Hammer-beam Truss.

FlG.243.

Increased stress in curved members.

m = s.x

(129)

Where m = the (cross) bending-moment, in pounds-inch, existing in a curved strut or tie at its centre due to longitudinal strain.

Where s = the calculated longitudinal strain that would come on the strut or tie if it were straight, in pounds, per square inch.

Where x = the length, in inches, of the longest versed Bine of the curve (at the centre.)

If the longest versed sine is not at the centre, the bending-moment m will be at the point where the longest versed sine is located. Of course, the curved member must be designed to resist both the longitudinal strain and the additional strain due to the bending-moment.

Fig. 248.

Example I.

A Georgia pine strut is 10 feet long

(measured on a straight line) but is curved, the versed sine at centre being 20 inches. It resists a compression strain of 15000 pounds.

What size strut is required?

The bending-moment at centre will be Formula (129) m= 15000.20 = 300000 pounds-inch.

The safe modulus of rupture (k/f) for

Georgia pine (Table IV) is 1200 pounds.

Inserting these values in Formula (18) we have the required moment of resistance to this bending-moment: r=300000/1200 =250.

Example of curved strut.

From Table I, Section No. 2, we have for a rectangular cross-section r = b.d2 /6therefore b.d2/6= 250, and b.d2 = 1500

If now we make d = 12 inches, we should require a width b=1500/144 = 10,4 inches.

It should be noted that d must always be assumed in the direction of the versed sine, that is resisting the bending-moment.

We must now add sufficient material to resist the longitudinal compression. As the strut will evidently be nearly square, 12 inches will be our least diameter, inserting therefore values in Formula (3) we should have (assuming the ends to be secured by bolts only) :

15000= a.750/1+1202.00067/12 or, a = 36 square inches.

Now as a = 6.d and d having been fixed at 12 inches we should have, b =36/12 = 3 inches.

Adding this to the above we should require a strut,

12 inches X (10,4+ 3) = 12 X 13,4

In practice we should probably make it 12 inches X 12 inches.

Example II.

The lower chord of a wrought-iron arched truss is made of a channel iron. At one panel the length between bearings is 5 feet; the tension 36000 pounds, the versed sine of the curve 6 inches. What size channel is required?

Example of curved tie.

The bending-moment m will be m = 36000.6 = 216000 pounds-inch. Inserting values in Formula (18) we have required moment of resistance r = 216000/12000 = 18 The additional area required to resist the direct tension will be a = 36000/12000 = 3 square inches.

We now consult Table XXI. We take the neutral axis normal to web, as this will, of course, be the position of channel in the truss and in resisting the bending-moment.

We select for a trial the 10 1/2 inch-105 pounds per yard channel; its area a is 10,5 square inches, and its moment of resistance r = 24,64. Now of the area 3 square inches resists tension leaving us 7,5 square inches or 7,5/10,5 = 5 of the whole amount to resist cross-bending. The amount of re available to resist cross-bending will, therefore, be = 5/7.r = 5/7 .24,64 = 17,6.

This is not quite equal to the required r (18) but is near enough for all practicable purposes, we should, therefore, use a 10 1/2 inch - 105 pounds per yard channel.

We will now return to our hammer-beam truss. Assuming then that the wall is capable of resisting the thrust we should analyze our truss somewhat similarly to the way we did with the scissors truss.

On the left side of truss (Figure 244) we omit the circle corresponding to 0 W entirely, and draw the one corresponding to O V straight. We now can assume that we can divide the truss horizontally at P O which will give us a truss above P 0 similar to Figure 213 and a truss below it consisting of an inclined trussed strut running from bottom joint X to joint C D with a horizontal strut P O and an outward thrust O X to be resisted at its foot. To obtain this outward horizontal thrust 0 X we will temporarily consider the inclined trussed strut as a straight line running from joint X to CD. The load will evidently be three-quarters of the load coming on the entire rafter, if then in Figure 247 we make ax = this load or = 3 in our case, and draw x o horizontally and o a parallel to the straight line (representing the inclined strut in Figure 244) we shall have the amount of the horizontal thrust x o. We can now make separate strain diagrams for the upper and lower trusses, in which case we will find that for the upper truss there exists a tension just equal to o x in the member P 0 and for the lower truss there exists a compression just equal to o x in the member P 0. In other words there is no stress in P 0. This is confirmed by drawing the combination diagram (Figure 245) where p and o fall on the same point.