June 25: Design of slab and beams

— David Wagner 2007/06/24 20:18

Slab Design

The one-way roof slabs span 15'=180” center-to-center, with an estimated clear span of 140”. Longitudinal wind loads are borne by shear walls and not considered to effect the slab.

Total wD: 219 psf, wL = 10 psf, wL,rain = 31 psf, l=15'=180 in, ln=11.67'=140 in

Due to the sod roof, both L and Lrain may occur together. The wind load does not effect slab design, and wLr » wL right (9-3) controls.

  • (9-1) U = 1.4D → w_u=1.4*219=306.6
  • (9-3) U = 1.2D + 1.6L → w_u=1.2*219 + 1.6*41 =328.4 psf

The maximum negative moment is controlled by an interior span. The maximum positive moment and minimum slab depth t are controlled by the end span (to avoid deflection calculations).

  • M_{u+}={w_u {l_n}^2}/11={328.4*{11.67}^2}/143195 lb-ft = 38.3 kip-in
  • M_{u-}=-{w_u {l_n}^2}/11=-{328.4*{11.67}^2}/11=-4066 lb-ft = -48.8 kip-in
  • t_{min}=180/24=7.5, say t=8 in

d≈8-0.75-0.5=6.75

First, find top grade 60 steel to handle the larger negative moment.

A_{s,top} approx {48.8}/{0.9*60*0.9*6.75} approx0.149 in²/ft

Try #4 bars (the smallest permissible on top), designing a 1' strip of slab as a 1' wide beam, b=bw=12”.

s={12*A{bar}}/{A_s} = {12*0.20}/{0.146} =16.4→ #4@16”

A_s={12*0.20}/16 =0.15 in²/ft

d=d_t=8-0.75-{4/16}=7 in

A_{s,min}=max[3*sqrt{4000}*12*7/60000=0.2656,{200*12*7}/60000]=max[0.2656,0.28]=0.28 in²/ft

s= {12*0.20}/{0.28} =8.6→ #4@8”

A_s={12*0.20}/8 =0.3 in²/ft

a={0.3*60}/{0.85*4*12}=0.441

The β1 of 4000 psi concrete is 0.85.

epsilon_t = 0.003{{beta_1 d_t - a}/a} = 0.003*{{0.85*7 - 0.441}/0.441}=0.0374 > 0.005 → φ=0.9

phi M_n = 0.9*0.3*60*(7-{0.441/2})=110 kip-in > 48.8✔

Since the minimum amount of steel applies the same to the bottom mat of steel, the same analysis applies.

Slab reinforcing: #4@8” top and bottom

Beam Design

Considering the frame as pin-connected to the nasty gumbo soil around these parts seems a reasonable way to model it. This will mean the wind load can increase the maximum negative moment on the beam at the inside face of the column, though it will have negligible effect on the maximum positive moment near the center and little effect on the negative moment at the exterior column face produced by the cantilevered portion.

Assume a 10×8 beam.

h=8+8=16 in, bw=10 in

wW,max=10.9 psf

w_D=219*15 + 150*{10/12}*{8/12}=3285+83=3.368 kip/ft

w_L=41*15 =0.615 kip/ft

w_W=10.9*18=0.196 kip/ft

  • (9-1) U = 1.4D → w_u=1.4*3.368=4.715 kip/ft
  • (9-3) U = 1.2D + 1.6L + 0.8W → w_u=1.2*3.368 +1.6*0.615 + 0.8*0.196 =5.026 kip/ft down, and 0.157 kip/ft laterally. controls
  • (9-6) U= 0.9D + 1.6W = 0.9*3.368 + 1.6*0.196=3.032 kip/ft down, and 0.314 kip/ft laterally.

Ignoring more complicated wind effects, the negative moment on the beam at the exterior face of the column is simply the reaction to the cantilevered load, a much smaller value than the moment on the interior face.

