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 16.10P
To determine
To select the timber beam southern pine S4S for the given figure.
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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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- 1) In each of the following scenarios, based on the plane of impact (shown with an (n, t)) and the motion of mass 1, draw the direction of motion of mass 2 after the impact. Note that in all scenarios, mass 2 is initially at rest. What can you say about the nature of the motion of mass 2 regardless of the scenario? m1 15 <+ m2 2) y "L χ m1 m2 m1 בז m2 Farrow_forward8. In the following check to see if the set S is a vector subspace of the corresponding Rn. If it is not, explain why not. If it is, then find a basis and the dimension. X1 (a) S = X2 {[2], n ≤ n } c X1 X2 CR² X1 (b) S X2 = X3 X4 x1 + x2 x3 = 0arrow_forward2) Suppose that two unequal masses m₁ and m₂ are moving with initial velocities V₁ and V₂, respectively. The masses hit each other and have a coefficient of restitution e. After the impact, mass 1 and 2 head to their respective gaps at angles a and ẞ, respectively. Derive expressions for each of the angles in terms of the initial velocities and the coefficient of restitution. m1 m2 8 m1 ↑ บา m2 ñ Вarrow_forward
- The fallowing question is from a reeds book on applied heat i am studying. Although the answer is provided, im struggling to understand the whole answer and the formulas and the steps theyre using. Also where some ov the values such as Hg and Hf come from in part i for example. Please explain step per step in detail thanks In an NH, refrigerator, the ammonia leaves the evaporatorand enters the cornpressor as dry saturated vapour at 2.68 bar,it leaves the compressor and enters the condenser at 8.57 bar with50" of superheat. it is condensed at constant pressure and leavesthe condenser as saturated liquid. If the rate of flow of the refrigerantthrough the circuit is 0.45 kglmin calculate (i) the compressorpower, (ii) the heat rejected to the condenser cooling water in kJ/s,an (iii) the refrigerating effect in kJ/s. From tables page 12, NH,:2.68 bar, hg= 1430.58.57 bar, hf = 275.1 h supht 50" = 1597.2Mass flow of refrigerant--- - - 0.0075 kgls 60Enthalpy gain per kg of refrigerant in…arrow_forwardstate the formulas for calculating work done by gasarrow_forwardExercises Find the solution of the following Differential Equations 1) y" + y = 3x² 3) "+2y+3y=27x 5) y"+y=6sin(x) 7) y"+4y+4y = 18 cosh(x) 9) (4)-5y"+4y = 10 cos(x) 11) y"+y=x²+x 13) y"-2y+y=e* 15) y+2y"-y'-2y=1-4x³ 2) y"+2y' + y = x² 4) "+y=-30 sin(4x) 6) y"+4y+3y=sin(x)+2 cos(x) 8) y"-2y+2y= 2e* cos(x) 10) y+y-2y=3e* 12) y"-y=e* 14) y"+y+y=x+4x³ +12x² 16) y"-2y+2y=2e* cos(x)arrow_forward
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