Homework #4 Due 6/12/07

Homework #4 Due 6/12/07

Problem 1.

Using the cross-section for problem 2.27 in your textbook find Mn and φMn. Make sure the ACI code is satisfied.

For the following problems assume #4 stirrups and minimum cover requirements

[The textbook specifies fy=60,000 psi and f'c=4,000 psi and the answer should be Mn=688.2 ft-k.]

b=b_w=16”, h=30”, d_t=30-2.5=27.5”, A_s=6#9=6.00 in²

d={(30-2.5-{1.128/2})+(30-5)}/2=d= 25.968”

$A_{s,min}>={3 sqrt{f prime_c} b_w d}/f_y={3*sqrt{4000}*16*25.968}/60000=$ 1.3139 in²

$A_{s,min}>={200b_w d}/f_y ={200*16*25.968}/60000=$ 1.385 in²

$right A_{s,min}=$ 1.4 in², 6.00>1.4✔

$a={A_s f_y}/{0.85 f{prime}_c b}={6*60}/{0.85*4*16}=$ 6.6176 in

beta_1(f prime_c=4000)= 0.85

$epsilon_t={0.003(beta_1 d_t -a)}/a = {0.003*(0.85*27.5-6.6176)}/6.6176={0.003*17.2}/6.6176=$ 0.00835

epsilon_t=0.00835 >= 0.005 right phi= 0.9

$M_n=A_s f_y (d-{a/2})=6*60*(25.968-{6.6176/2})=$

M_n=8237 kip-in

phi M_n = 0.9*8338=

φM_n=7413 kip-in

Find the area of steel needed to carry the maximum moment for the beam in problem 4.10 in your textbook and an assumed cross-section of 14 in x 28 in (width x height, always). Ignore the given value of ρ but use the same material properties and unit weight of concrete given. HINT: Use Method A

[“In Problems 4.10 to 4.22 […] Assume concrete weighs 150lb/ft³, fy=60,000 psi, and f'c=4,000 psi.”]

P_L=30 k, l=24', b=b_w=14”, h=28”, d=d_t≈h-2.5=25.5” (assume single row)

$w_D=3+0.150*{14/12}*{28/12}=$ 3.41 k/ft

For U=1.4D

$M_u = 1.4({w_D l^2}/8)=1.4({3.41*24^2}/8)=1.4*245.5=$ 343.7 kip-ft

For U=1.2D + 1.6L

$M_u = 1.2({w_D l^2}/8) + 1.6({P_L l}/4)=1.2({3.41*24^2}/8) + 1.6({30*24}/4) = 1.2*245.5+1.6*180 = 294.6+288 =$ 582.6 kip-ft

→M_u=582.6 kip-ft = 6991 kip-in

Assume φ=0.9 and M_n=M_u/φ=7768 kip-in

$R=M_n/{b d^2}=7768/{14*25.5^2}=$ 853 psi

ρ≈0.017 (from chart)

A_s=rho b d = 0.017*14*25.5= 6.07 in²

Assume two rows: d≈h-3.5=24.5”

$R=M_n/{b d^2}=7768/{14*24.5^2}=$ 924 psi

ρ≈0.018 (from chart)

A_s=rho b d = 0.018*14*24.5= 6.17 in²

$a={A_s f_y}/{0.85f prime_c b}={6.17*60}/{0.85*4*14}=$ 7.78 in

beta_1(4000)= 0.85

$epsilon_t = {0.003(beta_1 d_t -a)}/a={0.003(0.85*25.5 -7.78)}/7.78=$ 0.00536 > 0.005 → phi=0.9

$M_n=A_s f_y(d-{a/2})=6.17*60*(24.5-{7.78/2})=$ 7630 kip-in

Not quite enough… Try quadratic, one row.

d=d_t≈h-2.5=25.5” (assume single row)

$a={A_s f_y}/{0.85f prime_c b}$

$M_n=9940=A_s f_y (d-{a/2}) = A_s f_y (d-{ {A_s f_y}/{2*0.85 f prime_c b} }) = A_s*60*(25.5-{ {A_s*60}/{2*0.85*4*14} })$ $right 1530 A_s - 37.8{A_s}^2-7768=0={A_s}^2 -40.5A_s +205.5$

$right A_s={40.5 pm sqrt{40.5^2-4*205.5}}/2 = {40.5 pm 28.6}/2 = 34.6,5.95$

A_s=5.95 in²

4#11=6.24 in² will fit in 13.8 in, and the larger area will more than offset the slight reduction in d…

Problem 3.

