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- ✓ Compute the interaction formula. (Express your answer to three significant figures.) -Select- 1.0 This member is -Select-
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- ✓ Compute the interaction formula. (Express your answer to three significant figures.) -Select- 1.0 This member is -Select-
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter10: Plate Girders
Section: Chapter Questions
Problem 10.7.9P
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Transcribed Image Text: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- ✓
Compute the interaction formula.
(Express your answer to three significant figures.)
-Select- 1.0
This member is -Select-
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