Concept explainers
(a)
The yield strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation.
The ultimate strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation.
The percentage increase in yield strength.
The percentage increase in ultimate strength.
(a)
Answer to Problem 10P
The yield strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation is
The ultimate strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation is
The percentage increase in yield strength is
The percentage increase in ultimate strength is
Explanation of Solution
Write the expression for work load factor and area.
Here, the original area is
Write the expression for true strain.
Here, the true strain is
Calculate the new yield strength after cold work.
Here, the yield strength is
Calculate the new ultimate strength after cold work.
Here, the new ultimate strength after clod work is
Calculate the percentage increase in yield strength.
Here, the percentage increase in yield strength is
Calculate the percentage increase in ultimate strength.
Here, the percentage increase in ultimate strength is
Conclusion:
Refer to Table A-22 “Results of Tensile Test of Some Metals” for hot rolled AISI 1212 steel.
Obtain the yield strength as
The steel undergoes
Substitute
Substitute
Substitute
Thus, the yield strength is
Substitute
Thus, the percentage increase in yield strength is
Substitute
Thus, the ultimate strength is
Substitute
Thus, the percentage increase in ultimate strength is
(b)
The ratio of ultimate and yield strength before cold-work operation.
The ratio of ultimate and yield strength after cold-work operation and explain the ductility of part with the help of result.
(b)
Answer to Problem 10P
The ratio of ultimate and yield strength before cold-work operation is
The ratio of ultimate and yield strength after cold-work operation is
Explanation of Solution
Calculate the ratio of ultimate strength to yield strength before cold work.
Here, the ratio of ultimate strength to yield strength is
Calculate the ratio of ultimate strength to yield strength after cold work.
Here, the ratio of ultimate strength to yield strength after cold work is
Conclusion:
Substitute
Thus, the ratio of ultimate strength to yield strength before cold work is
Substitute
Thus, the ratio of ultimate strength to yield strength after cold work is
From the calculated value of
Want to see more full solutions like this?
Chapter 2 Solutions
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
- Which R134a table is referring to? My reference as attached. The answer result should be different?arrow_forwardAnswer Qb(i) and (ii) with detailed explanationsarrow_forward1- A large slab of steel at a uniform tempering temperature of 200°C and suddenly quenching, and it surface temperature lowered to 70°C. calculate the required time when the temperature at a depth 4.0 cm has dropped to 120°C? Given: steel [k =45 W/m°C, a=1.4x10 m²/s, c=0.46 kJ/kg]arrow_forward
- can you answer this too please with free body diagram and complete solutioinarrow_forward3.2 High-strength steel is required for use in building structures and equipment (e.g., cranes). It is produced by heat treating quench-hardened steel in a process called tempering that reduces brittleness and imparts toughness. In a production facility, alloy steel plates (k = 50 W/m K, c = 460 J/kg K, and ρ = 7865 kg/m3) of thickness 3.0 cm have to be tempered in a convective oven by heating them to 550°C. If the plates are initially at 40°C and the air inside the heat treating oven is at 700°C with a convective heat transfer coefficient of 45 W/m2 K, determine how long the plate has to remain in the oven.arrow_forwardPick the best soultion for the multiple choices. Modulus of resilience is: Slope of elastic portion of stress – strain curve Area under the elastic portion of the stress –strain curve Energy absorbed during fracture in a tension test Energy absorbed during fracture in an impact test Slope of the plastic portion of the stress- strain curve ( ) Fatigue failure occurs under the condition of: High elastic stress High corrosivity High stress fluctuations High temperature High rate of loading ( ) Creep failure of a material occurs most rapidly when the operating temperature is: Cryogenic temperature Equal to its melting point Grater than 0.4 times its melting point in K Greater than 200 oF Close to boiling point ( ) Young’s modulus of a material is indicative of its: Tensile strength Yield Strength Ductility Stiffness Corrosion Resistance…arrow_forward
- How do you get this equation?arrow_forwardDesign an efficient riser system for thecasting shown in Figure 9-31. Be sure toinclude a sketch of the system, along withappropriate dimensions.arrow_forward5-16 A small hand sander for home-workshop use (see Fig. 5-43) is arranged to be ventilated by attaching a vacuum-cleaner hose to an opening in the frame of the unit. Sandpaper is attached to an oscillating pad on the bottom of the machine, and an opening between this pad and the frame of the machine serves as the hood. Take this opening to be 1.0 cm above the plane of the surface being sanded. Assume the sander makes 10 strokes/s, that each stroke is 6 mm long, and that it is desired to collect all particles of 100 um diameter or smaller. (a) Estimate the pulvation distance for a particle of 100-um diameter. Assume that the particle will travel in the plane of the sandpaper. (b) Determine the flow rate of ventilation air required. For a velocity of 10 m/s in the vacuum-cleaner hose, what diameter should the hose be? Handle Vacuum-cleaner hose 2.0 Sandpaper Frame Blood opening Oscillating pad (8 x 15 cm) FIGURE 5-43arrow_forward
- What is the final temperature when a 200-gram copper with specific heat of 0.1 cal/gm-°C is at a temperature of 80°C when it is added to a 200 gram of water at 20°C that is held in a 50-gram aluminum cup having a specific heat of 0.22 cal/gm-°C. COMPLETE FBD SOLUTION AND REQUIREMENTS PS. THIS IS A HEAT TRANSFER PROBLEMarrow_forwardCan you elaborate on answer ? Also what would be the answer for question b? I don't understand your explanation. Thank you.arrow_forward2. The dameter of a bar abricaled from an allay wlth properties showa in figure below redaced fromm an inltial 1.2 in to a 1.0 In diameler, to a final 1.8 in. diameter. For parts a, b. and when using the graph, you can ignore the marked paints A. B. C, and O 0) Determine the % CW nfter the first step of the process b) Determine the yield stress (in ksi) after the first step of the process t) Determine the % CW after the second step of the proces 22 Tensile srengin (ks 20 18 Yicld strength s) 16 14 10 Elongatie () Percent cold ssokarrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning