Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
Question
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Chapter 3, Problem 3.8.1P
To determine

(a)

The section for loads given loads using load and resistance factor design (LRFD) method.

Expert Solution
Check Mark

Answer to Problem 3.8.1P

The section for loads given loads using load and resistance factor design (LRFD) method is WT5×11.

Explanation of Solution

Given data:

Length of the connection is 11in.

Spacing of truss in the roof system is 12feet6inches.

Snow load is 20psf of horizontal projection.

Weight of roofing is 12psf.

Section for the purlins is MC8×8.5.

Weight of the truss is 1000lb.

Calculation:

Calculate the snow load.

S=20psf×10ft×12.5ft=2500lb

Calculate the load due to purlins.

Dp=(8.5lb/ft)×(12.5ft)=106.25lb

Calculate the weight of the truss.

Dt=1000lb3=333.33lb

Calculate the slant height of the roof.

l=(3ft)2+(30ft)2=9ft2+900ft2=30.15ft

Calculate the weight of the roof.

Dr=(12psf)×(10ft)×(30.15ft30ft)×(12.5ft)=1507.5lb

Write the expression to calculate the total dead load.

D=Dt+Dr+Dp ...... (I)

Here, total dead load is D, load due to truss is Dt, load due to roofing is Dr and load due to purlins is Dp.

Substitute 333.33lb for Dt, 106.25lb for Dp and 1507.5lb for Dr in Equation (I).

D=333.33lb+106.25lb+1507.5lb=1947.08lb

Calculate the factored load using following load combination.

Pu=1.2D+1.6S ...... (II)

Here, factored load is Pu, total dead load is D and snow load is S.

Substitute 1947.08lb for D and 2500lb for S in Equation (II).

Pu=1.2×(1947.08lb)+1.6×(2500lb)=2336.5lb+4000lb=(6336.5lb)(1kips103lb)=6.34lb

Write the expression to calculate the exterior joint load.

Pux=1.2(Dt2+Dr2+Dp)+1.6(S2) ...... (III)

Here, load on the exterior joint is Pux.

Substitute 333.33lb for Dt, 1507.5lb for Dr, 106.25lb for Dp and 2500lb for S in Equation (III).

Pux=1.2(333.33lb2+1507.5lb2+106.25lb)+1.6(2500lb2)=1.2×1026.66lb+1.6×1250lb=3232lb×1kips103lb=3.23kips

Consider the free body diagram of the truss shown below.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 3, Problem 3.8.1P , additional homework tip  1

Figure-(1)

Write the expression to calculate the moment about point A.

MA=0(Pu×10ft)+(Pu×20ft)+(Pux×30ft)(Rbx×3ft)=0 ....... (IV)

Substitute 6.34kips for Pu and 3.23kips for Pux in Equation (IV).

(6.34kips×10ft)+(6.34kips×20ft)+(3.23kips×30ft)(Rbx×3ft)=063.4kipsft+126.8kipsft+96.9kipsft(Rbx×3ft)=0Rbx×3ft=287kipsftRbx=287kipsft3ft

Solve further.

Rbx=95.67kips

Consider joint B as shown in figure-(1).

Write the expression for summation of forces acting in the horizontal direction.

Fx=0RBxFBCcosθ=0 ...... (V)

Here, summation of all horizontal forces is Fx, horizontal reaction at point B is RBx and force in member BC is FBC.

Substitute 95.67kips for RBx in Equation (V).

95.67kipsFBC×3030.15=0FBC=(95.67kips×30.15)30FBC=96.15kips

Write the expression to calculate the required area.

Ag=FBC0.9fy ...... (VI)

Here, gross area is Ag and yield strength is fy.

Substitute 96.15kips for FBC and 36ksi for fy in Equation (VI).

Ag=96.15kips0.9×36ksi=2.96in2

Write the expression to calculate the required area.

Ae=FBC0.75fu ...... (VII)

Here, effective area is Ae and ultimate strength is fu.

Substitute 96.15kips for FBC and 58ksi for fu in Equation (VII).

Ae=96.15kips0.75×58ksi=2.21in2

Calculate the effective length of the truss.

l=10ft×30.15ft30ft=10.05ft

Calculate the radius of gyration.

rmin=l300 ...... (VIII)

Substitute 10.05ft for l in Equation (VIII)

rmin=(10.05ft)(12in1ft)300=0.402

Use section WT4×9.

Ag=3.24in2rmin=1.33in

Write the expression to calculate reduction factor.

U=1x¯l ...... (IX)

Here, reduction factor is U, distance between the centroid and the plane is x¯ and length is l.

Substitute 1.07 for x¯ and 11in for l in Equation (IX).

U=11.07in11in=10.0972=0.9027

Write the expression to calculate the effective area for the section.

Ae=UAg ...... (X)

Substitute 0.9027 for U and 3.24in2 for Ag in Equation (X).

Ae=0.9027×3.24in2=2.92in2

Conclusion:

Since the gross area, net area and radius of gyration for this greater than the calculated value,

Ag=3.24in2>2.96in2rmin=1.33in>0.402inAe=2.92in2>2.21in2

To determine

(b)

The section for loads given loads using allowable strength design (ASD) method.

Expert Solution
Check Mark

Answer to Problem 3.8.1P

The section for loads given loads using allowable strength design (ASD) method is WT5×11.

Explanation of Solution

Given data:

Length of the connection is 11in.

Spacing of truss in the roof system is 12feet6inches.

Snow load is 20psf of horizontal projection.

Weight of roofing is 12psf.

Section for the purlins is MC8×8.5.

Weight of the truss is 1000lb.

Calculation:

Calculate the ultimate load using the following load combination.

P=D+S ...... (XI)

Here, ultimate load is P, dead load is D and snow load is S.

Substitute 1947.08lb for D and 2500lb for S in equation (XI).

P=1947.08lb+2500lb=(4447.08lb)(1kips103lb)=4.45lb

Write the expression to calculate the exterior joint load.

Pux=Dt2+Dr2+Dp+S2 ...... (XII)

Here, load on the exterior joint is Pux.

Substitute 333.33lb for Dt, 1507.5lb for Dr, 106.25lb for Dp and 2500lb for S in Equation (XII).

Pux=(333.33lb2)+(1507.5lb2)+(106.25lb)+(2500lb2)=(2276.665lb)(1kips103lb)=2.27kips

Consider the free body of the truss as shown below.

Steel Design (Activate Learning with these NEW titles from Engineering!), Chapter 3, Problem 3.8.1P , additional homework tip  2

Figure-(2)

Write the expression to calculate the moment about point A.

MA=0(P×10ft)+(P×20ft)+(Pux×30ft)(Rbx×3ft)=0 ....... (XIII)

Substitute 4.45kips for Pu and 2.27kips for Pux in Equation (XIII).

(4.45kips×10ft)+(4.45kips×20ft)+(2.27kips×30ft)(Rbx×3ft)=044.5kipsft+89kipsft+68.1kipsft(Rbx×3ft)=0Rbx×3ft=201.6kipsftRbx=201.6kipsft3ft

Solve further.

Rbx=67.2kips

Consider joint B as shown in figure-(1).

Write the expression for summation of forces acting in the horizontal direction.

Fx=0RBxFBCcosθ=0 ...... (XIV)

Here, summation of all horizontal forces is Fx, horizontal reaction at point B is RBx and force in member BC is FBC.

Substitute 67.2kips for RBx in Equation (XIV).

67.2kipsFBC×3030.15=0FBC=67.2kips×30.1530FBC=67.54kips

Write the expression to calculate the required area.

Ag=FBC0.6fy ...... (XV)

Here, gross area is Ag and yield strength is fy.

Substitute 67.54kips for FBC and 36ksi for fy in Equation (XV).

Ag=67.54kips0.6×36ksi=3.13in2

Write the expression to calculate the required area.

Ae=FBC0.5fu ...... (XVI)

Here, effective area is Ae and ultimate strength is fu.

Substitute 67.54kips for FBC and 58ksi for fu in Equation (XVI).

Ae=67.54kips0.5×58ksi=2.33in2

Calculate the effective length of the truss.

l=(10ft)×(30.15ft30ft)=10.05ft

Calculate the radius of gyration.

rmin=l300 ...... (XVII)

Substitute 10.05ft for l in Equation (XVII).

rmin=(10.05ft)(12in1ft)300=0.402

Use section WT4×9.

Ag=3.24in2rmin=1.33in

Write the expression to calculate reduction factor.

U=1x¯l ...... (XVIII)

Here, reduction factor is U, distance between the centroid and the plane is x¯ and length is l.

Substitute 1.07 for x¯ and 11in for l in Equation (XVIII).

U=11.07in11in=10.0972=0.9027

Write the expression to calculate the effective area for the section.

Ae=UAg ...... (XIX)

Substitute 0.9027 for U and 3.24in2 for Ag in Equation (XIX).

Ae=0.9027×3.24in2=2.92in2

Conclusion:

Since the gross area, net area and radius of gyration for this greater than the calculated value,

Ag=3.24in2>3.13in2rmin=1.33in>0.402inAe=2.92in2>2.33in2

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ISBN:9781337094740
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