This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

Angle of Inclination of Rafters with Horizon. | Pressure or Load, in pounds, per square foot of Roof Surface. |

10° | 9 2/3 |

15° | 14 |

20° | 18J |

25° | 22 1/2 |

30° | 26 1/2 |

35° | 30 |

40° | 33 1/3 |

45° | 36 |

50° | 38 |

55° | 39 2/3 |

60o | 40 |

to | |

90° |

It will be noticed that approximately the pressures in pounds are about eighty per cent or four-fifths of the number of degrees of the angle of inclination.

Where the wind is taken separately the allowance for snow in the strain diagram for dead loads should be fifteen pounds per square foot of roof surface.

Having once ascertained the amount of load on each joint, the strains on the different members of the truss are found by the general methods given at the end of Chapter I (Strength Of Materials). (Pages 69 to 74, Vol. I.)

Particular attention is again called to the method of notation, and to the necessity of reading off the pieces in their proper order, and of reading around each joint in the same direction. The writer always uses the direction in which the hands of a watch would travel around each joint.

In calculating the strains each joint can be analyzed by a separate strain diagram, or all of the strain diagrams can be combined into one. As the latter method is much more convenient and less liable to error, it is the one always adopted.

In laying out the strain diagram we begin with the joint with the least number of members and this usually is at one of the reactions, where we have only one strut and one tie. Having found these strains we pass to one of the joints at their other ends, and so on. The reason for doing this is that it will be found impossible to draw the lines in the strain diagram representing any joint where there are more than two unknown strains. By beginning, therefore, with a joint of two members only, the strains on these can be found. We then can pass to joints of three members, containing at least one of the former strains and so on. Figures 213 and following ones give a large number of roof designs with their corresponding strain diagrams. We will analyze one or two of these and the student can puzzle out the rest.

Approximate Rule for Wind.

Allowance for Snow.

Drawing of strain diagram.

Only a few will offer any particular difficulty, and these will be taken up and explained later.

In all the Figures dotted lines mean that the dotted member is in tension and full lines that the member is in compression. The numbers in Figures 213 to 228 give the amount of strain, in pounds, on each member due to one pound of load at each joint for roofs with inclination angles as shown in Figures.

All that is necessary therefore, where roofs are designed similar to any of these Figures, and with same inclinations and angles, to ascertain the amount of load on the joints and then multiply the number or given strain on each member by the amount of load on each joint. This will give the actual amount of strain on each member.

Dotted lines in tension.

For instance, we will say we have designed a roof truss similar to Figure 213 with the principal rafter at an inclination of 26o30'; we will say the trusses are 10 feet apart and 48 feet span, and weight of roof including snow and wind 50 pounds per square foot measured on the slant. By scaling the rafters we find they measure 27 feet each in length, therefore load on each joint = 27/2 .10.50 = 6750 pounds. We now refer to Figure 213 and have the strains, as follows :

Compression on rafter | BG | = | 3,35.6750 | = | + | 22612 | pounds. |

Compression on rafter | CH: | = | 2,25.6 750 | = | + | 15187 | pounds. |

Compression on strut | GH | = | 1,1.6750 | = | + | 7425 | pounds. |

Tension on tie | GO | = | 3,0.6750 | = | - | 20250 | pounds. |

Tension on tie | HI | = | 1,0.6750 | = | - | 6750 | pounds. |

Had we drawn the strain diagram and made a b = ey=6750/2 = 3375 pounds at any scale, and at same scale made bc = cd = de = 6750 pounds, we should find that at the same scale the respective lines would measure:

bg | = | 22700 | pounds. |

c h | = | 15200 | pounds. |

gh | = | 7400 | pounds. |

go | = | 20300 | pounds. |

h i | = | 6750 | pounds. |

and to ascertain whether these strains were compression or tension we should follow the direction of each line at each joint and see whether it thrusts against or pulls away from the joint.

To draw the strain diagram we first select a convenient scale, by which we will measure all the loads and strains. We now draw our load line, making in (Figure 213) a 6 = the load on A B the foot of main rafter, we then make bc = the load on joint B C, the next one on main rafter, c d = the load on apex and so on ; as we know the reactions will each be just one-half the load, we locate o half way between f and a, in other words we make fo = reaction FO and o a - reaction O A.

To get the strains we begin at the joint A B G 0 A at foot of main rafter, for here there are only two unknown strains, namely, the compression on B G and the tension on GO.

In strain diagram draw b g parallel B G; and g o parallel G O till they intersect at g; then will bg be the amount of thrust on the joint or the compression in B G, and g o will be the amount of pull on the joint or tension in G O. To make sure of these we read off the lines following the proper succession, namely, A B, B G, G O, O A ; referring now to strain diagram we read a b, this is down or a vertical load ; next b g, this is down or towards the joint, therefore compression; next g o, this is to the right or pulling away from the joint, therefore tension; and finally o a (which brings us back to the point of starting a) and being upward is, of course, the direction of the reaction. As our strain diagram is a closed Figureabgoa, we know the joint is in equilibrium. Had we failed to get back to the point of starting a, we should have known there was some missing member to the truss at this joint.

Whereto begin.

On the other hand, if we had had two letters coming onto the same point - (which would be the case with letters F and G were we to draw strain diagram for Figure 212) - we should know that the member or line between these two letters was superfluous so Ear as aiding the general truss is concerned. We now (in Figure 213) pass to the next joint with only two unknown strains. This is evidently the joint at half the height of rafter, we have just found the amount of compression on G B - (note that we now read G B and not B G as before, for around this new joint the hands of a watch would travel in the direction G B) - and the only unknown strains are the compressions on CH and on H G. In strain diagram we draw ch parallel C H and h g parallel H G till they intersect; c h will then he the amount of compression on C H and h g the amount of compression on H G. We now read off the pieces in succession B C,

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