Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357112311
Author: Saeed Moaveni
Publisher: Cengage Learning US
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Chapter 13, Problem 31P
To determine
Calculate the electricity generated by a photovoltaic system.
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(5) The simply supported beam shown in figure supports uniformly distributed characteristicdead and imposed loads of 5 kNm-1 each, as well as a characteristic imposed point load of 30 kN at mid-span. Assuming the beam is fully laterally restrained and there is nominaltorsional restrain at supports, select a suitable UB section in S275 steel to satisfy bendingand shear considerations. Also, check the selected UB section against the deflection. (40 Marks)
A simply supported beam is subjected to the end couples (bending is about the strong axis) and the axial load shown
in the figure below. These moments and axial load are from service loads and consist of equal parts dead load and
live load. Lateral support is provided only at the ends. Neglect the weight of the beam and investigate this member as
a beam-column. Use Fy = 50 ksi. Suppose that P = 40 k and M = 68 ft-k.
For W10 x 33: Ix
=
=
171 in.4;
for L = 10 ft and C
for Lc = 10 ft:
= 1.0: фь Мп
=
134 ft-kips and Mn/b
=
89.3 ft-kips;
Pn = 330 kips and Pr/c = 220 kips.
W10 X 33
P
M
M
10'
a. Use LRFD.
Select the interaction formula:
A)
Pu 8
+
Mur
Muy
+
<1.0
Ферп 9
Фь Мих
Pu
Mux
Muy
B)
+
+
20c Pn
Фь Мих
ФоМпу
.)
<1.0
-Select-
Compute the interaction formula.
(Express your answer to three significant figures.)
-Select- 1.0
This member is -Select-
b. Use ASD.
Select the interaction formula:
Pa
A)
+
Pn/Sc
Max
Mnx/b
May
+
< 1.0
Mny
1/526
Pa
Max
May
B)
+
+
<1.0
2Pn/c Mnx/b
Mny/b
-Select- ✓…
5. Use the graph and data table below to determine:
strain(in/in)
0
Stress (psi)
30,000
25,000
20,00
15,000
stress (psi)
0
2,500
0.00025
5,000
0.0005
7,500
0.00075
9,600
0.00096
11,170
0.001117
13,500
0.00135
10,000
16,612
0.001875
15,430
0.0025
5,000
22,350
0.00312
26,800
0.0042
25,810
0.00472
Stress Strain Curve
☑
0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
0.004
0.0045 0.005
Strain (in/in)
a) Proportional limits: (stress in units of psi, strain in units of in/in)
b) Modulus of elasticity (units: ksi)
c) Ultimate strength (units: psi)
d) Rupture strength (units: psi)
Chapter 13 Solutions
Engineering Fundamentals: An Introduction to Engineering
Ch. 13.2 - Prob. 1BYGCh. 13.2 - Prob. 2BYGCh. 13.2 - Prob. 3BYGCh. 13.2 - Prob. 4BYGCh. 13.2 - Prob. 5BYGCh. 13.2 - Prob. BYGVCh. 13.4 - Prob. 1BYGCh. 13.4 - Prob. 2BYGCh. 13.4 - Prob. 3BYGCh. 13.4 - Prob. 4BYG
Ch. 13.4 - Prob. BYGVCh. 13.5 - Prob. 1BYGCh. 13.5 - Prob. 2BYGCh. 13.5 - Prob. 3BYGCh. 13.5 - Prob. 4BYGCh. 13.5 - Prob. 5BYGCh. 13.5 - Prob. BYGVCh. 13 - Prob. 1PCh. 13 - Prob. 2PCh. 13 - An elevator has a rated capacity of 2200 lb. It...Ch. 13 - Prob. 4PCh. 13 - Prob. 5PCh. 13 - Prob. 6PCh. 13 - Prob. 7PCh. 13 - Prob. 8PCh. 13 - Prob. 9PCh. 13 - Prob. 10PCh. 13 - Prob. 12PCh. 13 - Prob. 14PCh. 13 - Prob. 15PCh. 13 - Prob. 16PCh. 13 - Prob. 17PCh. 13 - Prob. 18PCh. 13 - Prob. 19PCh. 13 - Prob. 20PCh. 13 - Prob. 22PCh. 13 - Prob. 23PCh. 13 - Prob. 24PCh. 13 - Prob. 25PCh. 13 - Prob. 26PCh. 13 - Prob. 28PCh. 13 - Prob. 29PCh. 13 - Prob. 30PCh. 13 - Prob. 31PCh. 13 - Prob. 32PCh. 13 - Prob. 33PCh. 13 - Prob. 34PCh. 13 - Prob. 35PCh. 13 - Prob. 37PCh. 13 - Prob. 38P
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