The resultant w2 of the weight of the portion KLHG acts vertically through C2 the centroid of KLHG. The resultant of the load on GH is W, which is the resultant of w, and w. W2 cuts GH at c2 which is the centre of stress for the horizontal section GH.

2

Continuing in the same way, W3, the resultant of w1, w2, and w3, cuts EF at c, the centre of stress for EF, and W., the resultant of w1, w2, w, and w1, cuts BC at c, the centre of stress for BC.

A fair curve drawn through the centres of stress of the various horizontal sections is called the line of resistance for the cross section of the dam.

Reservoir full. When the reservoir is full the gravity forces which have just been considered will still act but there will now be in addition the pressure of the water on the inner face.

The intensity of the pressure of the water per square foot at any depth h feet from the upper or free surface of the water is equal to h multiplied by the weight of a cubic foot of water or say 62-3h.

4

Still considering a portion of the dam one foot long. The resultant force on the horizontal section BC is the resultant of W, and P1, where P, is the resultant of the pressure of the water on the face of the one foot length of dam from A to B. P, acts at two-thirds of AB from A and at right angles to AB. The magnitude of P, is the area of the face AB, one foot long, multiplied by half the intensity of the water pressure at B. Let P, and W, intersect at O. R, the resultant of P. and W, acts through O in a direction determined by the triangle of forces rpw. The line of action of R, cuts BC at 8, which is the ⚫ centre of stress for BC.

4

In a similar manner 8, 82, and 8, the centres of stress for EF, GH, and KL may be determined. A fair curve through these centres of stress of the various horizontal sections determined as above is the line of resistance for the cross section of the dam when the reservoir is full. If each of the horizontal lines BC, EF, etc. be divided into three equal parts then each of the middle parts is called a middle third. If the corresponding extremities of the various middle thirds be joined by fair curves these curves will enclose the middle third of the cross section of the dam, and if a dam is properly designed the lines of resistance should fall within this middle third whether the reservoir is empty or full.

1

If R is the resultant force on any horizontal section XX, (Fig. 189) of the dam, then S the vertical component of R causes a normal stress on XX, and the horizontal component T causes a tangential or shear stress on XX1.

a

R

a

FIG. 189.

The normal stress produced by S on XX, is not uniformly distributed but varies uniformly from q- p at X which is furthest from S to q+p at X1.

1 If the contour of the cross section of the dam is made up of straight lines these fair curves will become a series of straight lines.

Let C be the middle point of XX,. Let XX, = 2a, and let the

distance of S from C be x, then, still considering 1 foot length of the

dam, q =

a

3

If x =

then p

S

2a

a

then p = q and there is no stress at X. If x is greater than

is greater than q and there is a tension in the dam at X which should be avoided.

The maximum compressive stress p + q should not exceed 7 tons per square inch.

foot.

The average weight of masonry dams is about 150 lb. per cubic

Exercises VIIIa

1. The horizontal distance between the axles of a bicycle is 3 feet 6 inches, and its weight is 28 lb. Assuming that the weight on the saddle is vertically above a point 9 inches in front of the rear axle, and that a greater pressure than 200 lb. must not be brought on either wheel, find by a funicular polygon, and mark distinctly, the greatest weight the bicycle will bear. Scale of weight, 60 lb. to 1 inch. Scale of length, th full size. [B.E.] 2. Using a funicular polygon, determine the resultant of the five parallel forces given in Fig. 190. The magnitudes of the forces are given in pounds. Use a force scale of 1 inch to 20 lb.

3. Determine the parallel forces, which, acting through the points L and M (Fig. 191), should balance the given forces. The magnitudes of the forces are given in pounds.

In reproducing the above diagrams take the small squares as of 05 inch side. 4. Six forces are given in Fig. 192, the magnitudes being in pounds. Find the resultant of these forces.

5. Find the resultant of the following three given forces, that is, find and measure r, 0, and a in the vector equation

Ꭱ Ra=121140 +433-10 +8161150 pounds. Employ scales of inch to 1 foot, and 1 inch to 10 pounds. 6. Five coplanar forces P, Q, R, S, T are given as follows:

[B.E.]

(To explain this manner of defining a force see Art. 91, p. 91.)

Find the resultant of the given system of forces. Measure its intercept, direction, and magnitude.

Find also and measure the resultant of Q, R, and S.

Employ a linear scale of full size, and a force scale ofinch to 10 lb. [B.E.]

7. Find the resultant of the forces given in Fig. 174, p. 91. The intercepts are in inches and the magnitudes of the forces are in pounds.

8. Determine r, x, and 0 in the equation

R == 01650 +0-7355285 +4410330 +2.73232600 + 0162140 +4-3205240°* Force unit, 1 pound. Linear unit, 1 inch.

9. Given the equation of equilibrium of five coplanar forces

090+10=1200305+3150270° + 8500315"

find r, s, and e. The unit of force is the pound and the unit of length is the foot. Employ a linear scale of inch to 1 foot, and a force scale of 1 inch to 100 pounds.

