Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E= 200 GPa. It has a diameter of 100 mm. Prob 12-21
Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E= 200 GPa. It has a diameter of 100 mm. Prob 12-21
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter11: Columns
Section: Chapter Questions
Problem 11.9.13P: A steel pipe column with pinned ends supports an axial load P = 21 kips. The pipe has outside and...
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Transcribed Image Text:Figure 12-20 Full Alternative Text
12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of
steel having E = 200 GPa. It has a diameter of 100 mm.
Prob. 12-21
6 kN.m
A
-X
1.5 m
8 kN
с
1.5 m
B
6 kN.m
![← Mechanics of Materials 11th Edition
C₂ = 0
Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (4).
EI = 2x² + 6x - 13.5
du
(2x² + 6x - 13.5)
Calculate the Equation of the elastic curve u
Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (5).
Elu = 2+3x²+(-13.5) x +0
Elu = 2² + 3x² - 13.5x
(+3x² - 13.5x)
Calculate the moment of inertia (I) using the formula:
I = d+ (7)
Substitute 100 mm for d in Equation (7).
4
=
U =
¹001) × 2 = 1
mm x
= 4.9087 x 10-6 m4
Umax =
103 m
1mm
200 GPax-
(6)
Calculate the maximum deflection of the solid circular shaft (Umax):
Substitute 1.5 m for x, 4.9087 x 10-6 m² for I, and 200 GPa for E in Equation (6).
[2015+3(1.5)²-13.5(1.5)]
106 kN/m²
1 GPa
= -0.011456 mx
Chapter 12.2, Problem 21P
-x4.9087x10-6
10¹ mm
1m
= -11.5 mm
Hence, the maximum deflection of the solid circular shaft (Umax) is -11.5 mm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2F30f64566-053a-4ffa-a6c1-df789c51de54%2Foftf4ck_processed.png&w=3840&q=75)
Transcribed Image Text:← Mechanics of Materials 11th Edition
C₂ = 0
Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (4).
EI = 2x² + 6x - 13.5
du
(2x² + 6x - 13.5)
Calculate the Equation of the elastic curve u
Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (5).
Elu = 2+3x²+(-13.5) x +0
Elu = 2² + 3x² - 13.5x
(+3x² - 13.5x)
Calculate the moment of inertia (I) using the formula:
I = d+ (7)
Substitute 100 mm for d in Equation (7).
4
=
U =
¹001) × 2 = 1
mm x
= 4.9087 x 10-6 m4
Umax =
103 m
1mm
200 GPax-
(6)
Calculate the maximum deflection of the solid circular shaft (Umax):
Substitute 1.5 m for x, 4.9087 x 10-6 m² for I, and 200 GPa for E in Equation (6).
[2015+3(1.5)²-13.5(1.5)]
106 kN/m²
1 GPa
= -0.011456 mx
Chapter 12.2, Problem 21P
-x4.9087x10-6
10¹ mm
1m
= -11.5 mm
Hence, the maximum deflection of the solid circular shaft (Umax) is -11.5 mm.
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