Find the area of steel needed to carry the maximum moment for the beam in problem 4.20 and an assumed cross-section of 16 in x 32 in. Ignore the given value of ρ but use the same material properties and unit weight of concrete given. HINT: Use Method A

[“In Problems 4.10 to 4.22 […] Assume concrete weighs 150 lb/ft³, fy=60,000 psi, and f'c=4,000 psi.”]

w_D=2 k/ft, l=14, b=16”, h=32”

Assume one layer: d_t=d≈h-2.5=29.5”

L » D → U = 1.2D + 1.6L

$w_D=2+0.150*{16/12}{32/12}=2.53$

$M_u=1.2({2.53*14^2}/2) + 1.6(20*7 + 10*14) = 1.2*248+1.6*280 =298+448=$ 746 kip-ft = 8946 kip-in

Assume φ=0.9 and M_n=M_u/φ = 9940 kip-in

$R={M_n}/{bd^2} = 9940/{16*29.5^2}=0.714=$ 714 psi → ρ≈0.011

A_s=rho b d = 0.011*16*29.5= 5.19 in² ✔one layer OK

beta_1(4000)=0.85

$a={A_s f_y}/{0.85 f prime_c b} = {5.19*60}/{0.85*4*16}=$ 5.724

$epsilon_t={0.003(beta_1 d_t -a}/a={0.003*(0.85*29.5-5.724}/5.724=$ 0.01014 > 0.005 → φ=0.9✔

$M_n=A_s f_y(d-{a/2})=5.19*60*(29.5-{5.724/2})=$ 8295 kip-in ✘Not enough.

Try A_s=6 in² ✔one layer OK

$a={A_s f_y}/{0.85 f prime_c b} = {6*60}/{0.85*4*16}=$ 6.6176

$epsilon_t={0.003(beta_1 d_t -a}/a={0.003*(0.85*29.5-6.6176}/6.6176=$ 0.008367 > 0.005 → φ=0.9✔

$M_n=A_s f_y(d-{a/2})=6*60*(29.5-{6.6176/2})=$ 9430 kip-in ✘Still not enough.

Try A_s=6.35 in² ✔one layer OK

$a={A_s f_y}/{0.85 f prime_c b} = {6.35*60}/{0.85*4*16}=$ 7.0037

$epsilon_t={0.003(beta_1 d_t -a}/a={0.003*(0.85*29.5-7.0037}/7.0037=$ 0.00774 > 0.005 → φ=0.9✔

$M_n=A_s f_y(d-{a/2})=6.35*60*(29.5-{7.0037/2})=$ 9905 kip-in ✘Still not enough.

(This method sucks. The chart is useless; it doesn't result in a correct answer. Forget it and use some well-written software. Solving this by exhaustive enumeration of all acceptable reinforcing bar combinations is trivial for modern computers and a waste of time for human beings.)

Fine. We'll go quadratic on it. Hmm… Need to derive the formula first…

$a={A_s f_y}/{0.85f prime_c b}$

$M_n=9940=A_s f_y (d-{a/2}) = A_s f_y (d-{ {A_s f_y}/{2*0.85 f prime_c b} }) = A_s*60*(29.5-{ {A_s*60}/{2*0.85*4*16} })$ $right 1770 A_s - 33.1{A_s}^2-9940=0={A_s}^2 -53.5A_s +300$

$right A_s={53.5 pm sqrt{53.5^2-4*300}}/2 = {53.5 pm 40.8}/2 = 47.15,6.35$

A_s=6.35 in²

I guess it was close enough after all…

Problem 4.

