Optional Homework (due July 9)

— David Wagner 2007/07/01 11:01

Problem 1.

Solve problem 12.16 in your textbook (Assume a 25-in footing with a d = 20.5 in). Do not forget to check development length.

“For Problems 12.1 to 12.30, assume reinforced concrete weighs 150 lb/ft3, plain concrete 145 lb/ft3, and soil 100 lb/ft3.”

“12.16 Design a footing limited to a maximum width of 7 ft 0 in. for the following: 15 in. x 15 in. interior column, D = 180 k, L = 160 k, f'c = 4000 psi, fy = 60,000 psi, qa = 4 ksf, and a distance from top of backfill to bottom of footing = 5 ft.”

$q_e=4000-{25/12}*150-(5-{25/12})*100 = 4000-312.5-291.7=$ 3396 psf = 3.396 psf

$A_{req}={180+160}/{3.396}=$ 100.1 ft²

L=100.1/7= 14.3 ft, Try 15'x7', A=105 ft²

$q_u={1.2*180+1.6*160}/{105}={472}/{105}=$ 4.495 ksf

b_0=4*(15+20.5)= 142 in = 11.83 ft = 142 in

$V_{u2}=( A - ({b+d}/12)^2 )*q_u=(105 - ({15+20.5}/12)^2 )*4.495=$ 432.6 k

$min[2+{4/beta_c},{alpha_s d}/b_0 +2,4] = min[2+{4/1},{40*20.5}/142 +2,4] = min[6,7.775,4] =$ 4

$V_c = 4* sqrt{f prime_c} b_0 d = 4*sqrt{4000} * 142*20.5=$ 736.4 kip

phi V_n = phi V_c = 0.75*736.4 = 552.3 kip > 432.2 ✔

One-way shear

G={L/2}-{15/{12*2}}-d={15/2}-{15/{12*2}}-{20.5/12}=7.5-0.625-1.708= 5.167 ft

V_u=q_u W G = 4.495*7*5.167= 162.6 kip

$phi V_c = phi 2 sqrt{f prime_c} b_w d = 0.75*2*sqrt{4000}*7*12*20.5=$ 163.4 kip > 162.6✔

Check critical sections for bending. W=7 ft, L=15 ft

Long direction: F = 15/2 - 15/24 = 6.875 ft

M_u=q_u W F^2/2 = 4.495*7*6.875^2 /2= 743.6 ft-kip = 8923 in-kip

$A_s approx {M_u}/{phi f_y 0.95 d} = {8923}/{0.9*60*0.95*20.5} =$ 8.485 Try 11#8 As=8.69 in²

$A_{s,min}=0.0018bh=0.0018*7*12*20.5=$ 3.1 in < 8.69✔

$a={A_s f_y}/{0.85 f prime_c b} = {8.69*60}/{0.85*4*7*12}=$ 1.826 in

$epsilon_t = {0.003*(beta_1 d_t-a)}/{a} = {0.003*(0.85*20.5-1.826)}/1.826=$ 0.0256>0.005✔

$phi M_n = phi A_s f_y(d-{a/2})=0.9*8.69*60*(20.5-{1.826/2})=$ 9191 in-kip >8923✔

$l_d={15*12-15}/2-3=$ 79.5 in

$s_c = {7*12-2*3.5}/{11-1} - 1=$ 6.7 in

c=min[3.5,{6.7+1}/2]=min[3.5,3.85]= 3.5

${c+k_{tr}}/d_b = {3.5+0}/1=$ 3.5 >2.5 → c=2.5

$l_d={d_b 3 f_y psi_t psi_e psi_s lambda}/{40 sqrt{f prime_c} {{c+k_{tr}}/d_b} }=$ {1* 3 *60000* 1*1*1*1}/{40 sqrt{4000} *2.5 }= 28.4 < 79.5✔

Use 11#8 spaced uniformly across the long direction.

