Prelim #1 and solutions
contact | homepage | spring 2008 index for ARCH 264/564 | current index for ARCH 264/564 | homework index | textFebruary 14, 2008
Show all calculations and write all answers in the exam booklet.
1. Write the following number on the front cover of your exam booklet, top-left corner, above your name:
2. [10 points] Answer the following questions:
The LRFD (strength) design method uses safety factors on both loads and resistance (true or false).
True. Hence, the name: "load resistance factor design."
Heavy building elements are useful in resisting wind forces (true or false).
True. The downward-acting building weight counteracts the horizontal force of the wind.
Heavy building elements are beneficial because they reduce seismic forces (true or false).
False. Heavy building elements increase the inertial forces that must be resisted in seismic design.
What are the units of wind speed used in the equation to determine wind pressure?
Miles per hour (or MPH). See Table A-2.3 in text.
Which of the following is NOT considered in the calculation of seismic base shear?
a) soil; b) importance; c) proximity to bodies of water; d) building weight; e) period of vibration
c) proximity to water bodies is relevant in wind design, but not seismic design.
3. [10 points] Live load reduction: Find the "reduced" live load, in pound units, on an interior second-floor column assuming an unreduced live load, L = 75 psf on all floors (but not including the roof). Dimension "A" = 25 feet and dimension "B" = 30 feet. The column is supporting a single floor (of live load).
The live load reduction coefficient = 0.25 + [15 / (sq.rt. 4 x 750)] = 0.52. In this equation, KLL = 4 (for ordinary columns), and AT = 25 x 30 = 750 sq.ft. for 1 floor of live load.
The reduced live load, L = 75(0.52) = 39.29 psf
The load on the column = 39.29 x 750 = 29,467#.
4. [15 points] Load combinations: A column supports the following loads: L = 80 k; D = 120 k; LR = 30 k; and S = 35 k. Using load combinations for Allowable Stress Design, and assuming a steel structure, what is the governing load on the column?
The relevant load combinations are:
D + L = 120 + 80 = 200
D + 0.75L + 0.75S = 120 + 60 + 26.25 = 206.25 k
The governing load is: 206.25 k.
5. [15 points] Wood tension analysis: Find the capacity of the 6x10 Hem-Fir No.1 tension chord member in the roof truss shown to the right. Assume that the governing load consists of dead and snow loads. The member is bolted through the short (5.5") dimension with two lines of 5/8"-diameter bolts.
Ag = 52.25 in2 [Table A-4.1 in text]
An = 52.25 - 2(3/4 x 5.5) = 44.0 in2
Ft = 525 psi [Table A-3.1 in text, in the "beams and stringers" size group]
F't = 525(1.15) = 603.75 psi [since snow is the load of least duration, use CD = 1.15]
The capacity, P = F't x An = 603.75 x 44.0 = 26,565#
6. [25 points] Wood tension design: Find the smallest acceptable 4x member to support a tension force, P = 16,000#. Use 5/8" diameter bolts; assume 1 line with 3 bolts per line, as shown. Wood is Douglas Fir-Larch No. 2. CD = 1.0 (live and dead loads only). After finding a provisional answer based on assumptions, check the "next smallest size" if necessary.
The adjusted allowable stress, F't = 575(1.0) = 575 psi, where the size factor, CF is assumed to be 1.0 (we will check this value later).
The required net area, An = P / F't = 16,000 / 575 = 27.83 in2.
The required gross area, Ag = An + (bolt hole area) = 27.83 + 1(3/4 x 3.5) = 30.46 in2.
Select a 4x10 provisionally, with actual Ag = 32.38 in2. Since the 4x10 has a size factor, CF = 1.1 which is larger than what was assumed, it will certainly work. The only remaining question is whether a smaller member, with a larger size factor, will also work.
Try a 4x8. The adjusted allowable stress, F't = 575(1.2) = 690 psi, where the size factor, CF is 1.2.
The required net area, An = P / F't = 16,000 / 690 = 23.19 in2.
The required gross area, Ag = An + (bolt hole area) = 23.19 + 1(3/4 x 3.5) = 25.82 in2 which is greater than the actual gross area, Ag = 25.38. Therefore, the 4x8 does not work, and the correct answer is to use the 4x10.
7. [25 points] Steel tension analysis: Find the capacity, P, of the W18x130 shown below. Use 3/4" diameter bolts; assume 2 lines with 3 bolts per line in each flange, as shown. Use A-992 steel.
The capacity must be determined twice, once for yielding on the gross area, and once for rupture on the effective net area.
Yielding on gross area:
Ag = 38.2 in2 [Table A-4.3 in text]
Ft = 0.6 x 50 = 30 ksi [Tables A-3.11 and A-3.12 in text]
P = 30 x 38.2 = 1146 k.
Rupture on effective net area:
The shear lag coefficent, U, is found based on the criteria listed in Table A-6.1 in the text:
W-section? yes
Connection to flange? yes
3 bolts per line yes
bf greater or equal to 2/3 d? From Table A-4.3 in the text, is 11.16 greater or equal to 2/3(19.25) = 12.83? no
Therefore, we must use U = 0.85
Ag = 38.2 in2 [Table A-4.3 in text]
An = 38.2 - 4(7/8 x 1.2) = 34.0 in2, where 4 represents 4 bolt holes in the failure plane, and 1.2 is the flange thickness from Table A-4.3.
Ae = An x U = 34.0 x 0.85 = 28.9 in2
Ft = 0.5 x 65 = 32.5 ksi [Tables A-3.11 and A-3.12 in text]
P = 32.5 x 28.9 = 939 k.
The governing capacity is the smaller of the two values: 939 k.
© 2008 Jonathan Ochshorn. First posted 18 February 2008. Last updated: 14 May 2008