In a similar manner, taking moments about the intersection of the top chord and the diagonal, leaves only the moment of the bottom chord stress to be determined, which must equal the sum of the moments of the loads about this point.

In Fig. 272 these top and bottom chord stresses are determined by taking sections through the truss at the left of each panel point. These top chord stresses will be worked out below.

Stress in ab:

77,400 X 6.17= + 476,000 6,450 X 6.17 = - 39,500

+ 436,500 ft. lbs. = Moment to be balanced by mo ment of stress in top chord.

Stress in ab = 436,500/6.75 = +64,700 lbs.

Stress in be:

77,400 X 6.17 X 2 = + 955,000

12,900 X 6.17 X 2 = - 159,000

M of bc = + 796,000

Stress in be = 796,000/6.75 = +118,000 lbs

Stress in cd:

77,400 X 6.17 X 3 = + 1,430,000

12,900 X 6.17 X 4.5 = - 357,000

M of cd = + 1,073,000

Stress in cd = 1,073,000/6.75 = +159,000 lbs.

Stress in de:

77,400 X 6.17 X 4 = + 1,910,000

12,900 X 6.17 X 8 = - 637,000

M of de = + 1 273,000

Stress in de = 1,273,000/6.75 = +188.500 lbs

Stress in ef:

77,400 X 6.17 X 5 = + 2,390,000

12;900 X 6.17 X 12.5 = - 995,000

M of ef = + 1,395,000

Stress in ef = 1,395,000/6.75 = + 207,000 lbs.

Stress in fg:

77,400 X 6.17 X 6 = + 2,860,000

12,900 X 6.17 X 18 = - 1,430,000

M of fg = + 1,430,000

Stress in fg = 1,43,000/6.75 = + 212,000 lbs

In explanation of the above, it will be noted that the moments of those forces causing right-handed rotation are designated "+" (plus), and those causing left-handed rotation are designated " - " (minus). Also note that the moment at any point consists of the moment of the reaction which for the left-hand reaction causes a positive moment and of the moment of the panel loads (including those over the end) which cause negative moment. As these panel loads are all equal, their moment can most easily be obtained by multiplying this panel load by the panel length and by the sum,of the number of panels detween the origin of moments and the loads. Take for example the stress in cd; there is one full panel load distant one panel length, and a half-panel load distant two panel lengths; combined, these equal one full panel load distant two panel lengths.

As a check on the moment at the center, it is well to calculate in a different manner. As this is the point of maximum moment, this moment is the sum of the maximum moments which each load can produce. Or it is the sum of the reaction of each panel load, multiplied by the distance from the reaction to the panel point. Therefore, as a check, we have:

M = 12,900 X 6.17 X 18 = 1,430,000 foot-pounds.

In a similar manner, the stresses in the bottom chord would be determined, taking moments about the top chord intersections with the diagonals.

There is a simpler way, however. If a section is taken along the line C D, and the portion to the right is removed as shown by Fig. 274, it will be seen that - just as was explained for the section A B - the forces acting along the lines of the members cut are the stress in these members necessary to maintain equilibrium. Since the forces along ah and ij are horizontal, and are the only horizontal forces acting upon the structure, then these two must be equal in order to fulfill the condition stated - that the sum of the horizontal forces equals zero. This determines all the bottom chord stresses from the top chord stresses.

Direction of Stress. A stress acting toward the portion of the truss not considered removed, is positive and is compression. A stress acting toward the portion considered removed, is negative and is tension. The direction in which the stress must act is determined by the direction of the resulting moment of the external forces. If these produce right-hand rotation, then the stress in the member must produce left-hand rotation in order that the algebraic sum of the moments shall be zero. Therefore, in the case of the top chord stresses previously' illustrated, since the resulting moment of the external forces is always positive, the moment of the stress in the chord must be negative or act toward the portion not removed, and the stress is therefore compression.

In the case of the bottom chord, this stress must act in the opposite direction to the stress in the top chord, and is therefore tension.

Stress in Verticals. This is determined by the condition that the algebraic sum of the vertical forces must be zero. Taking a section similar to C D, the only vertical force, aside from the loads acting on the truss, is the stress in the vertical member cut. This stress, therefore, equals the algebraic sum of the external forces on the left of this section, or the shear, and is opposite in direction or acts downward toward the portion of the truss not removed; the stress therefore is comnression.

Fig. 274.

 Stress in ah = 77,400 - 1,200 = + 76,400 " " bi = 77,400 - 8,850 = + 68,500 " " cj = 77,400 - 21,750 = + 55,650 " " dk = 77,400 - 34,050 = + 42,750 " " cl = 77,400 - 47,550 = + 29,850 " " fm = 77.400 - 60,450 = + 16,950 " " gn panel load = + 10,500

This latter stress in gn is obtained by taking the section around the panel point g, thus cutting only the top chord and the vertical. If the section was taken any other way through this vertical, it would cut a diagonal, and it would be necessary to determine the vertical component of this stress before the stress in the vertical would be known.

Stress in Diagonals. This is determined by taking sections similar to A B, and determining the vertical component of the stress in the diagonal. This vertical component must equal the algebraic sum of the vertical forces on the left, or the shear at the section. The relation of the actual stress in the diagonal to the vertical component, is the same as the relation between the length of the diagonal and the vertical depth. In this manner the stresses are worked out below:

 Stress in ai = 1.35 x 70,950 = - 96,000 " " bj = 1.35 x 58,050 = - 78,500 " " ck = 1.35 x 45,150 = - 61,000 " " dl = 1.35 x 32,250 = - 43,700 " " em = 1.35 x 19,350 = - 26,200 " " fn = 1.35 x 6,450 = - 8,750

The direction of stress in these diagonals will be understood from Fig. 273, which shows the vertical component acting in an opposite direction to the resultant external forces.