others as given. The depth should vary as the B. M., hence the outline

should be a parabola the same as the curve of B. M. 12. In questions 7 and 8, page 363, compare the resistance to bending of the two sections, per sq. inch of sectional area.

Ans. 1:1.24.

13. Find the moments of inertia of the sections given in questions 9, 10, 11 and 12 of the preceding chapter.

Ans. 968; 82; 1190; 1950 inch units.

14. In the preceding, find the weight each could carry at the centre of a 20 ft. span. Material in (9) cast-iron, tensile stress allowed 2 tons, compressive 8 tons. In I, II, and 12, wrought-iron, tensile 5 tons, compressive 3 tons.

Ans. 6.9; .89; 8; 14.2 tons.

CHAPTER XX

SHEARING AND TORSION

IN dealing with bending, the compound nature of the action is so evident that it is not necessary or useful to assign a bending strength to a material, but the strength of a beam is deduced from the values of the tensile and compressive strengths. (See note at end of ch. xxi.)

In the case of shearing, however, it is not SO apparent that there is a dual action, and hence it has been, and still is practically, treated as a single action of a nature different from either tension or compression, and a metal is said to have a shearing strength just as it has a compressive or tensile strength.

In

We have seen in chap. xiv. (Fig. 185) the nature of the action called shearing, and how to calculate the shearing force on any transverse section of a beam. the figure there given we have both bending and shearing, and it will be instructive to inquire what kind of forces are necessary to produce pure shearing, just as we have in the preceding chapter seen what produces pure bending.

In order to have no bending moment at a given section a force must be applied in the line A of the section, or indefinitely near to it on one side, as P1 (Fig. 288). Then such a force can

1

K

Fig. 288.

B

be balanced by an equal and opposite force P2 acting

2

indefinitely near to the section on the other side.

1

2

In

the figure we have drawn P1 and P, palpably out of line to show on which side of the section KK each is supposed to act, but each is supposed indefinitely near to KK, whence they are also indefinitely near each other, i.e. are in the same line and hence balance. The beam is now said to be in pure shear, and the shearing force is P1 or P2.

1

Then, considering the equilibrium of the right-hand

B

piece, it is in equilibrium as shown under P, (Fig. 289), and the total shearing stress on the section KK, shown by the arrows, and plainly differing from tension or compression, being a sort of frictional action between the two surfaces at KK resisting relative sliding, instead of a direct pull or push acting normally to the surfaces. Hence

Fig. 289.

Total shearing stress on KK=P1.

And assuming it to be uniformly distributed over the section, if

q=intensity, A= area of section,

Pure shear is extremely rare in practice, e.g. in the shearing machine, bars after being cut can be plainly seen to have bent; the reason is that it is impossible to bring the forces P1 P2 indefinitely near each other; the jaws of the machine have a certain clearance, and in this distance the bar bends.

If the bar be both bent and sheared, we still have, referring to Fig. 289,

Total stress on section = F the shearing force,

but the value of q is altered, as we shall explain farther on. Shear on Rivets.-The most usual example given of pure shear is that of a rivet connecting plates exposed

Fig. 290 shows a section and

to a longitudinal pull.
plan of a single riveted lap
joint. Looking at the plan
we see that the centre rivet
shown must support the piece
of plate between the dotted
lines, which are half-way

between that and the next
rivet on each side. Each
rivet has to support an
equal strip, and hence we
will consider the actions
on one such strip and its
rivets.

Let

о

Fig. 290.

pdistance between the rivet centres or the Pitch.

Then p is the width of the piece of plate also. Let

P=pull on the strip,
t=thickness of metal,
d=diameter of rivet,

q=shearing stress on rivet,

these letters we will use right through the work.

Then, if the rivet fit its hole tightly, it will be under nearly pure shear at KK, and hence, its sectional area being #d2/4,

π

4

Р

Why now must we insert the condition as to tight fitting? To answer this, consider how the forces P are applied to the rivet. They will be distributed over the rivet surfaces from A to K, and from B to K, so that the resultant forces P P will act roughly through the centres of AK and BK. But this being so, there will be a bending moment at KK, equal to P. AK/2 or P. BK/2, and consequently not pure shear. Now if the rivet bend it