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 that P40 k and M = 68 ft-k. For W10 × 33: I = 171 in.4; 10 ft and C₁ = 1.0: Mn = 134 ft-kips and Mn/₁ = 89.3 ft-kips; = 50 ksi. Suppose for Lb for Lc - = 10 ft: Pn 330 kips and Pr/c = 220 kips. W10 X 33 P M M 10' Pu A) + Ферп 9 a. Use LRFD. Select the interaction formula: Mur 84, Mnz Muy + <1.0 Фь Мпу Pu Mur Muy B) + + ≤ 1.0 20c Pn Фь Мих nx ФоМпу -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: Ра 8 Max May A) + + <1.0 Pn/c Mnx/b Mny/b 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 that P40 k and M = 68 ft-k. For W10 × 33: I = 171 in.4; 10 ft and C₁ = 1.0: Mn = 134 ft-kips and Mn/₁ = 89.3 ft-kips; = 50 ksi. Suppose for Lb for Lc - = 10 ft: Pn 330 kips and Pr/c = 220 kips. W10 X 33 P M M 10' Pu A) + Ферп 9 a. Use LRFD. Select the interaction formula: Mur 84, Mnz Muy + <1.0 Фь Мпу Pu Mur Muy B) + + ≤ 1.0 20c Pn Фь Мих nx ФоМпу -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: Ра 8 Max May A) + + <1.0 Pn/c Mnx/b Mny/b 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
Related questions
Question
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
that P40 k and M = 68 ft-k.
For W10 × 33: I = 171 in.4;
10 ft and C₁ = 1.0: Mn = 134 ft-kips and Mn/₁ = 89.3 ft-kips;
=
50 ksi. Suppose
for Lb
for Lc
-
= 10 ft: Pn 330 kips and Pr/c
=
220 kips.
W10 X 33
P
M
M
10'
Pu
A)
+
Ферп 9
a. Use LRFD.
Select the interaction formula:
Mur
84, Mnz
Muy
+
<1.0
Фь Мпу
Pu
Mur
Muy
B)
+
+
≤ 1.0
20c Pn
Фь Мих
nx
ФоМпу
-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:
Ра
8
Max
May
A)
+
+
<1.0
Pn/c
Mnx/b Mny/b
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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