### Problem Statement:Compound axial member ABC has a uniform diameter of \( d = 1.45 \, \text{in} \). Segment (1) is an aluminum \([E_1 = 10{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_1 = 85 \, \text{in} \). Segment (2) is a copper \([E_2 = 17{,}000 \, \text{ksi}]\) alloy rod with a length of \( L_2 = 121 \, \text{in} \). When an axial force \( P \) is applied, a strain gage attached to copper segment (2) measures a normal strain of \( \varepsilon_2 = 1800 \, \mu\text{in.} / \text{in.} \) in the longitudinal direction. What is the total elongation of member ABC?### Diagram Explanation:The diagram provided shows a compound member ABC, composed of two segments:- Segment (1) is the aluminum rod, labeled \( L_1 \), starting from point \( A \) and ending at point \( B \).- Segment (2) is the copper rod, labeled \( L_2 \), starting from point \( B \) and ending at point \( C \).Both segments have the same diameter \( d \) and are connected in series. An axial force \( P \) is applied, causing the entire member ABC to elongate.### Required Calculation:To find the total elongation \( \delta_{ABC} \):\[ \delta_{ABC} = \text{Total elongation of member ABC} \]The problem also provides buttons for additional study resources like eTextbook and Media, and a GO Tutorial.Additionally, there are options to save your progress and to submit the final answer, with a note indicating `Attempts: 0 of 5 used`.### Elongation Calculation:The total elongation \( \delta_{ABC} \) can be broken down as the sum of the elongations of each segment due to the applied force \( P \).#### Elongation of Segment (1) - Aluminum:\[ \delta_1 = \frac{P \cdot L_1}{A \cdot E_1} \]#### Elongation of Segment (2) - Copper:Given that the

Answered: Image /qna-images/answer/5b217381-3a2f-433f-862d-1334b4c9038b.jpg

Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a ength of L₁= 85 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 pin./in. in the longitudinal direction. What is the total elongation of member ABC? LA Aluminum B L2 Copper CP

Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a ength of L₁= 85 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 pin./in. in the longitudinal direction. What is the total elongation of member ABC? LA Aluminum B L2 Copper CP

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

A 45° strain gauge rosette placed on a stainless steel structure results in the following values being determined for the maximum and minimum principal strains: €1 = 1,625.9 ×10–6 and €2 = -753.4 x10-6. Assuming plane stess, determine the factor of safety against the Tresca criterion. For stainless steel, use Young's modulus E = 196 GPa, Poisson's ratio v = 0.29 and yield stress oy = 1120 M Pa. Give your answer to 2 decimal places.
5. Consider a cylindrical specimen of a steel alloy (Figure 6.22) 10 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied. Stress (MPa) 600 500 400 300 200 100 0 0.00 0.04 Stress (MPa) 500 400 300 200 100 0.000 0.002 0.004 0.006 0.08 Strain Strain 0.12 0.16 0.20 Figure 6.22 Tensile stress-strain behavior for a steel alloy.
A bar BD is held in a horizontal position when there is no load by the steel wire AB. When a weight Wis added at C, also hanging from a steel wire CE, then point C is displace down by 0.025in. The cross-sectional area of the wires is 0.002in². The wires are made of a steel alloy with E = 29000ksi and σy = 70ksi. Calculate the strain in both wires, and the weight W. B A N E Parameter ft L_1 4 L_2 1 L_3 2 L_4 4 The strain in AB is € AB = The strain in CE is ECE = The weight W = 30 W cc 080 BY NO SA 2021 Cathy Zupke Xlb D in./in. in./in.
Learning Goal: A material is homogenous and isotropic when it has a modulus of elasticity that does not vary with direction and it has uniform properties throughout. When a point in a material is subjected to only normal stresses in three dimensions, but they are not necessarily equal, the state is called triaxial stress. When the response of the material is linear elastic, the strains along the three axes can be calculated using a generalized form of Hooke's law. Each strain depends on all three stresses because of the Poisson effect. €1 Figure 1 [0₂ - V (O₂ +0₂)] a 1 of 1 A cube is subjected to stresses with magnitudes a = 16 MPa, b = 26 MPa, and c = 5 MPa (Figure 1). What is the strain in the y-direction? Let E = 200 GPa and v= 0.26. Express your answer to three significant figures. ► View Available Hint(s) Submit VE ΑΣΦ Part B - Change in volume f change in volume = vec The cube from Part A originally has side length 3.2 cm . What is the change in volume of the cube under the given…
Incorrect A thin square plate PQRS is symmetrically deformed into the shape shown by the dashed lines in the figure. Assume d = 255 mm, d₁=256.1 mm, and d₂ = 253.6 mm. For the deformed plate, determine (a) the normal strain of diagonal QS. (b) the shear strain Yxy at corner P. d₂ 12 d d₁ Undeformed Answer: dos = i 9810.3 R Calculate the deformation of diagonal QS. Deformed X mm
45-degree rosette *Please answer this analytically!*
The surface stresses of a pure aluminum device (E = 70 GP a, v = 0.33) that were tested on a space shuttle were obtained with stress sensors. The sensors were oriented as shown in Figure 3 and the unit strains obtained were εa = 1100 × 10−6, εb = 1496 × 10−6, and εc = −39.44 × 10−6.(a) What is the stress σxx in the x direction? Use Mohr's circle.
Rigid bar ABCD is loaded and supported. Steel [E=30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.625 in.2 and bar (2) has a cross-sectional area of 1.25 in. 2. After load P is applied, the strain in bar (2) is found to be 900 με. If there is a clearance of 0.05 in. in the pin connection at C, determine: (a) the stresses in bars (1) and (2).(b) the vertical deflection of point D.(c) the load P.
An initially rectangular element of a material is deformed into the shape shown in the figure. Deformed Undeformed 16.6 0.1992 mm 15.7 0.2 mm Normal strain in the x- direction * -4.0 x 10^-3 4.0 x 10^-3 10 x 10^-3 -10 x 10^-3 Normal strain in the y- direction O -4.0 x 10^-3 4.0 x 10^-3 O 10 x 10^-3 O -10 x 10^-3 0.1515 mm 0.15 mm