Having thus shown the basis of the graphical method of analyzing transverse strains, we will now give the actual method without wasting further space on proofs.

it there are three loads w, w1 and w11 on a beam A B (Fig. 151) we proceed as follows: at any convenient scale - to be known as the pounds-scale - layoff in pound-, dc = w11 also cb = w1 and ba = w. Let A B = l measured in inches-this scale being called the inch-scale. Now select pole x at random,

but at a distance (measured with pounds-scale) xy=. (k/f) = the safe modulus of rupture of the material. Draw x d, xc, xb and x a. Now begin at any point G of reaction q, draw G F parallel d x, till it intersects vertical w11 at F: then from F draw F E parallel cx to vertical to,: then draw ED parallel b x to vertical to; and then D C parallel ax to reaction p. From C draw C G, and through x draw xo parallel C G.

Strain Diagram.

We now have the following results:

Reactions.

 o d = reaction q (measured with pounds-scale.) ao = " p " " " "

any vertical through figure C D E F G C, (measured with inch-scale) gives the amount of r = required moment of resistance in inches, at point of beam where vertical is measured. The longest vertical passes through the point of greatest bending-moment in beam. Multiply any vertical (in inches) with x y (in pounds) to obtain amount of bending-moment at point of beam through which vertical passes.

or we should have: r=v

(92)

Moment of Resistance

Where r = the required moment of resistance, in inches, at any point of beam, provided pole distance x y = (k/f).

Where v= the length (measured with inch-scale) of the vertical through upper figure C D E F G C at point of beam for which r is sought. And further: m =v.xy

(93)

Bending-moment.

Where m = the bending moment at any point of beam in pounds-inch.

Where v = the same value as in Formula (92)

Where xy = the length, (measured with pound-scale) of distance of pole x from load line, in upper strain diagram x a d.

If now we draw horizontal lines through d, c, b and a; and through o the horizontal line for horizontal axis; and continue these lines until they intersect their respective load verticals w11, wl and w, the shaded figure 01 HIJKLMN 0 01 will give the vertical shearing strain along beam. Any vertical (as R, S) drawn through this figure to horizontal axis and measured with pounds-scale, gives the amount of vertical shearing at the point of beam (R) through which vertical is drawn. Or, s = v11 (94)

Vertical Cross-shearing.

Where s = the amount of vertical shearing strain in pounds, at any point of beam.

Where v11 = the length (measured with pounds-scale) of vertical through figure 01HIJ K L M N 0 01 dropped from point of beam for which strain s is sought.

We now divide G C into any number of equal parts - say twelve in our case - and begin with a half part, or G to 1 = 12 to C= 1/24. G C; also 1 to 2 = 2 to 3 = 3 to 4 = 4 to 5, etc. = 1/12. G C and make the new lower load line gc with inch-scale so that g to I = length of vertical 1 e

Deflection

Diagram

 further I to II = length of vertical 2f " II " III = " " " 3h " III " IV = " " " 4 i, etc. unti. " XI " c = " " " 12 k

Now select arbitrarily a pole z at any distance zj from load line g c. Now draw below the beam where convenient (say I. Fig. 151) beginning at g1, the line gt1 e1 parallel g z till it intersects the prolongation of 1 e (from above) at el; then draw e1fl parallel I z till it intersects vertical 2 f at f1; and similarly draw f1 h1 parallel II z; also h1 i1, parallel III z, etc., to ml k1 parallel XI z and finally k1 c1 parallel c z. The more parts (l1) we divide the beam into, the nearer will this line g1 e1fl m1 k1 c1 approach a curve. The real line to measure deflections would be a curve with the above lines as tangents to it; we need not, however, bother to draw this curve for practical work. Now draw c1 gl and parallel thereto z o. Divide g1 c1 at o11 so that1: g1 o11: c1 o11 = c o: go, then will o11 be the point of greatest deflection along beam. This will be further proven by

1 Note that the division of the line g1o11c1 is the reverse of the division of the line g o c.

the fact that the greatest vertical (in lower figure I) will pass through o11, if the real curve were drawn. The figure g1 e1 f1 h1 i1 m1 k1 c1 g1 will measure the amount of deflection of beam at all points of beam. The deflection at any point of beam being proportionate to length of its vertical through lower figure I. The amount of this deflection will be

Fig. 152.

δ = v1.l1.z.j. (k/f)/e.i (95)

Where δ = the deflection, in inches, at any point of beam, if pole distance of upper strain diagram (x y) = (k/f).

Where v1 = the length of vertical, in inches, dropped from said point through lower figure I (see Fig. 151)

Where l1 = the length, in inches, of each equal part 1 to 2, 2 to 3, 3 to 4, etc., into which beam was divided, [in our case l1= 1/12.l]

Where i = the moment of inertia, of cross-section at said point, in inches.

Where zj= the distance (measured with inch-scale) of pole z from load line in lower strain diagram.

Where (k/f) = the safe modulus of rupture, per square-inch, of the material.

Where e = the modulus of elasticity, in pounds-inch, of the material.

If we were to so proportion the beam that the moment of resistance at each point would exactly equal the required moment of resistance as found above, we should have:1

δ = v1.l1.zj. (k/f) (96) d

v. - . e 2

Where δ, δ, zj, (k/f) , e and l same value as in Formula (95).

Where v= length of corresponding vertical in upper figure C D G E C, (to vertical v1 of lower Fig. I) to be measured in inches.

Where d/2= one-half the total depth of beam, in inches. Had we not made x y = (k/f), we should have