Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 6, Problem 6.13P
To determine
The elastic modulus, upper yield point and lower yield point for 1045 steel.
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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.
In the figure shown below, determine:
1) The final temperature if the normal stress at
aluminium is Oal =
= -90 MPa and the initial
temperature 20°C.
2)
The final length of the aluminium member.
Aluminum
Bronze
A=1800mm2
A=1500mm2
E=105GPA
E=73GPA
a=23.2x10-6/°C
a=21.6x10-6/°C
Gap=0.5mm
0.35m
0.45m
A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find . The stress at elastic limit. • Young's modulus. . Percentage elongation. Percentage reduction in area. Ultimate tensile stress. .
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
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- 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_forwardIn the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is Og = -90 MPa and the initial temperature %3D 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? Bronze A=1500mm? E=73GPA E=105GPA a=23.2x10-6/°C a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45marrow_forwardThe following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300 MPa and ɛ = 0.003054 for o = 600 MPa. (a) Draw the stress-strain diagram and calculate the E for this alloy.arrow_forward
- 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_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_forwardA 19-mm reinforcing steel bar and a gauge length of 75 mm was subjected to ten- sion, with the results shown in Table P3.27. Using a computer spreadsheet pro- gram, plot the stress-strain relationship. From the graph, determine the Young's modulus of the steel and the deformation corresponding to a 150-kN load. TABLE P3.27 Load, kN Deformation, mm 54 0.084 163 0.168 284 0.336 330 1.428 366 3.360arrow_forward
- Question No.2 Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness I. Which metal is more ductile? Why? 900 -Metal A E 600 Metal B 300 0.00 a02 004 a.06 0.08 0.10 0.12 014 Strain, matm FIGURE P1.16 Strees, MPaarrow_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_forwardAn extruded polymer bear is subjected to a bending moment M. The length of the bear is L = 500 mm. The cross-sectional dimensions of the bear are b₁ = 39 mm, d₁ = 95 mm, b₂ = 23 mm, d₂ = 23 mm, and a = 8 mm. For this material, the allowable tensile bending stress is 18 MPa, and the allowable compressive bending stress is 9 MPa. Determine the largest moment M that can be applied as shown to the beam. M A Answer: M = L B N-m ↓ a d₂ ➡AK b₂ b₁ yakarrow_forward
- Sketch the stress–strain behavior of steel, and identify different levels of strength. What is a typical value for yield strength of mild steel? What is theeffect of increasing the carbon content in steel on each of the each of thefollowing items?a. Yield strengthb. Modulus of elasticityc. Ductilityarrow_forwardShow Stress/ strain curve for nitinol and steel (2 separate charts please)arrow_forwardThe following figure shows the tensile stress-strain curve for a plain-carbon steel. 600 80 500 MPa 600 10 psi 60 80 400 400 60- 300 40 40 200 200 20 20 100 0.005 0.05 0.10 0.15 Strain (a) What is this alloy's tensile strength? MPа (b) What is its modulus of elasticity? GPa (c) What is the yield strength? i MPa Stress (MPa) Stress (10 psi)arrow_forward
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