Final Drawings

— David Wagner 2007/07/01 22:50

First, the preliminary dimensions result in too much ductility for column analysis using the methods presented in this course so everything needs to be redesigned using much larger dimensions. Once the loading conditions are estimated, this project has few variables to find when restricted to required and suggested design constraints.

Member Dimensions

  • Slab thickness: tslab = 8”
  • Beam height: hbeam = 18”
  • Beam width and square column section size: b = hcol = 24”
  • Square footing width: w =
  • Footing thickness: tfoot =

Reinforcement

  • Slab top and bottom mat
  • Beam top and bottom bars
  • Beam stirrups
  • Beam-to-column hooks (if necessary)
  • Column bars
  • Column ties
  • Footing dowls (if necessary)
  • Footing bottom bars

Note the beam, columns, and footings of the two end frames may require less reinforcement.

The slab thickness is still fine at 8”, and a beam width and square column section size of 24” seems reasonable. The beam height will be 18” if the minimum beam thickness is increased by 10-20% and rounded up to the nearest even number as recommended.

Unfortunately, this changes the bending analysis, but there is enough valid information left to do the footing design. The tributary area transferring load to each intermediate column is 15'x18', the beam length supported is 15', and each column is 18'.

  • P_{slab}=0.150*15*18*{8/12}=27 kip
  • P_{beam}=0.150*15*{24/12}*{{18-8}/12}=3.75 kip
  • P_{col}=0.150*21*{24/12}*{24/12}=12.6 kip
  • P_D = 0.119*15*18 + 27+3.75+12.6=75.48 kip
  • P_L=0.031*15*18=8.37 kip

The lateral wind load results in a small live axial load on the columns which can be found by statics.

  • P_{lat}=0.0109*18*15=2.943 kip
  • 2.943*12 = 2*13*P_W right P_W=1.36 kip

Combinations

  • (9-1) U = 1.4D → P_u=1.4*75.48=106 kip controls
  • (9-3) U = 1.2D + 1.6L + 0.8W → P_u=1.2*75.48+1.6*8.37+0.8*1.36=90.58+13.92+1.09=105 kip
  • (9-6) U= 0.9D + 1.6W right P_u=0.9*75.48+1.6*1.36=67.93+2.18=70 kip

Try tfoot= h = 1.5b = 36”

  • W_{foot}=0.150*3=0.450 ksf
  • W_{soil}=0.120*3=0.360 ksf
  • W_{floor}=0.150*{4/12}=0.050 ksf

P_{u,net}=4-0.45-0.36-0.05-0.04=3.1 ksf

A_{req} >= {{P_D+P_L+P_W}/P_{u,net}}= {75.48+8.37+1.36}/{3.1} = 85.21/3.1=27.487 ft²

b=L=sqrt{27.487}=5.243 ≈ 6 ft = 72 in

A=6^2 =36 ft²

Assume #8: d=36-3-1=32 in

B=24+32=56 in = 4.667 ft

V_u=P_{u,net}*(A-B^2)=3.1*(36-4.667^2)=44.1 kip

beta_c={24/24}=1, αs=40, b0=4B=224 in

V_c=min [(2+{4/beta_c}) b_0 d sqrt{f prime_c},({{alpha_s d}/b_0}+2) b_0 d sqrt{f prime_c},4 b_0 d sqrt{f prime_c}]

min[2+{4/1},{{40*32}/224}+2 , 4]=min[6,5.7,4]=4

V_c=4 b_0 d sqrt{f prime_c}=4*224*32*sqrt{4000}=1813376 lb = 1813 kip

phi V_n = phi V_c = 0.75*1813=1360 kip > 44.1✔

Does φVn » Vu imply the footing is too thick? Try 2'?

G={L/2}-b_{col}-d={6/2}-2-{32/12}=-1.667 → There is no critical one-way shear.

Discussion


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