Homework #5 Due 6/13/07

— David Wagner 2007/06/12 19:26

Problem 1.

A cantilever walkway made up of slabs and frames uses the frame shown below. The frames are going to be spaced 30 ft apart (center to center). The dead load (including the weight of the elements) is 250 psf, the live load is 100 psf and the wind load is 24 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 column.

Applicable Load Factors

(9-1) U = 1.4D
(9-2) U = 1.2D + 1.6L
(9-3) U = 1.2D + (1.0L or 0.8W)*
(9-4) U = 1.2D + 1.6W* + 1.0L*
(9-6) U = 0.9D + 1.6W*

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

Walkway Loads

(Positive downward)

Contributing area A_walk = 18×30 = 540 sf, l=18.

7.5 k/ft (downward)
3 k/ft (downward)
~~-0.72 k/ft (upward)~~ 0

Factored Loads

(9-1)10.5 k/ft
(9-2)13.8 k/ft max
~~(9-3)<m>w_U = 1.2w_D + 1.0w_L=1.2*7.5+1.0*3=</m>12.0 k/ft~~
~~(9-3)<m>w_U = 1.2w_D + 0.8w_W=1.2*7.5-0.8*0.72=</m>8.4 k/ft~~
(9-4)12.0 k/ft
(9-6)6.75 k/ft min

Column Axial Loads

(Positive downward)

Assume the weight of the walls is not borne by the columns, but is carried straight down into the ground. Only the cantilevered walkway loads are borne by the columns.

Contributing area A_walk = 18×30 = 540 sf, l=24.

135 kip (downward)
54 kip (downward)
~~<m>P_W=24*30*18=</m>-12.96 kip (upward)~~

Factored Loads

(9-1)189 kip
(9-2)248 kip max
~~(9-3)<m>P_U = 1.2P_D + 1.0P_L=1.2*135+1.0*54=</m>216 kip~~
~~(9-3)<m>P_U = 1.2P_D + 0.8P_W=1.2*135-0.8*12.96</m>172 kip~~
(9-4)216kip
(9-6)121.5 kip min

Column Lateral Load

(positive leftward)

Contributing area A_col = 24×30 = 720 sf, l=24.

±0.72 k/ft (left/right)

(No other lateral loads are given.)

Factored Loads

(9-3)±0.576 k/ft
(9-4)±1.152 k/ft
(9-6)±1.152 k/ft max (absolute value)

Maximum Positive Beam Moment

The beam has a negative moment under all loading conditions since all loads are net vertically downward on this cantilevered beam. It's least negative moment is at the free end, where it is considered zero.

Zero

Maximum Negative Beam Moment

The maximum negative moment occurs at the point of attachment with the maximum downward load on the walkway.

$M_{u,max}={w_{u,max} l^2}/2={13.8 * 18^2}/2=$ 2236 kip-ft

-2236 kip-ft

Maximum Beam Shear

The maximum beam shear occurs at the point of attachment with the maximum downward load.

$V_{u,max}=w_{u,max}l=13.8*18=248$

248 kip

Maximum and Minimum Beam Axial Load

Axial loading of the beam is negligible since it is free at one end. The wind might blow on the section a little.

Negligible

Maximum Positive Column Moment

The maximum positive (least negative) moment occurs at the base of the column with the minimum downward load on the walkway and the maximum lateral wind load to the right.

$M_{u,max}={w_{u,walk,min} l_{walk}^2}/2 - {w_{u,col} l_{col}^2}/2 = {6.75 18^2}/2 - {1.152 24^2}/2 = 1094-332=762$

-762 kip-ft

Note this is still a negative moment (the column bends outward). Also, the moment at the 'free' top of the column is considered zero though it transitions around the corner to the location of maximum beam moment.)

Maximum Negative Column Moment

The maximum negative moment occurs at the base of the column with the maximum downward load on the walkway and the maximum lateral (wind) load to the left.

$M_{u,max}={w_{u,walk,max} l_{walk}^2}/2 + {w_{u,col} l_{col}^2}/2 = {13.8 18^2}/2 + {1.152 24^2}/2 = 2236+332=$ 2567 kip-ft

-2567 kip-ft

Maximum Column Shear

The maximum column shear occurs at the base with the maximum lateral load in either direction.

$V_{u,max}=w_{u,max}l=1.152*24=27.6$

28 kip

Maximum Column Axial Load

The maximum column axial load occurs with the maximum downward load.

248 kip (compression)

Minimum Column Axial Load

The minimum column axial load occurs with the minimum downward load.

121.5 kip (compression)

Problem 2.

