(a)
The new arm radius.
(a)
Answer to Problem 8.3P
Explanation of Solution
Given:
The radius of [resent control arm is
The length of the present control arm is
The maximum control force exerted on the arm is
The least ductility provided to sufficient fracture is
Formula used:
The expression for the weight of the material is given by
Here,
The expression for the new area is given as,
Here,
The expression for arm radius is given as
Here,
Calculation:
The weight of material is calculated as,
Substitute
The new area is calculated as,
Substitute
The arm radius is calculated as,
Substitute
Conclusion:
Therefore, the new arm radius is
(b)
The weight reduction that is provided by the new steel.
(b)
Answer to Problem 8.3P
Areduction of
Explanation of Solution
Formula used:
The expression for new weight of the material is given by,
Here,
Calculation:
The weight of the material is calculated as,
Substitute
The weight saving is calculated as,
Substitute
Conclusion:
Therefore, the weight reduction that is provided by this new steel is
Want to see more full solutions like this?
Chapter 8 Solutions
Materials Science And Engineering Properties
- a specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. it has expansion of 0.16mm under a load of 80kn and the load at elastic limit is 160kn.the maximum load is 180kn.the total expansion at failure is 56mm and diameter at neck is 18mmarrow_forwardQuestion 5 A steel component is subjected to a biaxial state of stress in which both stresses are tensile in nature, and having magnitudes 100 and 60 MPa. The plane at which the resultant stress has maximum obliquity with the normal (in degrees) isarrow_forward(b) (ii) A 20 mm diameter rod made from 0.4% C steel is used to produce a steering rack. If the yield stress of the steel used is 350 MPa and a factor of safety of 2.5 is applied, what is the maximum working load that the rod can be subjected to? 2arrow_forward
- Steel, Brass, and Copper rods are connected as shown in the figure. Initially, the temperature was 15 degrees Celsius and the stress on the bars is zero. Eventually, the temperature increased to 25 degrees Celsius. Determine the total deformation on the brass. Steel Brass Copper Est = 200 GPa 12(10-)/°C apr Ebr Ecu 17(10-)/°C 120 GPa 100 GPa %3D ast = 21(10-6)/°C acu Acu = 515 mm? |Ast 200 mm2 Abr = 450 mm2 300 mm -200 mm 100 mm O -0.0109mm O 0.0241mm O -0.0241mm O 0.0109mm oooOarrow_forward(b) (i) A tensile test specimen made from 0.4% C steel has a circular cross section of diameter d mm and a gauge length of 25 mm. When a load of 4500 N is applied during the test, the gauge length of the specimen extends to 25.02 mm. If the Young's Modulus of the steel is 199 GPa, calculate the diameter of the tensile test specimen used. 4arrow_forwardCalculate 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.arrow_forward
- d) No increase Which of the following is not a property of steel materials? C a) Homogeneous and isotropic b) Linearly elastic stress-strain behavior c) Recyclable d) Fire-resistantarrow_forwardProblem 1 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Aluminum Steel A=800mm2 A=1000mm2 Bronze A=700mm? 4P 2P 500mm 600mm 700mm 1. What is the maximum value of P that will not exceed the axial stress bronze of 105 MPa? 2. What is the maximum value of P that will not exceed the axial stress in aluminum of 90 MPa? 3. If P=10KN, what is the axial force to be carried by the aluminum in KN? 4. If P is 5KN, what is the axial stress of steel?arrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning