Question

Please remember all homework assignments are to be completed legibly on engineering paper or printed out Answer the following questions. Use lecture notes, text book and AlSC Steel Construction Manual for reference. This assignment is worth 5 points (2 points) A 50 ksiW10x60 tension member with no holes and Lb 14 ft is subjected to an axial service load Pd 40k and Pl 50k and to the service moments Mdk 20k-ft, Mix-40 k-ft, Mdy # 5 k-ft, Mly-8 k-ft. Is the member satisfactory if C-10? 1. 2. (1 point) Calculate the effective flange width of a composite beam given the girder layout below. Assume the distance to the edge of the girder is 4ft. 4 in Simple span 38 ft 9 f 9ft 9 ft 3. (2 points) Determine the flexural design capacity for the section shown, if 50 ksi steel and sufficient steel anchors are used to guarantee full composite action. Fc- 4ksi For an extra 0.5 points, verify the flexural capacity using the AISC Manual Table 3-19. b 84 in 4 in W18x 35 Page | 1

Answer 1

(1) W10X60 member

L_b=14 ft

P_u=1.2(D.L)+1.6(L.L)

P_u=1.2(40)+1.6(50)

P_u=128 kips

M_ux=1.2(M_dx)+1.6(M_lx)

M_ux=1.2(20)+1.6(40)

M_ux=88 kips-ft

M_uy=1.2(M_dx)+1.6(M_lx)

M_uy=1.2(5)+1.6(8)

M_uy=18.8 kips-ft

From the beam design charts, the capacity is 261.2 kips for Lb=14 ft and Cb=1

From column design tables for L_b=14 ft, we have design capacity

$\phi$ P_n=423 kips

P_u/ $\phi$ P_n=128/(423)=0.30>0.2

Apply AISC Equation H1-1(a)

P_u/ $\phi$ P_n+8/9[M_ux/ $\phi$ M_n+M_uy/ $\phi$ M_n] (The maximum B.M is on x direction, so y direction can be neglected)

0.30+8/9[88/261.2]=0.70<1 (This satisfies the AISC Equation)

(2)

Effective flange width=Span/4 or beam spacing (least of the above)

Span=38*12/4=114 in

Beam spacing=9*12=108 in

Effective flange width=108 in

(3)

b_e=84 in

Fy=50 ksi

fc=4 ksi

t=4 in

Calculate compressive force in both steel and concrete

C=A_sF_y

C=10.3x50=515 kips

Compressive force in concrete=0.85fcAc=0.85x4x(84x4)=1142.4 kips

So least compressive force=515 kips

a=C/(0.85fcb)

a=515/(0.85x4x84)

a=1.8 in

y=d/2+t-a/2

y=18/2+4-1.8/2

y=12.1 in

The design strength=0.9*C*y=0.9x515*12.1=5608 in-kips=467.3 ft-kips