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Assignment #8
Problems: all but 8A from Ambrose & Tripeny, Chapter 9, pgs 303, 304, 306, 322, and 243.
Note: Problems marked with a * have been altered with respect to the problem stated in the text.
Problem 9.4.D. USE US UNITS. A beam 30 ft [9 m] long has concentrated live loads of 9 kips
[40 kN] each at the third points and also a total uniformly distributed dead load of 20 kips [89
kN] and a total uniformly distributed live load of 10 kips [44 kN]. Design the beam for plastic
flexure. Use A36 steel and assume that least-weight members are desired. (LRFD beam design)
Partial answers to check with: Zx > 111 in3 including self weight
Problem 9.4.I.* (typographic error) USE US UNITS. A cantilever beam 12 ft [3.6 m] long has
a uniformly distributed dead live load of 600 lb/ft [8.8 kN/m] and a uniformly distributed dead
load of 1000 lb/ft. Design the beam for plastic flexure. Use A36 steel and assume that least-
weight members are desired. (LRFD beam design)
Problem 9.5.A*. USE US UNITS. A W shape steel is to be used for a uniformly loaded simple
beam carrying a total dead load of 27 kips [120 kN] and a total live load of 50 kips [222 kN] on
a 45-ft [13.7 m] span. Select the lightest weight shape for unbraced lengths of (1) 10 ft [3.05
m]; (2) 15 ft [4.57 m]; (3) 22.5 ft [6.90 m] with the Beam Design Moment chart provided for
steel with Fy = 50 ksi. Show inclusion of self weight in the design moment after selection for
case (c). (LRFD beam design and charts)
Partial answers to check with: (1) W 24x76 (2) W 24x84 (3) W 21x101
Problem 9.6.C. Compute the shear capacity (
vVn) for the following beam of A36 steel: W 10x33.
(LRFD shear)
Partial answers to check with: 60.95k
Problem 9.7.B. Find the maximum deflection in inches for the following simple beam of A36
steel with uniformly distributed load. Find the value using (1) the equation for deflection of a
uniformly distributed load, and (2) the curves in Figure 9.11. (beam diagrams and formulas,
W 16 x 36, span = 20 ft, total service load = 2.5 kips/ft [6 m, 36.5 kN/m]
Partial answers to check with: (1) 0.693 in (2) 0.7 in
Problem 9.10.C. USE US UNITS. Open web steel joists are to be used for a floor with a live
load of 50 psf [2.39 kN/m2] and a dead load of 45 psf [2.15 kN/m2] (not including the joist
weight on a span of 36 ft [11 m]. Joists are 2 ft [0.61 m] on center, and deflection is limited to
1/360 of the span under live load only and to 1/240 of the span under total load. Select (a) the
lightest possible joist and (b) the shallowest depth possible joist. (LRFD design and charts)
Partial answers to check with: (a) 24K4 (b) 20K7 (text answer is wrong)
8A) If the 36 ft joist of Problem 9.10.B is a 26K9 (having self weight) with 14 panels at 2.5 ft
for the top chord, use the method of sections to determine the member forces in the top and
bottom chord and the web for the section indicated in the figure using factored loads and self
weight. (load tracing)
Partial answers to check with:
Pend = 494.6 lb, Ppanel=706.6 lb
top chord = 18.5 or 18.7 k (C)
web = 2.04 k (C)
0.5 ft at end & 14 panels @ 2.5ft & 0.5 ft at end = 36 ft
26 inches
5087.5 lb
5087.5 lb
(20%)
(15%)
(15%)
(10%)
(10%)
(10%)
(20%)
and charts)
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