Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 16, Problem 16.3P
Rework Problem 16.1 given a load of 1.0 kip/ft and a span length of 30 ft.
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Select the wide flange steel girder for a simple span of 8m subjected to a concentrated load of 489 kN at the midspan. Use A36 steel and assume that beam is supported laterally for its entire length.
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Chapter 16 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 16 - Prob. 16.1PCh. 16 - A simply supported beam is to support a uniformly...Ch. 16 - Rework Problem 16.1 given a load of 1.0 kip/ft and...Ch. 16 - Select the lightest W shape to support a...Ch. 16 - Select the lightest W shape to support a...Ch. 16 - 16.6 A simply supported beam is to span 15 ft. It...Ch. 16 - 16.7 A simply supported beam is to span 24 ft. It...Ch. 16 - 16.8 Design a timber beam of hem-fir (S4S) to...Ch. 16 - Select simply supported timber beams (S4S) for the...Ch. 16 - 16.10 Select a southern pine (S4S) timber beam for...
Ch. 16 - 16.11 Select simply supported hem-fir (S4S) joists...Ch. 16 - Design simply supported timber beams (S4S) for the...Ch. 16 - For the following computer problems, any...Ch. 16 - 16.16 Select the lightest W shape to support a...Ch. 16 - 16.17 Select the lightest W shape for the beam...Ch. 16 - Select the lightest W shape for the cantilever...Ch. 16 - 16.19 Select the lightest W shape to support a...Ch. 16 - 16.20 Select the lightest W shape for the beams...Ch. 16 - 16.21 The structural steel floor system shown is...Ch. 16 - 16.22 The structural steel framing plan shown...Ch. 16 - 16.23 Select the lightest steel wide-flange...Ch. 16 - 16.24 Select the lightest steel wide-flange...Ch. 16 - 16.25 Design the lightest W-shape beams to support...Ch. 16 - 16.26 In Problem 16.18, assume that the 500 lb/ft...Ch. 16 - 16.27 Select a southern pine (S4S) simply...Ch. 16 - 16.28 A redwood beam is to support a uniformly...Ch. 16 - 16.29 A partial plan view for a residential floor...Ch. 16 - 16.30 For the floor framing of Problem 16.29,...Ch. 16 - 16.31 Select a Douglas fir (S4S) beam for the...Ch. 16 - 16.32 Select southern pine (S4S) simply supported...Ch. 16 - 16.33 Rework Problem 16.32 using joists spaced 12...Ch. 16 - 16.34 Select Douglas fir (S4S) simply supported...Ch. 16 - 16.35 Select southern pine (S4S) simply supported...Ch. 16 - 16.36 A 15-ft-span simply supported hem-fir (S4S)...Ch. 16 - 16.37 Select a timber beam (S4S) of Southern pine...Ch. 16 - 16.38 A series of 14-ft-long Douglas fir (S4S)...Ch. 16 - 16.39 A cantilever beam 3 m long is to be made...Ch. 16 - 16.40 Select an eastern white pine (S4S) beam for...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Calculate the bending moment for the fire design in the center of 5 m span if the strip of slab is 1 m wide. The factored load for fire design is 4.2 kN/m2. Select one: O a. 13.12 kNm O b. 12.75 kNm O c. 8.14 kNm Od. 15.86 kNmarrow_forwardExercise 38 Section 7.2 (Lesson 14) МeсМovies M9.1 Abeam with a rectangular cross section is simply supported and loaded at its center with a force P. The dimensions b, h, and L are shown in the figure. Given: Find: Determine (a) the maximum bending stress omax and (b) the maximum shear stress in this beam. (c) Then obtain the ratio Tmax/omax: Hint: The maximum stresses T max occur at different locations in the beam. Be sure to show sufficient work to justify your answers. Answers: (a) 3PL/(2bh²), (b) 3P/(4bh), (c) h/(2L)arrow_forward12 ok nces Required information For the simply supported beam shown, let a=6 ft, b=4 ft, and P= 800 lb. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the shear and moment as functions of position. Enter positive or negative numbers for the internal forces based on the standard sign convention for internal forces. The shear and moment are as follows: V= Ib and M=([ lb and M=( V= b)x for 0 sxs 6 ft Ib)(10 ft-x) for 6 ft sxs 10 ftarrow_forward
- A beam is of T section, as shown in fig. The beam is simply supported over a span of 4m and carries a UDL of 4.7 kN/m run over the entire span. Determine the maximum tensile and maximum compressive stressarrow_forwardAlaminated beam is composed of three planks, each 150mm. by 60mm, glued together to form a section 150mm wide by 180mm high. The allowable shear stress in the glue is 600kPa, the allowable shear stress in the wood is gook Pa, and the allowable flexure stress in the wood is 8 MPa. Determine the maximum uniformly distributed load which can be carried by the beam on a 2-m simple span.arrow_forwardA cantilever of length 2 m carries a uniformly distributed load of 1.5 kN/m run over the whole length and a point load of 2 kN at a distance of 0.5 m from the free end. Draw the S.F. and B.M. diagrams for the cantilever beam.arrow_forward
- load of 30kN/m. The second and fourth spans remain unloaded. Calculate the bending moments at B and C and draw the S.F. and B.M. diagrams.arrow_forwardFigure 14.20 Full Alternative Text 14.21 A solid rectangular simply supported timber beam 6 in. wide, 20 in. deep, and 10 ft long carries a concentrated load of 16,000 lb at midspan. Use nominal dimensions. a. Compute the maximum horizontal shear stress at the neutral axis. b. Compute the shear stress 4 in. and 8 in. above and below the neutral axis. Neglect the weight of the beam.arrow_forwardMethod of Sectionsarrow_forward
- A beam is of T section, as shown in fig. The beam is simply supported over a span of 4m and carries a UDL of (1.7+Your Roll No.) kN/m run over the entire span. Determine the maximum tensile and maximum compressive stress.arrow_forwardcalculate the forces on PQ and QG use sections method .arrow_forwardSelect a W- shape beam limited to a maximum depth of 23.50 in for a span of 40 ft. A 50 ksi steel is used, and the beam is assumed to have full lateral bracing for its compression flange. The nominal loads are a uniform dead load of 0.45 kip/ft and a uniform live load of 0.75 kip/ft.arrow_forward
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