204. If it is desired to take into account the forces due to the weights of the bars in a girder or roof-principal, we may as an approximation suppose half the weight of each bar to act at each end. This is an approximation to the true state of the case, because it neglects the stresses set up in the bar due to the bending. The error thus introduced is however very small.

205. In roof structures the engineer has also to take into account the stresses set up by the pressure of the wind on the roof and by the loading due to any likely fall of snow.

The wind pressure being normal to the roof has a component in a horizontal direction; it would not do therefore to have the framework simply supported freely at the ends as in § 203. There must be some constraint introduced at one end to prevent the roof moving bodily off.

206. Suppose the framework to be as sketched and the forces due to the weight of the bars, snow and roofing, and the wind pressure to act on the joints as indicated.

A and B are the points of support, and the reactions there must be such that they are in equilibrium with the forces F1, F2...

If we draw the vector and link-polygons for the forces F... their resultant can be found in magnitude, direction, sense and position.

The problem of finding the components of this resultant passing through two given points is an indeterminate one, but if we know the direction of one of the components, say at B, then the magnitudes of the components and their directions can be found. We know (§ 177) that, if three forces are in equilibrium, they must pass through a point; in the present case the point is determined by the known resultant and the known direction of the reaction at B; the reaction at A must be in the line joining the point so determined to the point A.

If the end at B is fitted with an iron shoe on rollers which rest on a horizontal iron plate, then the direction of the reaction at B is vertically upwards.

Suppose then the reactions at A and B have been found by the above method, then the vector-polygon for all the forces must be a closed one. In the figure 0, 1, 2, 5, 6, 10, 11, 0 is the vector-polygon for the external forces.

To determine the stresses in the bars we may now proceed as in the previous example.

Start with A and draw the vector-polygon for the forces acting there, it is 01230 and hence 23 is in compression and 03 in tension. Next take the point C where there are only two unknowns (instead of F where there are three). The polygon is 25432, hence 54 is in compression, and so are 43 and 32.

The rest of the construction offers no new difficulty.

and (2).

EXERCISES XII.

Determine the stresses in the bars of the following frames (1)

Determine the reaction at the supports and the stresses in the bars of the following frames (3) to (10).

(3) The Warren Girder of 9 equilateral triangles, having loads 2, 3, 3 and 4 tons on the lower boom.

(4) The Warren Girder of 7 equilateral triangles, having a load of 10 tons at the first junction of the lower boom.

(5)

3 tons 3 tons 3 tons

(6)

3 tons each.

8'

(8)

(7) Girder as in (6) but loaded with 3, 0, 5, 1, 4 tons from left to right.

(11) Calculate the various stresses in the bars in examples 1—7.

(12) Weights of 12, 16, 20, 15, 15 and 18 lbs. are to be attached to the vertices of a string polygon, the horizontal distances between the weights are to be 4, 3, 2, 4 and 5 inches. Find the form of the polygon and the tensions in each of the strings for the following cases. (The point A may be supposed known in each case and also the vertical through B.)

(a) The directions of the strings at A and B make an angle of 30° with the horizontal.

(b) The reactions at A and B to be 5 and 6 lbs.

(c) The string between the 2nd and 3rd lines to be horizontal and at B to make an angle of 60° with the vertical.

(d) The string between the 3rd and 4th lines to be horizontal and to have a tension equal to 3 lbs.

(13) A suspension bridge has a dip of 20 feet and a span of 250 feet. The total load is 300 tons and the vertical equidistant tie-rods are 5 feet apart. Draw the shape of the supporting chains and find the stresses in them.

Find the stresses in the roof girders (14) and (15) and determine which bars are in compression.

(16) In case (14) if A be pin-jointed and B on rollers and at C and D ́act in addition horizontal forces of magnitude 5, determine the reaction at A in magnitude and direction and determine also the tensions and compressions set up.

(17)

Find the compressions and tensions in (17) and (18).

2

(18)

(19) Find the reactions at A and B and the compressions and tensions in the bars.

2.5

3

B

207. Reciprocal Figures. In the examples given we have two figures. The first is the frame together with the lines of action of the applied forces, the second is the stress-diagram. Between these exists a remarkable reciprocal relation which is expressed by the notation used and which makes this notation not only possible but of great use in so far as it is a sure guide in the drawing of the stress-diagram.

By the construction we have to each line in the frame a corresponding line in the stress-diagram. These are parallel and denoted by the same two numbers, which in the one name the sides, in the other the ends of the line. These numbers denote in the frame spaces bounded by the lines, but in the other they denote points. Also each number occurs but once in each figure. All lines therefore which have in their name one number in common lie in the frame round an area or form a closed polygon like 8 in Fig. 131. But in the stress-diagram