10. Find the centre of gravity of a piece of wire bent to the form shown at (a) Fig. 193.

In reproducing the above diagrams take the small squares as of 0.5 inch side. 11. Determine the centroids of the figures (b), (c), (d), (e) and (f) Fig. 193. 12. The triangle ABC (Fig. 194) is a diagram showing the intensity of earth pressure on a retaining wall, its area representing the total pressure P to the same scale that the area of the cross section of the wall represents the weight W of the wall. Determine the lines of action of P and W, the face AB of the wall being vertical. Find the centre of pressure on the base of the wall, this being the point where the resultant of P and W intersects the base.

Also find the centres of pressure on the sections D, E, and F.

[B.E.]

13. A circle 4 inches in diameter is subjected to pressure which varies uniformly from 10 pounds per square inch at one end of a diameter to 5 pounds per square inch at the other end of that diameter. Find the centre of pressure and the magnitude of the resultant pressure.

D

Copy double

this

size

F

EA

B

14. A rail section is given in Fig. 195. Find the centroid of this section. Through the centroid draw a line parallel to the bottom line of the section; this is the "neutral axis" of the section. This section is subjected to stress which is normal to the section and varies uniformly from the top to the bottom, being zero at the neutral axis where it changes sign. Find the centres of stress of the portions of the section above and below the neutral axis. If the stress at the lower edge of the section is 4 tons per square inch what is the stress at the upper edge, and what are the magnitudes of the resultants of the stresses on the portions of the section above and below the neutral axis?

FIG. 194.

15. A section of a large dam is given in Fig. 196. The two curved faces are plotted from the vertical line shown. The weight of the masonry composing the dam is 125 lb. per cubic foot. The weight of the water may be taken as 62.5 lb. per cubic foot.

Draw the lines of resistance for this dam, (a) when the reservoir is empty, and (b) when the top surface of the water is 10 feet below the top of the dam.

Draw also the "middle third lines of the section of the dam, and the diagrams showing the distribution of the vertical stress on the lowest horizontal section when the reservoir is empty and when the reservoir is full.

Linear scale, say, 10 feet to 1 inch.

[U.L. modified.]

99. Moment of a Force.-The moment of a force about a point or about an axis perpendicular to its line of action, is the measure of its turning power round that point or axis. The magnitude of the moment (generally called the moment) is the product of the magnitude of the force and the perpendicular distance of its line of action from the point or axis. For example, the moment of the force AB (Fig. 197) about the point M is equal to the magnitude of the force AB multiplied by MN, the perpendicular distance of M from the line. AB. If the unit of force is the pound, and the unit of distance is the inch, then the unit of moment is the inch pound or pound-inch. Other units of moment in common use are the foot-pound or pound-foot, the foot-ton or ton-foot, and the inch-ton

or ton-inch.

h

The construction shown in Fig. 197 is a very convenient one for M determining graphically the moment of a force about a point. AB is the line of action of the force, and M is the point. The construction is as follows. Draw ab parallel to AB, and make the length of ab to represent the magnitude of the force. Through M draw a'Mb' parallel to

b

AB. Choose a pole o. Join oa and ob. Take any point o' in AB.

Draw o'a' parallel to oa to meet a'Mb' at a', and draw o'b' parallel to ob to meet a'Mb' at b'. Then a'b' measured with a suitable scale will be the magnitude of the moment of the force AB about the point M.

The

Draw oh perpendicular to ab, and o'h' perpendicular to a'b'. triangles oab and o'a'b' are obviously similar, and ab: a'b' :: oh : o'h'. Hence ab xo'h' = a'b' × oh. But ab is the magnitude of the force AB, and o'h', which is equal to MN, is the perpendicular distance of M from AB. Therefore ab × o'h' is equal to the moment of AB about M, and therefore a'b' × oh is equal to the moment of AB about M.

If oh is made equal to the linear unit, then a'b' measured with the force scale will give the moment required. For example, if oh is 1 inch and a'b' measures 20 lb. on the force scale, then the required moment is 20 inch-pounds. It is not always convenient to make oh equal to the unit of distance, but it should be made a simple multiple or sub-multiple of it.

The following is the simple rule for determining the moment scale. Let oh be m times the linear unit, and let the force scale be n units of force per inch. Then the moment scale will be mxn units of moment per inch. For example, let the linear unit be one foot, and suppose that oh, measured with the linear scale, is 4 feet. Let the force scale be 100 lb. per inch, then the moment scale will be 100 × 4 = 400 foot-pounds per inch.

It may be pointed out that the figure a'o'b' is the link polygon of the force AB with reference to the pole o.

The following problems are important.

(1) An unknown force P acting through the point A (Fig. 198)

N

P

has a moment m, about the point O, and a moment m2 about the point O2: it is required to determine the force P.

Join 0,02. Draw O̟M1 and OM, at right angles to 0,02. Make O,M1 = m to any convenient scale and make OM, m2 to the same scale. OM, and OM, are on the same or on opposite sides of 0,02 according as m, and m, have the same or opposite signs. Join M,M, and produce it if necessary to cut 0,02, or 0,0, produced at B. Then B is another point in the line of action of P. The sense of P is