Table of Contents

Homework #1 Due 6/6/07

— David Wagner 2007/06/05 23:24

Problem 1

For the following beam and loads find the ultimate maximum positive and/or negative moments using two different methods (use all load combinations that apply):

  1. Factor the loads first to obtain the factored load wu and Pu and then find Mu
  2. Find the moments for the unfactored loads and then combine the results to obtain Mu.

Illustration for problem 1

  • D = 2.5 k/ft (includes beam weight)
  • L = 5.0 k/ft
  • PL = 12 kips

Method 1

Assume dead loads control. U=1.4(D+F), F=0.[Text pg. 81]

right U=1.4D

omega_u = 1.4D = 1.4*2.5 =3.5 k/ft

M_u={omega_u l^2}/2 = {3.5*10^2}/2 = 350/2=

175 kip-ft

Assume live loads control. U=1.2(D+F+T) + 1.6(L+H) + 0.5(L_r or S or R)[Text pg. 81], F=T=H=L_r=S=R=0

right U=1.2D + 1.6L

omega_u = 1.2 omega_D + 1.6 omega_L = 1.2*2.5 + 1.6*5.0 = 3+8=11 k/ft

P_u = 1.2 P_D + 1.6 P_L = 1.2*0 + 1.6*12 =19.2 kips

M_u={omega_u l^2}/2 + P_u d = {{11*10^2}/2} + 19.2*4 = 550 + 76.8=

626.8 kip-ft

Method 2

Use the magic distributive property of multiplication. M_L = {omega_L l^2}/2 = {5*100}/2 =250 kip-ft

M_D = {omega_D l^2}/2 = {2.5*100}/2 =125 kip-ft

M_P_L = P_L l = 12*4=48 kip-ft

Assume dead loads control. U=1.4D

M_u = 1.4*M_D=1.4*125=

175 kip-ft

Assume live loads control. U=1.2D + 1.6L

M_u = 1.2(M_D) + 1.6(M_L+M_P_L)=1.2*125 + 1.6*(250+48)=150+476.8=

626.8 kip-ft

Unfactored Loads

M_D={omega_D l^2}/2 = {2.5*100}/2=250/2=125 kip-ft

M_L={omega_L l^2}/2 = {5*100}/2=500/2=250 kip-ft

M_{PL}=4*12=48 kip-ft

Sigma M=M_D + M_L + M_{PL} = 125+250+48= 423 kip-ft

Problem 2

If in problem 1 instead of PL the concentrated load is a dead load PD what values of wu and Pu will you use?

Assume dead loads control. U=1.4(D+F) : F=0. [Text pg. 81]

right U=1.4D

omega_u = 1.4 omega_D = 1.4*2.5 =3.5 k/ft

P_u=1.4 P_D = 1.4*12=16.8 kips

M_u={omega_u l^2}/2 + P_u d= {3.5*10^2}/2 + 16.8*4 = 175 + 67.2= 242.2 kip-ft

Assume live loads control. U=1.2(D+F+T) + 1.6(L+H) + 0.5(L_r or S or R) : F=T=H=L_r=S=R=0[Text pg. 81]

right U=1.2D + 1.6L

omega_u = 1.2 omega_D + 1.6 omega_L = 1.2*2.5 + 1.6*5.0 = 3+8=

11 k/ft

P_u = 1.2 P_D + 1.6 P_L = 1.2*12 + 1.6*0 =

14.4 kips

M_u={omega_u l^2}/2 + P_u d = {{11*10^2}/2} + 14.4*4 = 550 + 57.6=

607.6 kip-ft

Since these values of ωu and Pu produce the largest moment, use them.

Problem 3

For the following frame and loads find all load combinations that apply :

  • D = 3 k/ft (includes weight of structural elements)
  • L = 2.5 k/ft
  • wind load=0.8 k/ft (on walls)

Illustration for problem 3

Assume the top element is not a 'roof' and has no rain loading.

  • ACI eq. 9-1: U = 1.4(D+F)
  • ACI eq. 9-2: U = 1.2(D+F+T) + 1.6(L+H) + 0.5(L_r or S or R)
  • ACI eq. 9-3: U = 1.2(D+F+T) + 1.6(L_r or S or R) + (1.0L or 0.8W)
  • ACI eq. 9-4: U = 1.2(D+F+T) + 1.6W + 1.0L + 0.5(L_r or S or R)
  • ACI eq. 9-6: U = 0.9D + 1.6W +1.6H

: F=T=H=S=R=L_r=0 right

  • ACI eq. 9-1: U = 1.4D right omega_U=1.4*omega_D=1.4*3=4.2 k/ft down
  • ACI eq. 9-2: U = 1.2D + 1.6L right omega_U=1.2*omega_D + 1.6*omega_L = 1.2*3 + 1.6*2.5 = 3.6+4=7.6 k/ft down
  • ACI eq. 9-3: U = 1.2D + 0.8W right
    • omega_U=1.2*omega_D = 1.2*3= 3.6 k/ft down
    • omega_U=1.2*omega_W = 1.2*0.8= 0.96 k/ft ←
  • ACI eq. 9-4: U = 1.2D + 1.6W + 1.0L
    • omega_U=1.2*omega_D +1.0 omega_L= 1.2*3 + 1.0*2.5=3.6+2.5= 6.1 k/ft down
    • omega_U=1.6*omega_W = 1.6*0.8= 1.28 k/ft ←
  • ACI eq. 9-6U = 0.9D + 1.6W
    • omega_U=0.9*omega_D = 0.9*3= 2.7 k/ft down
    • omega_U=1.6*omega_W = 1.6*0.8= 1.28 k/ft ←

Problem 4

For the following beam and loads find the ultimate maximum positive (Mu+) and/or negative (Mu-) moments using two different methods( use all load combinations that you think apply):

  1. Factor the loads first to obtain the factored load wu and then find Mu
  2. Find the moments for the unfactored loads and then combine the results to obtain Mu.

