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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Textbook Question
Chapter 15, Problem 15.11P
For Problems 15.7 through 15.14, use the formula method.
15.11 A
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Refer to the previous problem.
Calculate the resulting maximum positive moment (kN-m).
O 77.4
O 96.8
O 54.4
O 108.8
SITUATION.
A simply supported beam has a span of 12 m. It carries a total uniformly distributed load of 21.5
kN/m. To prevent excessive deflection, a support is added at midspan. Calculate the reaction (kN) at
the added support.
O 96.75
O 161.25
80.62
O 48.38
PLS ANSWER ASAP THANKS
3. Calculate the slope and deflection at the 60-kNm couple on the structure shown in the
accompanying illustration.
Use: a. Moment Area Method,
b. Conjugate Beam Method
4 kN/m
Fixed
Hinge
60 kNm
5 m
-5 m
5 m
-5m
I=1.46 x 10° mm". E=200000MPa
Activate Windows
so to Settings to activate W ndows.
For the beam and loading shown, use discontinuity functions to compute:
(a) the deflection VA of the beam at A, and
(b) the deflection Vmidspan of the beam at midspan (i.e., x = 2.45 m).
Assume a constant value of El = 1270 kN-m² for the beam; M₁ = 9 kN-m, wo = 19.8 kN/m, LAB = 1.1 m, LBc = 2.7 m.
MA
A
Answer:
(a) VA =
(b) Vmid
i
LAB
i
Wo
B
LBC
mm.
mm.
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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- On a formatted bond paper, copy and solve the problem. Show your neat and detailed solution. Use three (3) decimal places for your answers and enclosed it in a box. Compute the deflection and slope at a section 8 ft from the wall for the beam shown in the figure using Double Integration Method. Assume that E = 28 x 103 psi and 1= 30.75 1200 lb in4. A 800 lb/ft -8 ft -8 ft Barrow_forwardA beam of uniform rectangular section 100 mm wide and 240 mm deep is simply supported at its ends. It carries a uniformly distributed load of 9.125 kN/m run over the entire span of 4 m. Estimate the deflection at the centre if E = 1.1 x 104 N/mm2.arrow_forwardFor the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at C. Assume a constant value of El = 40 x 10^6 Ib-ft^2 for the beam; wo= 6200 Ib/ft, LAB = 3.0 ft, LBC = 6 ft, LCD = 3 ft.arrow_forward
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