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Homework #2 Due 6/7/07

— David Wagner 2007/06/07 23:24

Illustration for problem 1.

Problem 1

For the beam cross-section shown, for each case find compression forces and centroids two different ways (when possible), As to satisfy equilibrium and the corresponding nominal moment. Assume that the section reaches ultimate capacity (In other words, there is no need to check for minimum or maximum amount of steel).

a) f’c = 3500 psi, fy = 60,000 psi, #3 stirrups and 4#10 as tension

A_s=2.00 in²

Assume steel fails at 85% concrete failure.

C=T=A_s f_y = 2.00 * 60000 =

C= 120 kip

C=0.85f^prime_c A_c right A_c=C/{0.85f^prime_c}=120000/{3500*0.85}=40.336 in²

A_top=4*(5+8+5)=72 in² ∴ b=18 (Compression area is entirely within the top of the channel.)

A_c=ab right a=A_c/b = 40.336/18 =2.24 in

c=a/2=2.24/2 =1.12 in

d=(4+16.5+3.5)-1.5-0.375-1.27-0.5=24-3.645=20.355 in

M_n=C(d-{a/2})=120*(20.355-{2.24/2})=120*19.235=

Mn=2308.2 kip-in

Now, redo it using a=β1c, β1=0.85 for f'c <4000. Assume whitney block model applies.

First, find c, the distance from the compression face to the neutral axis at the centroid.

c={(2*18*4)+(2*14*5*20)}/{(18*4)+2*(5*20)}={144+2800}/{72+200}=2944/272= 10.8235 in

a=beta_1 c = 0.85*10.8235=9.2 in > 4” → Find ybar, the centroid of Ac from the compression face.

overline{y}={sum{}{}(A_i overline{y_i})}/{A_c} = {2*(4*18)+(4+{5.2/2})*(10*5.2)}/{(4*18)+(10*5.2)} = {144+(6.6*52)}/{72+52} = {144+343.2}/{124}= 487.2/124 =3.929 in

M_n=C(d-overline{y})=120*(20.355-3.929)=120*16.426=

Mn=1971 kip-in

b) f’c = 6500 psi, fy = 60,000 psi, #3 stirrups and 4#10 as tension.

A_s=2.00 in²

Assume steel fails at 85% concrete failure.

C=T=A_s f_y = 2.00 * 60000 =

C= 120 kip

C=0.85f^prime_c A_c right A_c=C/{0.85f^prime_c}=120000/{6500*0.85}=21.719 in²

A_top=4*(5+8+5)=72 in² ∴ b=18 (Compression area is entirely within the top of the channel.)

A_c=ab right a=A_c/b = 21.719/18 =1.2066 in

c=a/2 = 1.2066/2=0.603 in

d=(4+16.5+3.5)-1.5-0.375-1.27-0.5=24-3.645=20.355 in

M_n=C(d-{a/2})=120*(20.355-{1.2066/2})=120*19.7517=

Mn=2370.2 kip-in

Now, redo it using a=β1c

beta_1 = 0.85-0.05({f{^prime}_c -4000}/1000) = 0.85-0.05*({6500-4000}/1000)= 0.85-.125 = 0.725

a=beta_1 c = 0.725*10.8235=7.847 in > 4” → Find ybar, the centroid of Ac from the compression face.

overline{y}={sum{}{}(A_i overline{y_i})}/{A_c} = {2*(4*18)+(4+{3.847/2})*(10*3.847)}/{(4*18)+(10*3.847)} = {144+(5.92*38.47)}/{38.47+72} = {144+227.9}/{110.47}= 371.8/110.47 =3.366 in

M_n=C(d-overline{y})=120*(20.355-3.366)=120*16.989=

Mn=2039 kip-in

Illustration for problem 2.

Problem 2

Determine the nominal flexural strength of the rectangular section shown for

  • f’c = 4000 psi
  • fy = 60,000 psi
  • a = 8.14 in

Assume that the section reaches ultimate capacity (In other words, there is no need to check for minimum or maximum amount of steel).

A_s=2.79 in²

Assume steel fails at 85% concrete failure.

C=T=A_s f_y = 2.79 * 60000 =

C= 167.400 kip

C=0.85f^prime_c A_c right A_c=C/{0.85f^prime_c}=167400/{4000*0.85}=49.235 in²

A_top=7x10=70 in² ∴ b=10 (Compression area is entirely within the top of the channel.)

A_c=ab right a=A_c/b = 49.235/10 =4.9235 in

However, a is given as 8.14 in ∴ A_c=70+16*(8.14-7)=78.24 in²

This means the area is the entire top 10×7 rectangle (with its centroid 3.5 in from the compression face) and a strip 16×1.14 (with its centroid 7.57 in from the compression face).

The distance from the top compression face to the centroid of A_c is {{(3.5*10*7)+(16*1.14*7.57)}/78.24} = {245+138.0768}/78.24 =4.896 in

d=23+7-2.5=27.5 in

M_n=C(d-4.896)=167.4*(27.5-4.896)

Mn=3783.9 kip-in

Illustration for problem 3a.

Problem 3

For the beam cross-sections shown, calculate the nominal and design moments. Do all necessary checks

a) Assume #3 stirrups, fy = 60,000 psi and f’c = 4,000 psi.

As=3#7=1.80²

d_t=d=22-1.5-{3/8}-{7/{2*8}}=19.7 in

A_s >= delim{lbrace}{matrix{2}{1}{{{{3 sqrt{{f^prime}_c}}/f_y}b_w d ={3*sqrt{4000}*12*19.7}/60000={44858/60000}=0.74756} {{200 b_w d}/f_y ={200*12*19.7/60000 = 0.788=A_smin} }} }}{}

a={A_s f_y}/{0.85 {f^prime}_c b } = {1.8*60}/{0.85*4*12} = 108/40.8=2.647 in

epsilon_t = {0.003(beta_1 d_t - a )}/a = {0.003*(0.85*19.7-2.647)}/2.647 = 0.0423/2.647 = 0.016 >= 0.005 right phi=0.9

M_n = A_s f_y(d-{a/2})=1.8*60*(19.7-{2.647/2})=

1985 kip-in

phi M_n=0.9*1985=

1787 kip-in

c) illustration for problem 3b.Assume #3 stirrups, fy = 40,000 psi and f’c = 4,500 psi

As=3#5=0.93 in²

d_t=d=22-1.5-{3/8}-{5/{2*8}}=19.813 in

A_s >= delim{lbrace}{matrix{2}{1}{{{{3 sqrt{{f^prime}_c}}/f_y}b_w d ={3*sqrt{4500}*10*19.813}/40000={39873/40000}=0.997=A_smin} {{200 b_w d}/f_y ={200*10*19.813/40000 = 0.991} }} }}{}

As < Asmin ∴ this beam is unacceptable by current standards.

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