Note: the live load does not have to be on every span.

Illustration for problem 4

  • D = 3.5 k/ft (includes weight of the beam)
  • L = 5.0 k/ft

First, the maximum positive moment occurs between the supports when the live load is only over the 30' span. The variable x is the distance from the rightmost support, l=30', a=10'. (The graphs appear reversed left-to-right, and in this section are scaled for a unit load.)

M_x_D30={{omega_D x}/{2l}}(l^2-a^2-xl) = {{omega_D x}/60}(800-30x)=omega_D({40/3}x - {1/2}x^2)

JpGraph Function Plot

M_x_L30={{omega_L x}/2}(l-x)={{omega_L x}/2}(30-x)=omega_L(15x-{1/2}x^2)

JpGraph Function Plot

The maximum negative moment occurs at the support when the live load is only over the 10' span.

M_D10={omega_D a^2}/2= 50 omega_D

M_L10={omega_L a^2}/2= 50 omega_L

Method 1

Assume dead loads control. U=1.4(D+F) : F=0. [Text pg. 81]

right U=1.4D

omega_u = 1.4D = 1.4*2.5 =3.5 k/ft

M_u(x)=M_x_D(x)=omega_u({40/3}x - {1/2}x^2)={140/3}x -{7/4}x^2

JpGraph Function Plot

Differentiate and set to zero to find the maximum positive moment.

{140/3}-{14/4}x=0 right {14/4}x={140/3} right x={40/3}=13.33'

M_{u+}={140/3}x -{7/4}x^2={140/3}*{40/3} -{7/4}*{40/3}*{40/3}={5600/9}-{11200/36}=622-311=

311 kip-ft

The maximum negative moment is M_{u-}=M_D10= -50 omega_u = -50*3.5 =

-175 kip-ft

Assume live loads control. U=1.2(D+F+T) + 1.6(L+H) + 0.5(L_r or S or R) : F=T=H=L_r=S=R=0[Text pg. 81]

right U=1.2D + 1.6L

omega_u = 1.2 omega_D + 1.6 omega_L

However, adding these at this point will not work since the maximum positive moment depends on the relative values of ω used in the two formulas above. In other words, the maximum moments produced by the dead and live loads occur in different locations along the beam, and so are not additive. The maximum positive moment (located between the live and dead maximum moments) may be found by summing the diagrams separately.

M(x)=M_D30(x)+M_L30(x)=omega_u_D({40/3}x - {1/2}x^2) + omega_u_L(15x-{1/2}x^2) =1.2*3.5({40/3}x - {1/2}x^2) + 1.6*5.0(15x-{1/2}x^2) ={168/3}x -2.1x^2 + 120x - 4x^2 = 176x - 6.1x^2

JpGraph Function Plot

Differentiate and set to zero to find the maximum positive moment. 176-12.2x=0 right x=14.426' M_{u+}=176x - 6.1x^2 = 176*14.426-6.1*14.426*14.426=2539-1269.5=

1255 kip-ft

The formulas for the maximum negative moment are linear so the sum of ωD and ωL may be used.

omega_u = 1.2 omega_D + 1.6 omega_L = 1.2*3.5 + 1.6*5 = 4.2+8 = 12.8

M_{u-} = - 50 omega_u = -50*12.8 =

-640 kip-ft

Method 2

The maximum positive moment produced by the dead load is around 13' from the right support. M_{DL30}={omega_D/{8l^2}}(l+a)^2 (l-a)^2={3.5/{8*30^2}}(30+10)^2(30-10)^2={3.5/7200}*1600*400=311

The maximum positive moment produced by the live load placed only between the supports is halfway (15') from the right support.

M_{LL30}={{omega_L l^2}/8}={{5.0*30^2}/8}=562.5

The maximum negative moments produced by both dead and live loads are both at the left support.

M_{DL10}={{omega_D a^2}/2}={{3.5*10^2}/2}=175

M_{LL10}={{omega_L a^2}/2}={{5*10^2}/2}=250

Assume dead loads control. U=1.4D

M_{u+} = 1.4*M_{DL30}=1.4*311=

435.4 kip-ft

M_{u-} = 1.4*M_{DL10}=1.4*175=

-245 kip-ft

Assume live loads control. U=1.2D + 1.6L

(Note the following really shouldn't be factored and added together as if they were collocated, but perhaps they're close enough.)

M_{u+} = 1.2 M_{DL30} + 1.6 M_{DL30}= 1.2*311+1.6*562.5 = 373.2+900=

1273.2 kip-ft

The next summation is OK since both maxima occur at the left support.)

M_{u-} = 1.2(175) + 1.6(250)L=210+400=

-610 kip-ft

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