This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

The wind blowing on one side of the building causes a compressive stress in the column on the leeward side (the side opposite that on which the wind blows) and a tensile stress in the column on the windward side (the same side on which the wind blows). It also creates a bending moment as mentioned above; and this, as well as the direct stresses, must be taken into account when the post is designed. The case is that of a member under direct compression and bending at the same time.

The stresses in the knee-braces and the columns, and the bending in the columns when the ends of the posts are not fixed, may be computed from the following formula?, in which,

W = Total wind load perpendicular to the roof;

Wh = Horizontal component of W;

Wv = Vertical component of W;

W1 = Total wind load on the side of the building; w = Unit wind load normal to the roof; w1 = Unit wind load normal to the side of the building; a = Distance between trusses, in feet.

These and other characters are shown in Fig. 43.

Fig. 43. Notation for Formulas, Ends Free.

w | = | wa | ||||

w1 | = | w1 | ah | |||

H1 | = | H2 | = | Wh | + | W1 /2 |

Sb'c'= - W1 h/2+Wh(h +r/2)+ Wv l /4 /e=V2

Sbc=+W1h/2+Wh(h+r/2)-Wv3l/4/e=V1

Sa'b'= - H2h/e

Sab = + H1h-W1 (h/2 / e

Bending moment at b = H1 - W1 ( h/2-m)

Bending moment at b' = H2 n.

The stresses in the truss caused by the wind are the same as if it were under the action of the normal wind load W, and in addition two concentrated loads equal in intensity and direction to the stresses in the knee-braces and at the same point of application, and two forces E1 and E2, which may be computed as follows:

E = H2n/ m

E2 = H1n-W1(h/2-m)/m

For the points of application for these loads and for their direction, see Fig. 44. The stresses can now be computed by the method of Statics.

The diagram for such a truss-bent is given in Fig. 45. The span is 60 feet, the rise ¼, the distance between trusses 16 feet; and the wind pressure is taken as 18 pounds per square foot normal to the roof surface, and 20 pounds per square foot normal to the sides. In this case, w = 18 pounds; a = 16 feet; r =60 ÷ 4 = 15 feet; w = 20 pounds; h = 20 feet; n = 14 feet; m = 6 feet; and l = 60 feet. The length of L0U4 is readily computed to be 33.5 feet; L0 L1 9.1 feet; and e = 5 feet. The values of the quantities and stresses are computed as follows (see Fig. 46):

w | = | 18 | X | 16 | _ | 9 650 pounds. | ||||||||

w1 | = | 16 | X | 20 | X | 20 | = | 6 400 pounds. | ||||||

Wh | = | (9 650 | 33.5) | X | 15 | = | 4 320 pounds. | |||||||

w. | = | (9 650 | 33.5) | X | 30 | = | 8 650 pounds. | |||||||

H1 | = | H2 | = | (4 320 | + | 6 400) | ÷ | 2 | = | 5 360 pounds. |

Fig. 44. Application and Direction of the Exterior forces.

S a b = + [5360 x 20 - 6 400 x 10] / 5= + 8640 pounds.

Sa'b'= - 5 360 X 20/5= - 21 440 pounds.

E1 = 5360 x 14/6 = 12520 pounds.

E2 = [5360 x 14 - 6 400 x4] / 6 = 8240 pounds.

The horizontal and vertical components of the stresses in the knee-bracing should now be computed. They are:

For ab: horizontal, 7 240; vertical, 4 760 pounds.

For a'b': horizontal, 17 900; vertical, 11 800 pounds.

As a check upon the computations, the sum of the values of E1, E2, and Wh should be equal to the sum of the horizontal components of the knee-braces. By summing up the above values, it will be seen that they check by SO pounds, which is less than 0.4 of one per cent and is a close enough check (see Figs. 4G and 47).

To obtain the vertical reactions, proceed as with a simple truss. For R2, take the center of moments at L0 (see Fig. 47). Then:

R2 = 1/60 { 8 650 x 15 + 4 320 x 7.5 + 4 760 x 9.1 - 11 800 x (60-9.1) }

= - 6 514 pounds.

The negative sign indicates that the reaction acts downward; that is, the truss must be riveted to the post at L8, or the end of the post would be lifted off the top of the column.

Fig. 45. Stress Diagram of Truss-Bent under Wind Load.

For R1, the center of moments is at L8, and the resulting equation is:

R1 = 1/60 { - 11 800x 9.1 + 8 650 (60-15) - 4 320 X 7.5 + 4 760 (60-9.1) }

= +8 180 pounds. The bending moment at b is:

Mb = 14 X 5 360 - 4 X 6 400 = 49 440 pound-feet; and the bending moment at b'is:

Mb' = 5 360 X 14 =75 040 pound-feet.

The forces in their proper direction are now placed on a diagram of the truss (Fig. 47), and the stresses are solved by the method of Statics. The stress diagram is given in Fig. 45, and the stress record in Table VII.

The above formulae are for cases when the columns are free at the lower end. When the columns are not free, they are called fixed; that is, they are supposed to be so tightly connected that they cannot move when the post bends as shown in Fig. 42. In such cases the result is the same as if the columns were shortened by an amonut n ⅓ 2, and the following formulae result (see Fig. 48):

Fig. 46. Position, Direction, and Intensity of Wind Forces, Ends Free.

Fig. 47. Position, Direction, and Intensity of Exterior Forces.

Member | Stress |

X-2 | -15 700 |

X-3 | -15 700 |

X-6 | -10100 |

X-7 | -10700 |

X-9 | +1500 |

X-13 | +15400 |

1-2 | +5200 |

2-3 | -2410 |

3-4 | +8500 |

4-5 | -7500 |

5-6 | +2700 |

6-7 | -2410 |

7-8 | +11000 |

8-9 | -7600 |

Member | Stress |

9-10,11-12 | +6500 |

12-13,14 | -15300 |

Y-4 | +6500 |

Y-8 | -1900 |

Y-12 | -9600 |

13,14-15 | -1300 |

1-Y | +8640 |

15-Y | -21440 |

A-1 | -8180 |

C-15 | +6514 |

b-c | -3457 |

b'-c' | - 5193 |

9-10-11 | 0 |

13-14 | 0 |

W = wa W1 = w1a (m + n/2)

H1 = H2=Wh+W1/2

Sab = { H1 (m+n/2)-W1 (m+n/2/2)}/e

Sa'b'= + {H2 (m/2 + n/4)} / e

Sbc= + - {W1 ( m/2 + n/4 ) + Wh (m + n/2 + r/2) - Wv, 3l/4}/l

Sb'c'= + {W1 (m/2 + n/4 ) + Wh (m + n/2 + r/2 ) + Wv l/4} /l

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