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.

At each of the next joints (4 and 5), there are three unknown forces, and the polygon for neither joint can be drawn. We might draw the polygons for the joints on the right side corresponding to 1, 2, and 3, but no more until the stress in one of certain members is first determined otherwise. If, for instance, we determine by other methods the stress in rk, then we may construct the polygon for joint 4; then for 5, etc., without further difficulty.

To determine the stress in rk, we pass a section cutting rk, qr, and eq, and consider the left part (see Fig. 44c). The arms of the loads with respect to joint 8 are 7.5, 15, 22.5, and 30 feet; and hence, assuming F1 to be a pull,

-F1 x 15 -1,900 X 7.5-1,900 X 15-1,900 X 22.5-950x30

+ 7,600 X 30 = 0; or,

F1= [- l,900x 7.5-1,900x15-1,900x22.5-950x30+7,600x30] /15

= 7,600 pounds

Since the sign of F1 is plus, the stress in rk is tensile.

Now lay off KR to represent the value of the stress in Kr just found, and then construct the polygon for joint 4. The polygon is KNORK, NO and OK representing the stresses in no and or. Next draw the polygon for joint 5; it is ONMCDPO, DP and PO representing the stresses in dp and po. Next draw the polygon for joint 6 or joint 7; for 6 it is PDEQP, EQ and QP representing the stresses in eq and qp. At joint 7 there is now but one unknown force, namely, that in qr. The polygon for the three others at that joint is ROPQ; and since the unknown force must close the polygon, QR must represent that force, and must be parallel to qr.

Fig. 45.

On account of the symmetry of loading, the stress in any member on the right side is just like that in the corresponding member on the left; hence, it is not necessary to draw the diagram for the right half of the truss.

Snow=Load Stress. The area of horizontal projection of the roof which is supported by one truss is (60 X 15 = 900 square feet; hence the snow load borne by one truss is 900 X 20 = 18,0000 pounds, assuming a snow load of 20 pounds per horizontal square foot. This load is nearly 1.2 times the dead load, and is applied similarly to the latter; hence the snow load stress in any member equals 1.2 times the dead load stress in it. We record, therefore, in the third column of the stress record, numbers equal to 1.2 times those in the second as the snow-load stresses.

Wind Load Stress. The tangent of the angle which the roof makes with the horizontal equals 15/30 or ½ ; hence the angle is 26° 34', and the value of wind pressure for the roof equals practically 29 pounds per square foot, according to Art. 19. As previously explained, the area of the roof sustained by one truss equals 1,006.2 square feet; and since but one-half of this receives wind pressure atone time, the wind pressure borne by one truss equals

503.1 X 29 = 14,589.9, or practically 14,600 pounds.

When the wind blows from the left, the apex loads are as represented in Fig. 45a, and the resultant wind pressure acts through joint 5. To compute the reactions, we may imagine the separate wind pressures replaced by their resultant. We shall suppose that both ends of the truss are fixed; then the reactions will be parallel to the wind pressure. Let R1 and R2 denote the left and right reactions respectively; then, with respect to the right end, the arms of R1 and the resultant wind pressure (as may be scaled from a drawing) are 16.77 + 36.89 and 36.89 feet respectively; and with respect to the left end, the arms of R2 and the resultant wind pressure are 16.77 + 36.89 and 16.77 feet respectively. Taking moments about the right en 1 we find that

- 14,600 X 30.89 + R1X (16.77 + 36.89) = 0;

R1=[14,600 X 36.89]/[16.77+36.89] =10,035 pounds. or,

Taking moments about the left end, we find that

14,600 X 16.77 - R2X (16.77 + 36.89) = 0; or, R2=14,600x16.77/16.77+36.89=4,565 pounds.

To determine the stresses in the members, we construct a stress diagram. In Fig. 45b, AB, BC, CD, DE, and EF represent the wind loads at the successive joints, beginning with joint 1. The point F is also marked G, H, I, and J, to indicate the fact that there are no loads at joints 9, 10, 11, and 12. JK represents the right reaction, and KA the left reaction.

We may draw the polygon for joint 1 or 12; for 1 it is KABLK, BL and LK representing the stresses in b1 and lk. We may next draw the polygon for joint 2; it is LBCML, CM and ML representing the stresses in cm and ml.

