This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.

-To determine the wind stresses, the frame diagram should be redrawn. The wind loads, considered as acting at each panel point, may be determined from Table VII, page 68, and are shown in Fig. 41. As the ends of the truss are secured against sliding, the reactions act parallel to the wind pressures. If, however, the left end of the truss is secured, and the right end rests on rollers (as is sometimes the case with iron or steel trusses to permit expansion), the right reaction, instead of being parallel to the direction of the wind, would be vertical. The wind-stress diagram for such a truss would, with this exception, be found similarly to that for a fixed-end truss.

To determine reactions R1 and R2, let R1 be the center of moments; then the perpendicular distance between the line of action of R2 and the point R1 is 71.22 ft., obtained by extending the left-hand rafter to intersect the line of action of R2 at y'. Regard Ay' as a beam, and calculate reactions R1 and R2 by the method given for beams. Taking moments about R1 the reaction R2 is found to be 3,516 lb.; and R1 = 11,200 lb., the total load, -3,516 lb. = 7,684 lb.

Proceed with the wind-stress diagram, Fig. 42, by drawing the load line a f parallel to the direction of the wind in the frame diagram. Lay off to any scale the forces a b, b c, c d, etc., equal to AB, BC, CD, etc., respectively. Then, from a lay off a z, equal to reaction ZA, or R1 . If the loads have been laid off accurately, f z should be equal to R2.

The first joint to analyze is A B K Z. Draw b k parallel to BK; and from z, z k parallel to KZ; they intersect at k. The polygon of forces is from a to b, b to k, k to z, and z to the starting point a. Joint B CL K is analyzed similarly.

Fig. 42.

To analyze joint KLMZ: k l being known, the next member is LM; from I draw I m, parallel to LM. As the next member is MZ, to which m z is parallel, the point m is located where I m intersects m z; this completes this joint, the polygon of forces being from k to I, I to m, m to z, and z to k. The stresses at the other joints may be found in the same way as those explained. The members shown in dotted lines do not sustain wind stresses when the wind blows upon the left side of the truss.

The final joint is EFQP, at which there is only one unknown force - the stress in FQ. A line drawn from f parallel to FQ should pass through q. This is always a test of the accuracy of the work, and if the last line does not close on the proper point, when drawn parallel to the member it represents, the stress diagram should be redrawn, to determine whether the loads and reactions have been laid out correctly, and whether any joint or member has been omitted.

The two diagrams completed, scale, in round numbers, the stresses in each member, indicating compressive and ten-sile stresses by plus and minus signs, respectively. Tabulate results as follows:

Member. | Vertical Load. Pounds. | Wind Load. Pounds. | Total. Pounds. |

BK | + 27,000 | + 12,000 | + 39,000 |

CL | + 23,500 | + 10,000 | + 33,500 |

DN | + 19.500 | + 7,600 | + 27,100 |

EP | + 16,000 | + 5,680 | + 21,680 |

KZ | - 24,000 | - 13,300 | - 37.300 |

MZ | - 21,000 | - 10,000 | - 31,000 |

OZ | - 17,500 | - 7.000 | -24,500 |

LK | + 4,000 | + 3.500 | + 7,500 |

NM | + 5,000 | + 4,400 | + 9,400 |

PO | + 6,500 | + 5,400 | + 11,900 |

ML | - 1,600 | - 1,500 | - 3.100 |

ON | - 3,500 | - 3,000 | - 6,500 |

QP | - 10,600 | - 4,500 | - 15.100 |

FQ | + 16,000 | + 6,600 | + 22,600 |

Continue to: