251-300 Flashcards
- A steel rod 3 m long has a cross-sectional area of 0.26 m2 and is stretched 0.05 cm by a load suspended from one end. The stress produced by the load is 350 kPa. Young’s modulus is 210 X 106. What is the load?
a. 91000 KN
b. 910 KN
c. 91000 Pa
d. 91000 N
e. 910 kPa
d. 91000 N
- A steel rod has a factor of safety of 4.3. The maximum stress on the rod is 5431 kPa. What is the ultimate strength of the steel rod.
a. 23353.3 Pa
b. 23353.3 kPa
c. 23353.3 MPa
d. 23353.3 N
e. 23353.3 KN
b. 23353.3 kPa
- A steel rod is 100 cm long and is subjected to a load of 10,000 N. While subjected to this load the length of the bar remains the same. Young’s modulus is 306 GPa. Determine the strain on the bar.
a. 10
b. 5
c. 1
d. 0
e. None of the above
d. 0
- A steel rod is 5 m long and has a cross-sectional area of 0.26 m2. When subjected to a load of 1421 N the rod is stretched to .0012 m. What is Young’s modulus for this material?
a. 22.772 GPa
b. 22.772 MPa
c. 22.772 kPa
d. 22.772 MN
e. 22.772 GN
b. 22.772 MPa
- If the factor of safety of a piston under a working load of 800 KN is 7.85 what is the diameter of the piston if the ultimate strength of the material is 800,000 kPa.
a. 10 cm
b. 10 mm
c. 5 cm
d. 5 mm
e. .05 m
a. 10 cm
- A 2.5 cm diameter bolt is in double shear and carries a load of 30 KN. What is the shear stress in the bolt?
a. 30,559 MPa
b. 61,115 MPa
c. 61,115 Pa
d. 30,559 kPa
e. 61,115 kPa
d. 30,559 kPa
- Hook’s Law applies to the portion in the stress-strain diagram where the curve representing the relationship between the stress and strain is:
a. Slightly curved upward
b. Linear
c. Slightly curved downward
d. Horizontal
e. Exponential
b. Linear
- According to Hooke’s Law, the stress in an elastic body is directly proportional to the strain if:
a. The yield point of the material is exceeded
b. The elastic limit of the material is exceeded
c. Young’s Modulus changes proportionally with stress
d. The factor of safety is not exceed
e. The elastic limit is not exceeded
e. The elastic limit is not exceeded
- In the equation where, E is known as the modulus of elasticity or __________.
a. the modulus of rigidity
b. the plastic modulus
c. the Wong’s modulus
d. the Young’s modulus
e. the modulus of elongation
d. the Young’s modulus
- The yield point for a material is the point at which it undergoes large deformation at a/an____________.
a. Accelerated stress
b. Maximum stress
c. Variable stress
d. Constant stress
e. Minimum stress
d. Constant stress
- The ultimate stress of a material is determined by the experimental test. The ultimate stress of a material can be calculated according to the following:
a. Multiplying the maximum load by the original cross-sectional area of the test sample.
b. Multiplying the maximum load by the new cross-sectional area of the test sample.
c. Dividing the maximum load by the new cross-sectional area of the test sample.
d. Dividing the breaking load by the new cross-sectional area of the test sample.
e. Dividing the maximum load by the original cross-sectional area of the test sample.
e. Dividing the maximum load by the original cross-sectional area of the test sample.
- If a material is stressed beyond the proportional limit and does not totally regain its original shape when the stress is removed, the amount of deformation is called:
a. Permenant strain
b. Plastic set
c. Strain
d. Maximum set
e. Permenant set
c. Strain
- A hole 12 mm in diameter is to be punched through a plate 18 mm thick. If the shear strength of the material is 300 N/mm2, find the load required on the punch.
a. 2036 N
b. 203.6 Kg
c. 203.6 Pa
d. 203.6 KN
e. None of the above
d. 203.6 KN
- The ratio of shear stress to shear strain is called the modulus rigidity (G). It is also known as the:
a. Modulus of elasticity
b. Shearing modulus of elasticity
c. Young’s modulus
d. Shear stress modulus of plasticity
e. Wong’s shear modulus
c. Young’s modulus
- A tie bar made of mild steel of tensile strength 462 MN/m2 is to carry a tensile load of 11.12 KN, find its diameter allowing a factor of safety of 12.
a. 19.17 mm
b. 19.17 cm
c. 1.917 m
d. 14.36 mm
e. 14.36 cm
a. 19.17 mm
- Safe working stress is determined by:
a. Interpolation
b. Extrapolation
c. Dividing the ultimate stress by the factor of safety
d. Multiplying the ultimate stress by the strain
e. The average value of the breaking stress determined by testing
c. Dividing the ultimate stress by the factor of safety
- A safe load of 20.31 KN is carried by a stud of 580.2 mm2 in cross-sectional area at the bottom of the thread. The bolt has a safe working stress of 35 MN/m2. Calculate the number of studs required to hold the cylinder cover of a diesel engine where the maximum pressure in the cylinder is 42 X 105 N/m2 and the diameter of the cover is 380 mm.
a. 23
b. 10
c. 20
d. 25
e. 24
e. 24
- The ratio of the output force to the input force is called the:
a. Actual mechanical advantage
b. Ideal mechanical advantage
c. Velocity ratio
d. Velocity proportion
e. Mechanical ratio
a. Actual mechanical advantage
- Velocity ratio is:
a. Ideal efficiency
b. Actual mechanical advantage
c. Actual efficiency
d. Actual mechanical ratio
e. Ideal mechanical advantage
e. Ideal mechanical advantage
- The efficiency of any apparatus is given by the ratio:
a. Input divided by output
b. Distance effort moves divided by distance load moves
c. Output divided by input
d. Load divided by effort
e. Load multiplied by effort
c. Output divided by input