Homework #7 Due June 19, 2007

— David Wagner 2007/06/17 13:58

Problem 1.

A walkway made up of slabs and frames uses the frame shown below. The frames are going to be spaced 25 ft apart (center to center). The dead load (including the weight of the elements) is 300 psf, the live load is 120 psf and the wind load is 30 psf. Live load may not be present all the time. Determine the maximum moment Mu (positive and negative), the maximum shear Vu, and the maximum and minimum axial load Nu for the beam and the columns.

Assume the dead load is distributed evenly over the roof. Assume the weight of the walls is not carried by the columns, though the wind-produced moment is.

l=25, l_r=28 ft, l_w=18 ft

$w_D=300 {lb/ft^2} * 25 ft =$ 7.5 kip/ft down

$w_L=120 {lb/ft^2} * 25 ft =$ 3 kip/ft down

$w_W=30 {lb/ft^2} * 25 ft =$ 0.75 kip/ft left or right

(9-1) U = 1.4D Controls roof min N
- $w_{u,r}=1.4 w_D=1.4*7.5=$ 10.5 kip/ft down
- $w_{u,w}=0$ laterally
~~(9-2) U = 1.2D + 1.6L~~ < (9-3) ∴ Cannot control.
(9-3) U = 1.2D + 1.6L_r + 0.8W* Controls roof max V and M, column max N
- $w_{u,r}=1.2 w_D + 1.6 w_l = 1.2*7.5 + 1.6*3=13.8$ kip/ft down
- $w_{u,w}=0.8 w_W=0.8*0.75=0.6$ kip/ft left or right
(9-4) U = 1.2D + 1.6W + 1.0L + 0.5L_r* Not appropriate for 1-story structures.
(9-6) U = 0.9D + 1.6W* Controls roof max N; column max V and M; and column min N
- $w_{u,r}=0.9 w_D = 0.9*7.5=$ 6.75 kip/ft down
- $w_{u,w}=1.6 w_W=1.6*0.75=$ 1.2 kip/ft left or right

*Assume notes 9.2(a) and 9.2(b) do not apply.

The shear and moment in the roof beam result from vertical loading on the roof only. The roof is equivalent to a “beam fixed at both ends–uniformly distributed loads”.

V_|max|=±wl/2 at the ends of the roof beams.
$|V_{r,max}|=={w_{u,r}l_r}/2 = {13.8*28}/2=$ 193.2 kip

Roof Beam Max V = 193 kip

M_|max|=-wl²/12 at both ends of the roof beams.
$M_{r,|max|}= {w_{u,r}{l_r}^2}/12 = {13.8*28^2}/12=$ 901.6 kip-ft

Roof Beam Max M = -902 kip-ft

The axial load in the roof beam results only from the wind blowing on the walls.

Roof Beam Min N = 0

$N_{r,max}= w_{u,w} l_w = 1.2 * 18=$ 21.6 kip

Roof Beam Max N = 22 kip C

The shear and moment in the column result from horizontal wind loading only. The left column is equivalent to a “beam fixed at one end, supported at the other–uniformly distributed loads”.

V_|max|=-5wl/8 at the fixed end (top) of the column.
M_|max|=-wl²/8 at the fixed end (top) of the column.

The right column is equivalent to a “cantilevered beam–uniformly distributed loads”. The shear and moment are greater than those in the left column.

V_|max|=-wl at the fixed end (top) of the right column.
$|V_{w,max}|=w_{u,w}l_w} = 1.2*18=$ 21.6 kip

Column Max V = 22 kip

M_|max|=-wl²/2 at the fixed end (top) of the right column.
$|M_{w,max}|={w_{u,w}{l_w}^2}/2 = {1.2*18^2}/2=$ 194.4 kip=ft

Column Max M = 194 kip-ft

$N_{w,max}={w_{u,r}l_r}/2 = {13.8*28}/2=$ 193.2

Column Max N = 193 kip C

$N_{w,min}={w_{u,r}l_r*14 - w_{u,w}l_w*9 }/28 = {6.75*28*14 - 1.2*18*9 }/28 = {2646-194.4}/28=$ 87.557

Column Min N = 88 kip C

Problem 2.

Design a cantilever concrete slab. Assume deflections will not be computed, service DL = 18 psf (not including slab weight), service LL = 100 psf, cover = 0.75 in, f’c = 5000psi, fy = 60,000psi, and span length equals to 10 feet. Sketch the solution.

LL»DL → U=1.2D + 1.6L

Approximate depth l/10 ≈ h ≈ 12”, design width b=1'=12”, approximate weight w_slab ≈ 0.15 k/ft, $beta_1 = 0.85-0.05*{1}=$ 0.80.

w_u = 1.2*(0.018*1+0.15) + 1.6*0.1 = 0.2016+0.16= 0.36 k/ft

$M_u = {w_u l^2}/2 = {0.36*10^2}/2=$ 18 kip-ft = 216 kip-in

Take d-a/2 ≈ 0.9h=10.8 in, and assume φ=0.9.

$216 = M_u = phi M_n = phi A_s f_y (d-{a/2}) right A_s = {216}/{0.9*60*10.8} =$ 0.370 in²

Try 2#4/12 in = #4@6”, A_s=0.40

$d=d_t=12-0.75-{4/16}=$ 11 in

$A_{s,min}= larger[{3*sqrt{5000}*12*11}/{60000}, {200*12*11}/{60000}] = larger[0.4667,0.44] =$ 0.4667 > 0.40✘

Try 5#3/12 in = #3@2.4 ≈ #3@3in which is actually 4#3/12 in, A_s=0.44 which won't work.

Try @5” → 2.4 bars/ft → 0.456 in²/ft / (2.4 bar/ft) = 0.19 in²/bar. Try #4@5”

A_s=0.20*2.4= 0.48 in² > 0.4667✔

$d=d_t=12-0.75-{4/16}=$ 11 in

$a={A_s f_y}/{0.85 f prime_c b} = {0.48*60}/{0.85*5*12}=$ 0.5647 in

$epsilon_t={0.003*(0.80*11-0.5647)}/0.5647=$ 0.04375 > 0.005 → φ=0.9

$phi M_n = 0.9*0.48*60*(11-{0.5647/2}) =$ 277.8 kip-in > 216 kip-in✔

h=12”, #4@5”, 3/4” from the top

Problem 3.

Design a simply supported concrete slab. Assume deflections will not be computed, service DL = 24 psf (not including slab weight), service LL = 120 psf, a linear load of 10 k/ft at the center of the slab, cover = 0.75 in, f’c = 5000psi, fy = 60,000psi, and span length equals to 18 feet. Sketch the solution. HINT: When you design for a one-foot width the linear load becomes a concentrated load of 10 kips applied at the middle of the slab)

Design b=1'=12”, h >=l/20≈10.8”, try h=12”, l=18'.

Assume the linear load is a live load.

Consider dead loads only: U=1.4D $M_u={w_u l^2}/8={1.4*w_D l^2}/8 = {1.4*(0.024+0.150)*18^2}/8 =9.8658$