Repeat problem 2 but do not use the assumed cross-section instead use εt = 0.01. HINT: Use Method B

[“In Problems 4.10 to 4.22 […] Assume concrete weighs 150lb/ft³, fy=60,000 psi, and f'c=4,000 psi.”]

b=b_w

M_u=582.6 kip-ft = 6991 kip-in

Assume φ=0.9 and M_n=M_u/φ=7768 kip-in

beta_1(4000)= 0.85

$A_{s,min}={3sqrt{f prime_c}b_w d}/fy={3sqrt{4000}bd}/60000=$ 0.003162 bd

$A_{s,min}={200bd}/60000=$ 0.003333 bd

$A_{s,min}=$ 0.003333 bd

$rho_{min} = {A_{s,min}}/bd =$ 0.003333

$rho = {{0.85 beta_1 f prime_c}/f_y}(epsilon_u/{epsilon_u+epsilon_t}) = {{0.85*0.85*4}/60} * (0.003/{0.003+0.01})={2.89/60} * (0.003/0.013)=rho=$ 0.0111 >0.003333✔

A_s=rho b d = 0.0111 bd

$a={A_s f_y}/{0.85 f prime_c b}={rho b d f_y}/{0.85 f prime_c b} = {0.0111*d*60}/{0.85*4}=$ 0.196d

$M_n=A_s f_y(d-{a/2})=0.0111*bd*60*(d-{{0.196d}/2}) = 0.666bd^2-0.0320bd^2 = 0.634bd^2 = 7768$ $right bd^2=12252 right d=sqrt{12252/b}$

b	10	12	14	16	18	20
d	35.0	32.0	29.6	27.7	26.1	24.75
d+2.5			32.1	30.2	28.6
h				30*

b=16”, h=30”, d≈27.5”

A_s=rho b d = 0.0111*16*27.5=</m>

A_s=4.884 in²*

*Choose bars with a total area slightly larger than A_s, for example 5#9=5.00 in², b_min=14.3”<16”✔.

Problem 5.

Repeat problem 3 but do not use the assumed cross-section instead use εt = 0.01. HINT: Use Method B

[“In Problems 4.10 to 4.22 […] Assume concrete weighs 150 lb/ft³, fy=60,000 psi, and f'c=4,000 psi.”]

Assume φ=0.9 and M_n=M_u/φ = 9940 kip-in

b_w=b

$A_{s,min}={3sqrt{f prime_c}b_w d}/f_y = 3*sqrt{4000}*bd/60000=0.00316bd$

$A_{s,min}=200b_w d/f_y = 200*bd/60000=$ 0.00333 bd ✔

$A_{s,min}=$ 0.00333 bd

$rho = {{0.85beta_1 f prime_c}/f_y}(epsilon_u/{epsilon_u=epsilon_t}) = {0.85*0.85*4*0.003}/{60*(0.003+0.010)} = 0.00867/0.780=$ 0.0111 > 0.00333✔

rho=0.0111

A_s = rho bd = 0.0111 bd

$a={A_s f_y}/{0.85 f prime_c b}={0.0111*b*d*60}/{0.85*4*b}=$ 0.196 d

$M_n=A_s f_y(d-{a/2})=9940=0.0111*bd*60*(d-{{0.196d}/2}) = 0.666bd^2-.065bd^2 = 0.601bd^2$ $right bd^2=16546 right d=sqrt{16546/b}$

b	10	12	14	16	18	20
d	40.7	37.1	34.4	32.2	30.3	28.8
d+2.5				34.7	32.8	31.3
h						32

b=20”, h=32”, d≈29.5”

A_s=rho b d = 0.0111*20*29.5=</m>

A_s=6.55 in²*

Single layer OK for various bar combinations.

Homework #4 Due 6/12/07

Problem 1.

Problem 2.

Problem 3.

Problem 4.

Problem 5.

Discussion

Views

Navigation

Personal Tools

Search

Toolbox