CIEG 407
Mid-term Exam Answer Key
1. Find the maximum vertical load using the strength load combinations at the end
of a 14 ft girder shown in the sketch below that supports a roof, 2nd floor and 1st
floor dead, live and snow loads. The givens are:
a. Neglect wind load
b. Snow load = 60 psf
c. Live load 2nd floor = 30 psf
d. Live load 1st floor = 50 psf
e. Roof weight = 26 psf
f. 2nd floor weight = 14 psf
g. 1st floor weight = 21 psf
h. Exterior wall weight = 11 psf
i. Interior wall weight = neglect
j. 2nd floor walls are 9 ft high
k. 1st floor walls are 9 ft high
l. House is 42 ft x 28 ft
m. Roof pitch is 6:12 and roof is framed with trusses
Answer:
Dead load on girder: Roof – 26 psf x 14 ft
2nd floor – 14 psf x 7 ft
1st floor – 21 psf x 7 ft
Walls – 11 psf x 18 ft
Total dead load – 364 lb/ft + 98 + 147 + 88 = 697 lb/ft
Total dead load on one end of girder = 697x14/2 = 4879 lb
Snow load = 60 psf x 14 ft = 840 lb/ft x 14/2 = 5880 lb
Live loads = (30 + 50)x7 = 560 lb/ft x 14/2 = 3920 lb
Load combination that governs:
2. 1.2D+1.6L+0.5S = 1.2x4879 + 1.6x3920 + 0.5x5880 = 15066.8 lb
3. 1.2D+1.6S+L = 1.2x4879 + 1.6x5880 + 3920 = 19182.8 lb
2. Find the highest wind pressures for the 10 regions of the MWFRS and 5 regions
of Components and Cladding using Method 2 from ASCE 7, Chapter 6 for the
following building and conditions:
a. building is 40 ft wide by 120 ft long
b. wind direction is perpendicular to the ridge
c. ridge direction is parallel to the long dimension
d. walls are 20 ft high
e. roof pitch is 3:12
f. Exposure C
g. Building is considered a critical facility
h. building is partially enclosed
i. effective wind area for C&C is 50sf
j. there are no overhangs
k. wind speed is 120 mph 3-second peak gust
l. assume Kd = 0.85 and Kzt = 1.0
Also find ‘a’, the dimension for C&C where higher pressures are applied
Answer:
q = .00256(0.85)(1.0)(1.15)(120)2
q = 33.15 psf
roof angle = 140 which equals a 3:12 roof pitch
mean roof height h = 22.5 ft
a = lesser of .10x40 ft or .4x22.5 ft so a = 4 ft
wind pressures on attached spreadsheet
3. Use the wind pressures determined for Problem 2 and the same building
dimensions for Problem 2 to find:
a. Is the building subject to overturning about the narrow dimension if the
building (including foundation) weighs an average of 30 psf? Justify the
answer using ASD load combinations. Assume the negative wind pressure
on Building Face 2 of the MWFRS helps to keep the building stable.
b. Do you get a different result if you neglect the negative wind pressure on
Building face 2?
c. What would be the equivalent seismic base shear V to overturn the
building? The overturning force is the force that occurs at the roof eave
(top of wall) which is a fraction of the base shear based on roof weight.
Assume the roof weight is 53% of the total weight.
d. From the seismic base shear V, what is the equivalent short period
response Sds? The response modification factor R = 3.25 and F = 1.0.
Answer:
a. yes the building will turn over from wind
building weight = 30 psfx40x120 = 144,000 lb
Resisting moments = 144,000x20x0.6 + 41.11 psfx5x120x22.5 ft + other
face 2 pressures = 2,294,325 ft-lbs
Overturning moments = (34.08+30.63)x20x120x10 + 32.69x5x120x22.5 +
other face 3 pressures = 5,837,785 ft-lbs
b. no, the result is not different, the resisting moment is reduced to 1,728,000
ft-lbs with the overturning moment = 2,817,385 ft-lbs.
c. Resisting moment provided by weight = 1,728,000 ft –lbs.
From the ASD load combination 7: 0.6D + 0.7E would govern so
resisting moment/.7/20ft = F at the roof eave.
F at roof eave = 123,429 lbs.
Since this is 53% of the base shear V, V = 123,429/.53 = 232,884 lb
d. V = FSdsW/R and solving for Sds = VR/FW = 232,884x3.25/1x144,000
Sds = 5.25
4. Will a 20 ft high wall (ft) that is 8” CMU reinforced with #5 bars at 24” o.c.
(using the working stress design method) be acceptable to support 2500 lb/ft at
the top of the wall and resist 25 psf of wind pressure across the entire wall height?
Yes or No – answer must be supported with all of the appropriate checks for
allowable stresses and moments. Wall will need to be checked for buckling.
