Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 6, Problem 1CQ
To determine
The result obtained by dividing the tensile force with the original cross-sectional area of specimen.
Expert Solution & Answer
Answer to Problem 1CQ
The result obtained by dividing the tensile force with the original cross-sectional area of specimen is Engineering Normal stress.
Explanation of Solution
Consider a specimen having a cross-sectional area as
The result of this tensile force over the cross-sectional area over the specimen is known as engineering normal stress.
Thus, the result obtained by dividing the tensile force with the original cross-sectional area of specimen is Engineering Normal stress.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A 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.
Calculate the stress and strain at each force interval. Plot a graph of the stress-strain curve. Estimate the yield point of the steel and note its location on the curve. Estimate the ultimate strength of the steel and note its location on the curve.
With the help of a diagram, contrast the stress-strain relationship with respect to ceramic and metalsWith the help of a diagram, contrast the stress-strain relationship with respect to ceramic and metals
Chapter 6 Solutions
Materials Science And Engineering Properties
Ch. 6 - Prob. 1CQCh. 6 - Prob. 2CQCh. 6 - Prob. 3CQCh. 6 - Prob. 4CQCh. 6 - Prob. 5CQCh. 6 - Prob. 6CQCh. 6 - Prob. 7CQCh. 6 - Prob. 8CQCh. 6 - Prob. 9CQCh. 6 - Prob. 10CQ
Ch. 6 - Prob. 11CQCh. 6 - Prob. 12CQCh. 6 - Prob. 13CQCh. 6 - Prob. 14CQCh. 6 - Prob. 15CQCh. 6 - Prob. 16CQCh. 6 - Prob. 17CQCh. 6 - Prob. 18CQCh. 6 - Prob. 19CQCh. 6 - Prob. 20CQCh. 6 - Prob. 21CQCh. 6 - Prob. 22CQCh. 6 - Prob. 23CQCh. 6 - Prob. 24CQCh. 6 - Prob. 25CQCh. 6 - Prob. 26CQCh. 6 - Prob. 27CQCh. 6 - Prob. 28CQCh. 6 - Prob. 29CQCh. 6 - Prob. 30CQCh. 6 - Prob. 31CQCh. 6 - Prob. 32CQCh. 6 - Prob. 33CQCh. 6 - Prob. 34CQCh. 6 - Prob. 35CQCh. 6 - Prob. 36CQCh. 6 - Prob. 37CQCh. 6 - Prob. 38CQCh. 6 - Prob. 1ETSQCh. 6 - Prob. 2ETSQCh. 6 - Prob. 3ETSQCh. 6 - Prob. 4ETSQCh. 6 - Prob. 5ETSQCh. 6 - Prob. 6ETSQCh. 6 - Prob. 7ETSQCh. 6 - Prob. 8ETSQCh. 6 - Prob. 9ETSQCh. 6 - At the ultimate tensile strength. (a) The true...Ch. 6 - Prob. 11ETSQCh. 6 - Prob. 12ETSQCh. 6 - Prob. 13ETSQCh. 6 - Prob. 14ETSQCh. 6 - Prob. 15ETSQCh. 6 - Prob. 16ETSQCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Compare the engineering and true secant elastic...Ch. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - An iron specimen is plastically deformed in shear...Ch. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10PCh. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Estimate the elastic and plastic strain at the...Ch. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.1DPCh. 6 - Prob. 6.2DP
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
- Estimate the elastic and plastic strain at the ultimate tensile strength in the low-carbon steel specimen in Figure 6.16.arrow_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 tensile force P Newton is applied gradually on a prismatic bar of length L m and areas of cross- section A m². The strain energy stored in the bar will be equal toarrow_forward
- The shown figure represents the stress-strain relations of metals A and B during tension tests until fracture.Determine the following for the two metals (show all calculations and units):a. Proportional limitb. Yield stress at an offset strain of 0.002 in./in.c. Ultimate strengthd. Modulus of resiliencee. Toughnessf. Which metal is more ductile? Why?arrow_forwardTesting a round steel alloy bar with a diameter of 15 mm and a gauge length of 250 mm produced the stress–strain relationship shown in Figure Determinea. the elastic modulusb. the proportional limitc. the yield strength at a strain offset of 0.002d. the tensile strengthe. the magnitude of the load required to produce an increase in length of 0.38 mmf. the final deformation, if the specimen is unloaded after being strained by the amount specified in (e)g. In designing a typical structure made of this material, would you expect the stress applied in (e) reasonable? Why?arrow_forwardThe stress in an elastic material is: A Inversely proportional to the force acting B Inversely proportional to the materials yield strength C Proportional to the displacement D Inversely proportional to the strainarrow_forward
- Q2c) Listed in the table below is the tensile stress-strain data for different grades of steels. Utilizing the data given answer the three queries given below. Material Yield Tensile Strain at Fracture Elastic StrengthStrengthFractureStrengthModulus (MPa) (MPa) (MPa) (GPa) A 410 1440 0.63 265 410 В 200 220 0.40 105 250 C 815 950 0.25 500 610 D 800 650 0.14 720 210 E Fractures before yielding 650 550 1) Which will experience the greatest percent reduction in area? Why? 2) Which is the strongest? Why? 3) Which is the stiffest? Why?arrow_forwardThe radius of Mohr's circle of stress of a strained element is 20 N/mm² and minor principal tensile stress is 10 N/mm². What is the value of the major principle stress?arrow_forward4. A 2.0 m steel is being subjected into a tensile test. At force 30 kN, the length of the steel is now 2.2 m, but after the force has been released, the steel returned to 2.1 m. The state of the steel is at?arrow_forward
- Within elastic limit, stress is O(A) Inversely proportional to strain O(B) Directly proportional to strain OCC) Square root of strain O(D) Equal to strainarrow_forwardAn aluminum alloy [E = 74 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 = 265 mm, a cross-sectional area of A = 5300 mm², and a length of L= 4.2 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 = 57°C, the longitudinal normal strain in the plate is found to be 2920 με. 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_forwarda13 botharrow_forward
arrow_back_ios
SEE MORE QUESTIONS
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
Experimental Testing of a Real Scale Flat Slab Building for Gravity and Lateral Loading; Author: American Concrete Institute;https://www.youtube.com/watch?v=t3jybLy7ev8;License: Standard Youtube License