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 15, Problem 15.16P
For Problems 15.15 through 15.26, use the moment-area method.
15.16 A
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A simply supported beam 10 m long has an overhang of 1.1 m at the left support. If a highway uniform load of 12.02 kN/m and a concentrated load of 194 kN, passes thru the beam, compute the maximum positive shear (kN) based on influence line for maximum shear at mid span.
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Compute the initial deflection of the beam at midspan under service loads with the following specifications: f'c = 4000 psi, 36-inch height, depth of rebar assumed to be 3 inches less than the height, 16-inch width, 4 #9 bars (tension), Grade 60 rebar, 30' clear spans, service loads of: DL = 0.25k/ft, LL = 1.2k/ft.
The DL does NOT include self-weight of the beam or of the precast concrete deck planks that have a weight of 60 PSF. The beam picks up a tributary width of 12 feet. Also, note that this beam is continuous and is the middle beam of 5 equal spans.
Check the initial deflections against the ACI deflection requirements. Then calculate the long-term deflections and check those against the ACI requirements. For both situations, assume that finish materials will be attached to the beam.
Last: Instead of performing a structural analysis to determine the maximum deflection in the beam, conservatively figure that the maximum deflection will be 60% of what it would have been for a…
Chapter 15 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 15 - A 14 in.-diameter aluminum rod is bent into a...Ch. 15 - 15.2 Calculate the maximum bending stress produced...Ch. 15 - A 500 -mm-long steel bar having a cross section of...Ch. 15 - 15.4 An aluminum wire has a diameter of in....Ch. 15 - 15.5 A -in.-wide by in.-thick board is bent to a...Ch. 15 - 15.6 A Douglas fir beam is in. wide and in. deep....Ch. 15 - Prob. 15.7PCh. 15 - For Problems 15.7 through 15.14, use the formula...Ch. 15 - For Problems 15.7 through 15.14, use the formula...Ch. 15 - For Problems 15.7 through 15.14, use the formula...
Ch. 15 - For Problems 15.7 through 15.14, use the formula...Ch. 15 - For Problems 15.7 through 15.I4, use the formula...Ch. 15 - For Problems 15.7 through 15.14, use the formula...Ch. 15 - For Problems 15.7 through 15.14, use the formula...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - For Problems 15.15 through 15.26, use the...Ch. 15 - 15.27 Draw the moment diagram by parts for the...Ch. 15 - 15.28 Draw the moment diagram by parts for the...Ch. 15 - 15.29 Draw the moment diagram by parts for the...Ch. 15 - 15.30 For the beam shown, draw the conventional...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - For Problems 15.31 through 15.43, use the...Ch. 15 - 15.49 If the elastic limit of a steel wire is...Ch. 15 - 15.50 Calculate the bending moment required to...Ch. 15 - 15.51 A 6-ft-long cantilever beam is subjected to...Ch. 15 - 15.52 A structural steel wide-flange section is...Ch. 15 - 15.53 A simply supported structural steel...Ch. 15 - 15.54 A structural steel wide-flange shape is...Ch. 15 - A solid, round simply supported steel shaft is...Ch. 15 - Using the moment-area method, check the...Ch. 15 - 15.57 A 1-in.-diameter steel bar is 25 ft long and...Ch. 15 - 15.58 A 102-mm nominal diameter standard-weight...Ch. 15 - I 5.59 Compute the maximum deflection for the...Ch. 15 - An 8-in-wide by 12-in-deep redwood timber beam...Ch. 15 - 15.61 A solid steel shaft 3 in. in diameter and 20...Ch. 15 - 15.62 For the beam shown, draw the conventional...Ch. 15 - 15.63 Rework Problem 15.62 with concentrated loads...Ch. 15 - 15.64 A solid steel shaft 3 in. in diameter and 20...Ch. 15 - 15.65 A structural steel wide-flange section is...Ch. 15 - 15.66 A 6-in.-by-10-in, hem-fir timber beam (S4S)...Ch. 15 - 15.67 A simply supported structural steel...Ch. 15 - Calculate the maximum permissible span length for...Ch. 15 - 15.69 A structural steel wide-flange section 10 ft...Ch. 15 - 15.70 A structural steel wide-flange section...Ch. 15 - 15.71 Determine the deflection at point C and...Ch. 15 - 15.72 Calculate the deflection midway between the...Ch. 15 - 15.73 Derive an expression for the maximum...Ch. 15 - 15.74 Derive an expression for the maximum...
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- A timber beam is 120 mm wide and 200 mm deep and is used on a span of 4 metres. The simply supported beam carries a uniformly distributed load of 2.8 kN/m run over the entire length. Calculate the maximum bending stress induced.arrow_forwardDraw the SFD and BMD for the beams loaded as shown in Calculate the position and magnitude of maximum bending moment and locate points of contra-flexures if any.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
- Draw shear force digam and bending moment digram for the following beam 2.0 kips / ft A B. 15.0 ft - 6.0 ft-arrow_forwardSelect the lightest-weight wide-flange beam with the shortest depth from Appendix B that will safely support the loading shown. The allowable bending stress is sallow = 24 ksi and the allowable shear stress of tallow = 14 ksi.arrow_forward14.1 A simple supported beam AB has rectangular cross section where h=60 mm, b=40 mm. The length of the beam and load are as shown. Find the max moment on the beam? where is it? calculate the max compressive / tensile stress of the beamarrow_forward
- Question is for review, not for gradearrow_forwardCalculate the slope at C using ONE of these methods: double integration method, area-moment and conjugate beam method. Also, determine the deflection at C using EITHER virtual work method or Castigliano theorem method. Set P = 10 kN, w = 2 kN/m, support A is pin and support B is roller. ... 1 marrow_forwardQ8b. The beam section below is solid and homogenous and has the following dimensions: b, = 65 mm (top flange width), b2 = 65 mm (bottom flange width) and bz = 25 mm (web width). d;=175 mm (section depth), d2=25 mm (top flange depth) and d3= 30 mm (bottom flange depth). For the solid homogeneous beam section shown below, determine its centroid coordinate y relative to the origin (0, 0) of the x-y axes. Give your answer for y in millimetres (mm) to two decimal places. b1 d2 ba dil b2arrow_forward
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Solids: Lesson 53 - Slope and Deflection of Beams Intro; Author: Jeff Hanson;https://www.youtube.com/watch?v=I7lTq68JRmY;License: Standard YouTube License, CC-BY