Solve problem 4-24 in your textbook. Do not use the ρ value given instead assume εt = 0.008 to determine the amount of steel in the controlling section (Section with maximum moment either negative or positive) then change the amount of reinforcement to carry the moments in other critical sections. DO NOT DESIGN FOR THE LARGEST OR SMALLEST SECTION AS WAS DONE IN THE LECTURE NOTES.

”…design rectangular sections for the beams and loads shown in the accompanying illustration. Beam weights are not included in the given loads. fy=60000 psi and f'c=4000 psi. Live loads are to be placed where they will cause the most severe conditions at the sections being considered. Select beam size for the largest moment (positive or negative) and then select the steel required for maximum positive moment and for maximum negative moment. Finally, sketch the beam and show approximate bar locations.”

Estimate w_{beam}=0.3 k/ft.

(9-1) U = 1.4D → k/ft
(9-2) U = 1.2D + 1.6L →

Assume live loads control and (9-1) applies since 7.16>4.2. The maximum positive moment occurs with the live load extending only from support to support. Since this is symmetric, taking a section at the center where the positive moment is greatest allows solving for it directly.

$F_{y,support} = 22*1.2*3.3 + 12*1.6*2=87.1+38.4 =$ 125.5 k $M_{umax+} = 12*125.5-{22/2}*(1.2*3.3*22) - {12/2}*(1.6*2*12) = 1506.2-958.3-230.4 =$ 317.5 kip-ft

The maximum negative moment occurs with the live load only on the overhanging ends. Taking a section of just the overhang (at the support where the negative moment is greatest) allows solving for the maximum negative moment directly as a simple cantilever.

$M_{umax-}=-{w l^2}/2 = -{7.16*10^2}/2=$ -358 kip-ft controls

First, the top bars.

Assume φ=0.9 and M_n=M_u/φ=358/.9=398 kip-ft = 4773 kip-in

b=b_w, choose epsilon_t=0.01

$beta_{1,4000}=0.85$

$A_{s,min}={3sqrt{f prime_c b d}/fy=0.00316 bd$

$A_{s,min}={200 bd}/fy =$ 0.00333 bd← use this

$rho_{min} = {A_{s,min}}/{bd}=$ 0.00333

$rho={{0.85 beta_1 f prime_c}/f_y}(epsilon_u/{epsilon_u+epsilon_t}) = {{0.85 *0.85 *4}/60}(0.003/{0.003+0.01}) = 0.04816*0.2308=$ 0.011115 > 0.0033✔

A_s=rho b d = 0.011115 b d

$a={A_s f_y}/{0.85 f prime_c b}={0.011115*b*d*60}/{0.85*4*b}=0.19615 d$

$M_n=4773= A_s f_y(d-{a/2}) = 0.011115*b*d*60*(d-{0.19615/2}d)=0.6015bd^2$

$right d^2={4773}/{0.6015b} = right d= 89.1/{sqrt{b}}$

b	10	12	14	16
d	28.2	25.7	23.8	22.3
d+2.5		28.2	26.3	24.8
h				26

A_s approx rho b d = 0.011115*16*22.3 = 3.966 in²

A_s=4#9= 4.00 in²

Use #3 stirrups.

$d=d_t=26-1.5-0.375-{1.128/2}=$ 23.56 in

$a={A_s f_y}/{0.85 f prime_c b}={4*60}/{0.85*4*16}=4.412$

$epsilon_t={0.003*(beta_1 d_t-a)}/a = {0.003*(0.85*22.3-4.412)}/4.412=$ 0.00989 > 0.005✔ (max steel satisfied)

$M_n=A_s f_y(d-{a/2})=4*60(23.56-{4.412/2})$ =5125 kip-in > 4773✔

Now, for the bottom bars max+ moment: M_u317.5 kip-ft = 3810 kip-in.

Assume φ=0.9 and M_n=M_u/φ=3810 /.9=4233 kip-in

This is very close to the absolute value of the max- moment, so the bar requirement will be very similar. So, let's see how similar; try using the same #9 bars.

$A_s = M_n/{f_y(d-{a/2})} = 4233/{60*(23.56-{4.412/2})}=$ 3.304 in²

There really are not many good, simple bar combinations to get 3.3 square inches or so. (6#7 would be 3.6, but that's about it.) For ease of construction, just use 4#9 again. The analysis is the same as for negative-moment reinforcing. 5125 kip-in > 4233✔

Problem 3.

For the beam and cross section shown, determine the maximum service load PL that can be applied to it if the service uniform dead load (wD) is 1 k/ft (not including the beam weight) and the uniform service live load (wL) is 2 k/ft. Assume #4 stirrups, f’c = 4000 psi and fy = 60,000 psi.