A straight and piecewise uniform bar ABCD of total length L is in Case 1 fixed at A in case 1 and in Case 2 fixed at both ends A and D. AB = BC = CD = L/3. The cross section area of segment CD is half of the area of the segment ABC, i.e. AABC = ACB. A point load of P is applied at B. Assume that bar material is linear and elastic and the modulus of elasticity is E. ВCDВCP-LL-L3Figure 1: Case 1: bar is fixed at AFigure 2: Case 2: bar is fixed at A and DL = 90cm, E = 70GPA, AcD= 4 cm? , P = 10KNFor both Cases:a) Find the average normal stress on a perpendicular cross-section along the bar in termsof the plot:Average normal stressx Position of theAВcross sectionb) Find the new length of segment ABC after deformation(Note: Case 2 is a statically indeterminate case)1/3

considering case 1applying static equlibirium conditions∑Fx=0−R+10=0R=10 KNconsidering section…

Answered: A straight and piecewise uniform bar…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A straight and piecewise uniform bar ABCD of total length L is in Case 1 fixed at A in case 1 and in Case 2 fixed at both ends A and D. AB = BC = CD = L/3. The cross section area of segment CD is half of the area of the segment ABC, i.e. A_ABC = A_CB. A point load of P is applied at B. Assume that bar material is linear and elastic and the modulus of elasticity is E.

В
C
D
В
C
P
-L
L
-L
3
Figure 1: Case 1: bar is fixed at A
Figure 2: Case 2: bar is fixed at A and D
L = 90cm, E = 70GPA, AcD= 4 cm? , P = 10KN
For both Cases:
a) Find the average normal stress on a perpendicular cross-section along the bar in terms
of the plot:
Average normal stress
x Position of the
A
В
cross section
b) Find the new length of segment ABC after deformation
(Note: Case 2 is a statically indeterminate case)
1/3

Expert Solution

Step 1

$\begin{array}{l} considering case 1 \\ applying static equlibirium conditions \\ \sum F_{x} = 0 \\ - R + 10 = 0 \\ R = 10 K N \\ c o n s i d e r i n g \sec t i o n A B \\ σ_{A B} = \frac{F_{A B}}{A_{A B}} \\ = \frac{10 \times 10^{3}}{8 \times 10^{2}} \\ = 12.5 \frac{N}{m m^{2}} \\ there is no force acting beyond point B \\ σ_{B C} = σ_{C D} = 0 \\ (2) \\ New length of segment ABC after deformation \\ δ_{A B C} = δ_{A B} + δ_{B C} \\ = \frac{σ_{A B} . L_{A B}}{E} + \frac{σ_{B C} . L_{B C}}{E} \\ = \frac{12.5 \times 300}{70 \times 10^{3}} + 0 \\ δ_{A B C} = 0.0536 m m \\ L'_{A B C} = 600 + 0.0536 \\ L'_{A B C} = 600.0536 m m \end{array}$

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