Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 13, Problem 8ETSQ
To determine
The option that is not an advantage of superplastic forming.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The assembly shown consists of an aluminum shell (E,= 70 GPa, a, = 23.6 × 10-6rC) fully bonded to a steel core (Es = 200 GPa, as =
11.7 x 10-6rC) and the assembly is unstressed at a temperature of 20°C. Considering only axial deformations, determine the stress in
the aluminum when the temperature reaches 215°C.
200 mm
20 mm
Aluminum shell
Steel
50 mm
core
The stress in the aluminum is
MPa.
Narrow bars of aluminum are bonded to the two sides of a thick
steel plate as shown. Initially, at T₁ = 70°F, all stresses are zero.
Knowing that the temperature will be slowly raised to T₂ and then
reduced to T₁, determine (a) the highest temperature T₂ that does
not result in residual stresses, (b) the temperature T₂ that will
result in a residual stress in the aluminum equal to 58 ksi. Assume
aa = 12.8 x 10-6/°F for the aluminum and a = 6.5 × 10-6/°F for
the steel. Further assume that the aluminum is elastoplastic with
E = 10.9 × 106 psi and ay = 58 ksi. (Hint: Neglect the small
stresses in the plate.)
Fig. P2.121
The stress-strain diagram for a bar of steel alloy is shown. Determine:
a) The modulus of Elasticity for the material
b) The proportional limit
c) The ultimate stress
Show a complete solution
Chapter 13 Solutions
Materials Science And Engineering Properties
Ch. 13 - Prob. 1CQCh. 13 - Prob. 2CQCh. 13 - Prob. 3CQCh. 13 - Prob. 4CQCh. 13 - Prob. 5CQCh. 13 - Prob. 6CQCh. 13 - Prob. 7CQCh. 13 - Prob. 8CQCh. 13 - Prob. 9CQCh. 13 - Prob. 10CQ
Ch. 13 - Prob. 11CQCh. 13 - Prob. 12CQCh. 13 - Prob. 13CQCh. 13 - Prob. 14CQCh. 13 - Prob. 15CQCh. 13 - Prob. 16CQCh. 13 - Prob. 17CQCh. 13 - Prob. 18CQCh. 13 - Prob. 19CQCh. 13 - Prob. 1ETSQCh. 13 - Prob. 2ETSQCh. 13 - Prob. 3ETSQCh. 13 - Prob. 4ETSQCh. 13 - Prob. 5ETSQCh. 13 - Prob. 6ETSQCh. 13 - Prob. 7ETSQCh. 13 - Prob. 8ETSQCh. 13 - Prob. 9ETSQCh. 13 - Prob. 10ETSQCh. 13 - Prob. 11ETSQCh. 13 - Prob. 12ETSQCh. 13 - Prob. 13ETSQCh. 13 - Prob. 14ETSQCh. 13 - Prob. 15ETSQCh. 13 - Prob. 16ETSQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0×10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 260 mm, a cross-sectional area of A = 14720 mm2, and a length of L = 5.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1680 µɛ. % D Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. L P Answer: Ad = i mmarrow_forwardGiven your understanding of what initiates and controls failure in materials, which of the following will increase the failure strength or lifetime of a test piece or component and why? a. Decreasing the difference between the maximum and minimum stress values, as this effects the stress concentration factor b. Decreasing the temperature below the brittle-ductile transition temperature, to make it harder C. Polishing to reduce surface defects Od. Increasing its volume, to give a larger cross sectional area Oe. Increasing the grain size so there are less grain boundaries to initiate failurearrow_forward1. The most important mechanical properties of brittle materials is Tensile strength compressive strength O rigidity hardness Creeparrow_forward
- -6 The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 16°C. It is known thatE = 105 GPa and a = 20.9 × 10 °C for the brass core and E = 71 GPa and a = 23.8 × 10¯6°C for the aluminum shell. Considering only axial deformations, determine the stress in the aluminum when the temperature reaches 185°C. 25 mm σα MPa -63 mmarrow_forwardAn aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.arrow_forwardA steel 0.6 inch×1.2 inch steel 90 m long is subjected to a 45 KN tensile load along its lenght.If poison's ratio is 0.3 Find: A. The deformation along its lenght. B. The deformation along its thickness. C. The defirmation along uts width. D. The lateral strain.arrow_forward
- An aluminum alloy [E = 72 GPa; v = 0.33; a= 23.0 x 10-6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 245 mm, a cross-sectional area of A = 5500 mm², and a length of L = 6.0 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by AT = 69°C, the longitudinal normal strain in the plate is found to be 3340 μc. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L P Answer: (a) P = i (b) Δd = i KN mmarrow_forwardAn aluminum alloy [E = 69 GPa; v = 0.33; a = 23.0 x 10-6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 215 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 3.9 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by AT = 53°C, the longitudinal normal strain in the plate is found to be 2320 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L Answer: (a) P = i (b) Δd = = i d KN mmarrow_forwardAn extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 500 mm. The cross-sectional dimensions of the beam are b, = 35 mm, d = 115 mm, b2 = 21 mm, d, = 21 mm, and a = 7 mm. For this material, the allowable tensile %3D %3D bending stress is 15 MPa, and the allowable compressive bending stress is 14 MPa. Determine the largest moment M that can be applied as shown to the beam. b2 立 a 不 d2 d1 A Answer: N.m M= i Submit Answer Attempts: 1 of 3 usedarrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning
Materials Science And Engineering Properties
Civil Engineering
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning