Design the tension members of the roof truss shown in Figure P3.8-4. Use, double-angle shapes throughout and assume 3 8 -inch-thick gusset plates and welded connections. Assume a shear lag factor of U = 0.85. The trusses are spaced 25 feet. Use A572 Grade 50 steel and design for the following loads. Metal deck: 4 psf of roof surface Build-up roof: 12 psf of roof surface Purlins: 6 psf of roof surface (estimated) Snow: 18 psf of horizontal projection Truss weight: 5 psf of horizontal projection (estimated) a . Use LRFD. b . Use ASD.

(a)

The double angle section for a tension member of the roof truss using Load And Resistance Factor Design (LRFD) method.

Given:

Metal deck load is 4 psf of roof surface.

Build up roof load is 12 psf.

Purlins load is 6 psf of roof surface.

Snow load is 18 psf of horizontal projection.

Truss load is 5 psf of horizontal projection.

The trusses are spaced at 25 ft.

Thickness of gusset plate is 38 in.

A572 Grade 50 steel.

Calculation:

The properties of A572 Grade 50 steel from the AISC steel manual are as follows.

• The yield strength (Fy) is 50 ksi.
• The ultimate strength is (Fu) is 65 ksi.

The truss diagram is shown below.

Figure (1)

Write the expression to calculate the area on which the load will act on the truss.

A=2×h×lt ...... (I)

Here, the area on which the load will act is A, the hypotenuse length of the truss is h and the length of the truss is lt.

Calculate the value of the hypotenuse length of the truss.

h=82+402 ft=1664 ft=40.79 ft

Here, the hypotenuse height EH of ΔAEH is h.

Substitute 40.79 ft for h and 25 ft for lt in Equation (I).

A=2×(40.79 ft)×(25 ft)=2040 ft2

Write the expression to calculate the total dead load.

D=(MD+R+P)A+(Tw×PA) ...... (II)

Here, the total dead load is D, the metal deck load is MD, the roof load is R, the purlin load is P, the truss load is Tw and the projection area is PA.

Substitute 4 psf for MD, 12 psf for R, 6 psf for P, 2040 ft2 for A, 5 psf for Tw and 2000 ft2 for PA in Equation (II).

D=((4 psf+12 psf+6 psf)×(2040 ft2))+(5 psf×2000 ft2)=44880 lb+10000 lb=54880 lb

Write the expression to calculate the load per truss.

DL=DNT ...... (III)

Here, the dead load per truss is DL, the number of trusses is NT.

Substitute 54880 lb for D and 8 for NT in Equation (III).

DL=54880 lb8=6860 lb

Write the expression to calculate the snow load per truss.

S=SL×PANT ...... (IV)

Here, the snow load per truss is S and the snow load is SL.

Substitute 18 psf for SL, 8 for NT and 2000 ft2 for PA in Equation (IV).

S=(18 psf)×(2000 ft2)8=4500 lb

Write the expression to calculate the factored load by LRFD.

Pn=1.2DL+1.6S ...... (V)

Here, the factored load is Pn.

Substitute 6860 lb for DL and 4500 lb for S in Equation (V).

Pn=(1.2×6860 lb)+(1.6×4500 lb)=(15432 lb)(1 kips1000 lb)=15.43 kips

The free body diagram of join E is shown below.

Figure (2)

Here, the vertical reaction at support E is R and the force members are FEF and FDE.

Write the expression to calculate the value of R.

R=NS×Pn2 ...... (VI)

Here, the number of spans is NS.

Substitute 8 for NS and 15.43 kips for Pn in Equation (VI).

R=8×15.43 kips2=61.72 kips

Consider the forces in vertical equilibrium.

R−VHhFDE=0 ...... (VII)

Here, the vertical height is VH.

Substitute 8 ft for VH, 61.72 kips for R and 40.79 ft for h in Equation (VII).

61.72 kips−8 ft(82+402) ftFDE=0FDE=61.72 kips8 ft(82+402) ftFDE=314.7 kips

Consider the force in horizontal equilibrium.

FDE×(40 ft40.79 ft)−FFE=0

Substitute 314.7 kips for FDE.

(314.7 kips)×(40 ft40.79 ft)−FFE=0FFE=308

(b)

The double angle section for a tension member using Allowed Strength Design (ASD) method.