Short direction: F = 7/2 - 15/24 = 2.875 ft

M_u=q_u L F^2/2 = 4.495*15*2.875^2 /2= 557.3 ft-kip = 6688 in-kip

$A_s approx {M_u}/{phi f_y 0.95 d} = {6688}/{0.9*60*0.95*20.5} =$ 6.359 Try 11#7 As=6.60 in²

$A_{s,min}=0.0018bh=0.0018*15*12*20.5=$ 6.64 in > 6.60✘ Try 10#8 As=7.90 in²

$a={A_s f_y}/{0.85 f prime_c b} = {7.9*60}/{0.85*4*7*12}=$ 1.660 in

$epsilon_t = {0.003*(beta_1 d_t-a)}/{a} = {0.003*(0.85*20.5-1.66)}/1.66=$ 0.0285>0.005✔

$phi M_n = phi A_s f_y(d-{a/2})=0.9*7.9*60*(20.5-{1.66/2})=$ 8391 in-kip >6688✔

$l_d={7*12-15}/2-3=$ 31.5 in

$s_c = {15*12-2*3.5}/{10-1} - 1=$ 18.22 in

c=min[3.5,{18.22+1}/2]=min[3.5,9.6]= 3.5

${c+k_{tr}}/d_b = {3.5+0}/1=$ 3.5 >2.5 → c=2.5

$l_d={d_b 3 f_y psi_t psi_e psi_s lambda}/{40 sqrt{f prime_c} {{c+k_{tr}}/d_b} }=$ {1* 3 *60000* 1*1*1*1}/{40 sqrt{4000} *2.5 }= 28.4 < 31.5✔

Use 10#8 spaced uniformly across the short direction.

Check bearing strength.

$sqrt{A_2/A_1}=sqrt{{7^2}/{1.25^2}}=$ 5.6 > 2 → use 2

Footing: $phi P_n = phi 0.85 f prime_c A_1 sqrt{A_2/A_1}=0.65*0.85*4*15^2 *2=$ 994.5 kip

Column: $phi P_n = phi 0.85 f prime_c A_1 sqrt{A_2/A_1}=0.65*0.85*4*15^2=$ 497.3 kip

$P_u={1.2*180+1.6*160}/{105}=$ 472 kip < 497.3 → Use minimum dowels.

$A_{s,min}=0.005*15^2=$ 1.125 in² use 4#5

l_d=25-3-1-1-0.625= 19.38 in

$l_{d5}=max[{0.02 d_b f_y}/sqrt{f prime_c},0.0003f_y d_b,8]=$ max[{0.02*5*60000}/{8*sqrt{4000}},0.0003*60000*{5/8},8]= max[11.86,11.25,8] approx 12 in < 19.38✔

Column bar diameter not given to determine compression splice length.

Problem 2.

Solve problem 12.21 in your textbook. Do not forget to check splice length and development length.

“For Problems 12.1 to 12.30, assume reinforced concrete weighs 150 lb/ft3, plain concrete 145 lb/ft3, and soil 100 lb/ft3.”

“12.13 Design for load transfer from an 18” x 18” column with 6 #8 bars (D = 200 k, L = 350 k) to an 8-ft-0-in. x 8-ft-0-in. footing. f’c = 4 ksi for footing and 5 ksi for column. fy = 60 ksi. (Ans. 4 #6, 13 in. into footing, 12.5 in. into column)”

“12.21 Repeat Problem 12.13 if a lateral force Vu = 120 k acts at the base of the column. Use shear friction concept. Assume that footing concrete has not been intentionally roughened before column concrete is placed. Thus μ = 0.6λ = (0.6)(1.0) = 0.6. (Ans. six #8 dowels, 27 in. into footing, 24 in. into column)”

