A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has a diameter d 1 = 100 mm and segment BC has a diameter d 2 = 60 mm. Both segments have a length L / 2 = 0.6 m. A longitudinal hole of diameter d is drilled through segment AB in the middle of its length (distance L / 4 = 0.3 m). The bar is made of plastic with a modulus of elasticity. E = 4.0 GPa. Compressive loads P = 110 kN act at the ends of the bar. (a) If the shortening of the bar is limited to 8.0 mm, what is the maximum allowable diameter d max of the hole? (See figure part a.) (b) Now, if d max is set to d 2/2, at what distance b from the end C must the load P be applied to limit the shortening of the bar to 8.0 mm? (See Figure part b.) (C) Finally, if the loads P are applied at the ends and d max = d 2/2, what is the allowable length x of the hole if shortening is limited to 8.0 mm? (See figure part c.)

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has a diameter d 1 = 100 mm and segment BC has a diameter d 2 = 60 mm. Both segments have a length L / 2 = 0.6 m. A longitudinal hole of diameter d is drilled through segment AB in the middle of its length (distance L / 4 = 0.3 m). The bar is made of plastic with a modulus of elasticity. E = 4.0 GPa. Compressive loads P = 110 kN act at the ends of the bar. (a) If the shortening of the bar is limited to 8.0 mm, what is the maximum allowable diameter d max of the hole? (See figure part a.) (b) Now, if d max is set to d 2/2, at what distance b from the end C must the load P be applied to limit the shortening of the bar to 8.0 mm? (See Figure part b.) (C) Finally, if the loads P are applied at the ends and d max = d 2/2, what is the allowable length x of the hole if shortening is limited to 8.0 mm? (See figure part c.)

dmax
B
dz
P
P
2
(c)
Transcribed Image Text:dmax B dz P P 2 (c)
dmax
B
dz
C
P
P
titil
(а)
d2
dmax =
B
dz
P
P
b
titil
(b)
Transcribed Image Text:dmax B dz C P P titil (а) d2 dmax = B dz P P b titil (b)
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