M_u={w_{u,down}*l^2}/2 = -{5.026*5^2}/2=-62.825 kip-ft = -754 kip-in

By using the frame diagrams, ignoring the small reduction in moment caused by the cantilever, and assuming the column is also 10×16 (I1=I2) and is pinned at the ground level results in the following analysis.

h = 18' = 216”, L = 26' = 312”

k={I_2 h}/{I_1 L} = 18/26=0.6923

(9-6) Analysis

H={w_{u,down} L^2}/{4h(2k+3)}={3.032*26^2}/{4*18*(2*0.693+3)}={2050}/{226}=9.07 kip

M_c=M_d=-Hh=-9.07*18=-163 kip-ft = -1960 kip-in

H_b={{w_{u,lat} h}/{8}} {{6+5k}/{2k+3}} = {{0.314*18}/{8}} * {{6+5*0.6923}/{2*0.6923+3}} = 0.7065*{{9.4615}/{4.3846}}=1.524 kip

H_a=w_{u,lat}h-H_b = 0.314*18-1.524=4.128 kip

M_c=H_a h - {w_{u,lat} h^2}/2=4.128*18 - {0.314*18^2}/2 = 74.304-50.868=23.436 kip-ft = 281 kip-in

M_d=-H_b h = -1.524*18=-27.432 kip-ft = -329 kip-in

(9-3) Analysis

H={w_{u,down} L^2}/{4h(2k+3)}={5.026*26^2}/{4*18*(2*0.693+3)}={3398}/{226}=15.03 kip

M_c=M_d=-Hh=-15.03*18=-271 kip-ft = -3247 kip-in

H_b={{w_{u,lat} h}/{8}} {{6+5k}/{2k+3}} = {{0.157*18}/{8}} * {{6+5*0.6923}/{2*0.6923+3}} = 0.35325*{{9.4615}/{4.3846}}=0.762 kip

H_a=w_{u,lat}h-H_b = 0.157*18-0.762=2.064 kip

M_c=H_a h - {w_{u,lat} h^2}/2=2.064*18 - {0.157*18^2}/2 = 37.152-25.434=11.718 kip-ft = 141 kip-in

M_d=-H_b h = -0.762*18=-13.716 kip-ft = -165 kip-in

Controlling Moments

Thus, the maximum negative moment in the beam is at the interior column face.

Mu- = -3247 -165 = -3412 kip-in

The maximum positive moment occurs in the middle of the span. (Wind loading has negligible effect.)

M_s=M_c=M_d=-3247 kip-in

M_g={w_{u,down} L^2}/8 = {5.026*26^2}/8 =425 kip-ft = 5096 kip-in

M_{u+}=M_g-M_s=5096-3247=1850 kip-in

Mu+ = 1850 kip-in

d approx h-2.5=16-2.5=13.5

b_e=min[26*12/4,10+16*8,15*12]=min[78,138,180]=78 in

A_s approx {1850}/{0.9*60*0.9*13.5} approx2.82 in

Try As= 3#9 = 3.00 in², bw >= 9.8 in✔

d=d_t=16-1.5-0.5-{9/16}=13.4375 in

A_{s,min}=max[{3*sqrt{4000}*10*13.4375}/{60000}, {200*10*13.4375}/{60000}] = max[0.425,0.448]=0.448 < 3.00✔

a={3.00*60}/{0.85*4*78}=0.6787

epsilon_t=0.003*{{0.85*13.4375-0.6787}/{0.6787}}=0.0475 > 0.005 → φ=0.9

phi M_n = 0.9*3*60*(13.4375-{0.6787/2})=2122 kip-in > 1850✔

Beam bottom reinforcement: 3#9

d approx h-2.5=16-2.5=13.5

A_s approx {3412}/{0.9*60*0.9*13.5} approx5.2 in

With an effective flange width of 78”, these can be spread out quite a bit. Try 7#8 = 5.53 in²

d=16-1.5-0.5-{8/16}=13.5 in

a={5.53*60}/{0.85*4*10}=9.7588

phi M_n = 0.9*5.53*60*(13.5-{9.7588/2})=2574 < 3412✘

A_s=3412/{0.9*60*(13.5-{9.7588/2})}=7.33 in²

Use 10#8@6” in the flange

A_s=7.90

phi M_n=0.9*7.9*60*(13.5-{9.7588/2})=3678 > 3412✔

Discussion


Personal Tools