P_u=1.2*200+1.6*350= 800 kip

Column Base: phi M_n = phi 0.85 f prime_c A_1 = 0.65*0.85*5*18*18= 895 kip >800✔

$sqrt{A_2/A_1}=sqrt{{8*8*12*12}/{18*18}}=sqrt{9216/324}=$ 28.4 > 2 → use 2

Footing: $phi M_n = phi 0.85 f prime_c A_1 sqrt{A_2/A_1} = 0.65*0.85*4*18*18*2=$ 1432 kip >800✔ → use min dowels

$A_{s,min} = 0.005*18*18=$ 1.62 in² → Use 4#6 As=1.76 in²

$l_d = max[{0.02d_b f_y}/{sqrt{f prime_c}},0.0003d_b f_y,8]=$

Footing: max[{0.02*6*60000}/{8*sqrt{4000}},0.0003*{6/8}*60000,8=max[14.23,13.5,8]= → at least 15 in into footing

Column: max[{0.02*6*60000}/{8*sqrt{5000}},0.0003*{6/8}*60000,8]=max[12.73,13.5,8] → at least 14 in into column

V_n=V_u/phi = 120/0.75= 160 kip

$A_{vf}=V_n/{ f_y mu} = 160/{60*0.6} =$ 4.444 in² > 1.76✘ → Use 6#8

min[phi 0.2 f prime_c A_c,phi 800 A_c]= min[0.75*0.2*4*18*18,0.75*0.800*18*18]= min[194.4,194.4]= 194.4 > 120✔

Assume ${c+K_{tr}}/{d_b}$ =2.5 (c»3.5)

$l_d={d_b 3 f_y psi_t psi_e psi_s lambda}/{40 sqrt{f prime_c} {{c+k_{tr}}/d_b} }=$

Footing: {6* 3 *60000*1*1*1*1}/{8*40 sqrt{4000}*2.5 }= 21.4 in into footing

Column: {6* 3 *60000*1*1*1*1}/{8*40 sqrt{5000}*2.5 }= 19 in into column

Column lap: max[0.0005 f_y d_b,12]=max[0.0005*60000*1,12]= 30 in into column

6#8 30 in into column, 22 in into footing

Problem 3.

Solve problem 12.22 in your textbook (Assume a 22-in footing)

“For Problems 12.1 to 12.30, assume reinforced concrete weighs 150 lb/ft3, plain concrete 145 lb/ft3, and soil 100 lb/ft3.”

“In Problems 12.22 and 12.23, design plain concrete wall footings of uniform thickness.”

Problem	Reinforced concrete wall thickness	D	L	f'c	qa	Distance from bottom of footing to final grade
12.22	12”	8 k/ft	10 k/ft	4 ksi	4 ksf	4'

$q_e=4-{22/12}*0.145-{{48-22}/12}*0.100=4-0.2658-0.2167=$ 3.518 ksf

W={8+10}/3.518= 5.117 ft say 6 ft

$q_u={1.2*8+1.6*10}/6=25.6/6=$ 4.267 ksf

$M_u={4.267*({6-1}/2)^2}/2=$ 13.33 ft-kip = 160 in-kip

Assume cast on dirt: d=22-2=20

$S={bd^2}/6 = {12*20^2}/6 =$ 800 in³

$phi M_n = phi 5 sqrt{f prime_c} S = 0.55*5*sqrt{4000} *800=$ 139.1 in-kip <160✘

Assume cast on concrete filler. d=22

$S={bd^2}/6 = {12*22^2}/6 =$ 968 in³

$phi M_n = phi 5 sqrt{f prime_c} S = 0.55*5*sqrt{4000} *968=$ 173.9 in-kip >160✔

$V_u=({6-1}/2)*4.267=$ 10.67 kip

$phi V_n=phi{4/3}sqrt{f pirme_c} b h = 0.55*{4/3}*sqrt{4000}*12*22=$ 16.7 kip > 10.67✔

6' wide, 22 in deep, not cast directly on soil

Optional Homework (due July 9)

Problem 1.

Problem 2.

Problem 3.

Discussion

Views

Navigation

Personal Tools

Search

Toolbox