STRESSES. | ||||||

MEMBER. | ||||||

Dead Load. | Snow Load. | Wind Left. | Wind Right. | Resultant. | Resultant. | |

bl | -14,700 | -17,600 | - 16,400 | 0 | - 48,700 | -32,300 |

cm | -13,700 | -16,400 | -15,900 | 0 | - 46,000 | -30,100 |

dp | -12,600 | -15,100 | -15,400 | 0 | - 43,100 | -28,000 |

eq | -11,600 | -13,900 | -14,900 | 0 | -40,400 | -26,500 |

lm | - 1,650 | - 2,000 | - 3,700 | 0 | - 7,350 | - 5,350 |

mn | + 1,650 | + 2,000 | + 3,700 | 0 | + 7,350 | + 5,350 |

no | - 3,300 | - 4,000 | - 7,400 | 0 | -14,700 | -10,700 |

op | + 1,850 | + 2,200 | + 4,100 | 0 | + 8,150 | + 5,950 |

pq | - 1,650 | - 2,000 | - 3,700 | 0 | - 7,350 | - 5,350 |

rq | + 5,000 | + 6,000 | +11,000 | 0 | +22,000 | +16,000 |

ro | + 3,400 | + 4,100 | + 7,400 | 0 | +14,900 | +10,800 |

kl | +13,300 | +16,000 | +18,300 | + 6,100 | +47,600 | +31,600 |

kn | +11,300 | +13,600 | +14,200 | + 6,100 | +39,100 | +25,500 |

kr | + 8,000 | + 9,600 | + 6,100 | + 6,100 | +23,700 | +17,600 |

kv | +11,300 | +13,600 | + 6,100 | +14,200 | +39,100 | +25,500 |

kx | +13,300 | +16,000 | + 6,100 | +18,300 | +47,600 | +31,600 |

ru | + 3,400 | + 4,100 | 0 | + 7,400 | +14,900 | +10,800 |

rs | + 5,000 | + 6,000 | 0 | +11,000 | +22,000 | +16,000 |

st | - 1,650 | - 2,000 | 0 | + 3,700 | - 7,350 | - 5,350 |

tu | + 1,850 | + 2,200 | 0 | + 4,100 | + 8,150 | + 5,950 |

uv | - 3,300 | - 4,000 | 0 | - 7,400 | -14,700 | -10,700 |

vw | + 1,650 | + 2,000 | 0 | - 3,700 | + 7,350 | + 5,350 |

vx | - 1,650 | - 2,000 | 0 | - 3,700 | - 7,350 | - 5,350 |

fs | -11,600 | -13,900 | 0 | -14,900 | -40,400 | -26,500 |

gt | -12,600 | -15,100 | 0 | -15.400 | - 43,100 | -28,000 |

hw | -13,700 | -16,400 | 0 | -15,900 | - 46,000 | -30,100 |

ix | -14,700 | -17,600 | 0 | -16,400 | - 48,700 | -32,300 |

We may draw next the polygon for joint 3 ; it is KLMNK, MN and N K representing the stresses in mn and nk. No polygon for a joint on the left side can now be drawn, but we may begin at the right end. For joint 12 the polygon is JKXIJ, KX and XI representing the stresses in kx and xi.

At joint 11 there are three forces; and since they are, balanced, and two act along the same line, those two must be equal and opposite, and the third must equal zero. Hence the point X is also marked "W to indicate the fact that XW, or the stress in xw, is zero. Then, too, the diagram shows that WH equals XI. Having just shown that there is no stress in xw, there are but three forces at joint 13. Since two of these act along the same line, they must be equal and opposite, and the third zero. Therefore the point W is also marked V to indicate the fact that WV, or the stress in wv, equals zero. The diagram shows also that VK equals XK. This same argument applied to joints 9, 15, 10, and 14 successively, shows that the stresses in st, tu, uv, ur, and sr respectively equal zero. For this reason the point X is also marked UTS and R. It is plain, also, that the stresses in sf and tg equal those in wh and xi, and that the stress in kr equals that in kv or kx. Remembering that we are discussing stress due to wind pressure only, it is plain, so far as wind pressure goes, that the intermediate members on the right side are superfluous.

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