Givens: a. Fs = 24,000 psi
b. f’m = 2000 psi
c. Deflection limit = 0.007h = .007*20*12 = 1.68 in
d. f’tn = 25 psi
e. f’tp = 50 psi
f. assume wall is pinned top and bottom so k = 1.0 for effective wall
height requirements
g. for 8” wall, wall weight is 55 psf, S = 93.2 in3, I = 355.3 in4, An =
51.3 in2
h. no eccentricity of the 2500 lb. vertical load
fa = P/An = 2500 lb/51.3 in2 = 48.7 psi
r = (I/A)1/2 = (355.3/51.3)1/2 = 2.63 in
Fa = .25f’m*(1-(kh/140r)2) = .25*2000*(1-(120*12/(140*2.63)2) =
500–3.69x10-6 h2
so Fa = 499.7 psi > 48.7 psi OK
Pcr = π2EIeff/(kh2) let Ieff for the reinforced wall = .25I = .25*355.3 = 88.8 in4
If Pcr = 4*applied load then Pcr = 4*2500 lb = 10,000 lb
Solve for Pcr equation for h: 10000 = (3.14172)(1.8x106)(88.8)/h2
h = 33 ft at Pcr load, > 20 ft so OK for buckling under axial load
P = C – T where T = AsFs = (.31/2)(24000) = 3720 lb
C = (Fs/n)(kd/d-kd)bkd/2 where n = 29x106/1.8x106 = 16
So C = (24000/16)(kd/7.625/2 – kd)24kd/2 = 18000(kd2/(3.81-kd))
And P = 2500 + wall self weight/2 = 2500+(55psf*h/2) = 2500+27.5*20 = 3050lb
M about the center of the wall = Pe + 27.5h(t/2)/12 + 25h2/8 = 0+8.73h+3.125h2
C = P+T = 3050+3720 = 6770 lb = 18000(kd2/(3.81-kd)) and solve for kd
kd = 1.0225 in
M = 8.73*20+3.125*202 = 174.625 + 1250 = 1424 ft-lb
fb = M/S = 1424 ft-lb*12 in/ft/93.2 in3 = 183.42 psi
Fb = 1/3f’m = 667 psi so fb < Fb
OK
Check compressive stress in masonry:
fm = (Fs/n)(kd/d-kd) = (24000/16)(1.0225/3.81-1.0225) = 550 psi < 667 psi OK
Fm = fa+Fb(1-fa/Fa) = 48.7+667(1-48.7/499.7) = 650 psi
fs = nFm(kd/d-kd) = 3825 psi < Fs so compression controls
by inspection fa/Fa + fb/Fb < 1, 48.7/499.7 + 183/667 < 1
check max allow Moment = Fmkjbd2/2
OK
j = 1-k/3 where k = [(nρ)2 + 2nρ]1/2 where ρ = As/bd
ρ = .31/24*3.81 = .0034 so k = [(16*.0034)2 + 2*16*.0034]1/2 = .334 and
j = 1-.334/3 so j = .889
allow M = (650 psi)(.334)(.889)(24 in)(3.81 in)2/2 = 33619 in-lb/12 = 2801 ft-lb.>
actual Moment of 1424 ft-lb OK
Answer: Yes 8” wall is acceptable
5. Find the maximum allowable lateral load at the top of the column shown on the
attached sketch. The column is 24” x 16” oriented as shown with b and t
dimensions of b = 24” and t =16”. Find the required reinforcing steel in the
column using the code requirement as a guide (min column steel = .0025An and
max steel = .04An).
Givens:
a. Pa = 20,000 lb
b. An = b x t = 384 in2
c. r = 0.29t
d. S = 1024 in3
e. I = 8192 in4
f. k = 2
fa/Fa + fb/Fb < 1.0
fa = P/A = 20,000/384 = 52 psi
Fa = .25f’m[1-(kh/140r)2] = (.25)(2000)[1-(2*18*12/140*4.64)2] = 278 psi
Fb = 1/3f’m = 667 psi
Set fa/Fa + fb/Fb = 1.0 and solve for max fb
52/278 + fb/667 = 1.0 and fb = 542 psi
fb = M/S = (Pe + self weight*t/2 + PLh)/S
Min e = 0.1t = 0.1*15.625 = 1.56 in
fb = (20000*1.56 + 8277*15.625/2 + PL*18*12)/1024 in3 where fb is 542 psi
Answer: solve for PL = 2125 lb
Min column steel = .0025An = .0025*384 = .96 in2
Max column steel = .04An = 15.36 in2
We know the max Moment from above by summing the forces x moment arms
since fb = M/S, solving for M = 555,008 in-lb
M = C(d-kd/3) where C = (Fs/n)*(kd/d-kd)(bkd/2)
n = Es/Em = 29x106/900*2000 = 16
555,008 = (24000/16)*(kd2/(11.81-kd))*24/2*(11.81-kd/3)
Solve for kd = 4.641 in
Plugging kd back into expression for C, C = 54080 lb
T = C – P = 54080 – 28277 = 25803 lb
As = T/Fs = 25803/24000 = 1.08 in2; > min, < max so OK
Answer: Use 6-#4 bars = 6*.2 in2 = 1.2 in2
Check deflection: Δ = PL3/3EI not to exceed .007h = 1.512 in.
Δ = 2125 lb*(18*12)3/3*(1.8x106)(8192 in4) = 0.48 in < 1.512 in
OK