201-250 Flashcards
- If we find the maximum load applied to a sample being subjected to an ultimate tensile strength test, and know the cross-sectional area of the sample prior to testing, we can calculate the ultimate tensile strength of the sample by:
a. Dividing the original cross-sectional area into the maximum load applied
b. Multiplying the original cross-sectional area by the maximum load applied
c. Dividing the final cross-sectional area by the maximum load applied
d. Dividing the original cross-sectional area by the maximum load applied and multiplying by the cross-sectional area found at the break
e. Dividing the original cross-sectional area into the minimum load applied
a. Dividing the original cross-sectional area into the maximum load applied
- If the factor of safety was increased on an object from 4 to 5, the load on the object operating at maximum safe working conditions:
a. Could be increased from the original
b. Could be increased providing new safety values were installed
c. Must be reduced from the original
d. Could be maintained at original
e. Must be increased to one times the value of the maximum working load
c. Must be reduced from the original
- If the operating temperature of a vessel operating at maximum allowable working stress was increased the safety factor value:
a. Remains unchanged
b. Will be increased
c. Will be reduced
d. Remains unchanged providing the vessel was fitted with high temperature alarms
e. Will be decreased to two times the original
c. Will be reduced
- The formula, strain equals change in length divided by original length, is used to find:
1. The strain in an object subjected to compressive loading 2. The strain in an object subjected to tensile loading 3. Young's modulus
a. 1
b. 2, 3
c. 1, 2
d. 1, 3
e. 1, 2, 3
c. 1, 2
- When calculating the ultimate strength of a material we should:
a. Multiple the maximum load by the original cross-sectional area
b. Multiply the maximum load by the new cross-sectional area
c. Divide the maximum load by the new cross-sectional area
d. Divide the maximum load by the original cross-sectional area
e. Multiply the maximum load by the original cross-sectional area
d. Divide the maximum load by the original cross-sectional area
- The ultimate strength of a material is:
a. Same as the elastic limit
b. Factor of safety multiplied by the safe working stress.
c. Same as the yield point
d. Inverse of Young’s modulus
e. The maximum load times the original area
b. Factor of safety multiplied by the safe working stress.
- Moment of a force is equal to force multiplied by:
a. Parallel distance to the pivot
b. Linear distance to the pivot
c. Perpendicular distance to the pivot
d. Slant distance to the pivot
e. Horizontal distance to the pivot
b. Linear distance to the pivot
- A simple beam is a beam that is supported at one end.
a. True
b. False
b. False
- A load carried by a beam will produce bending stress.
a. True
b. False
a. True
- If a beam has its supports arranged so that the beam is free to move on the supports and no additional forces occur the beam is said to be:
a. Simple supported
b. Amply supported
c. Singularly supported
d. Supported
e. Strongly supported
a. Simple supported
- When the sum of the clockwise moments equals the anti-clockwise moments then:
a. Both forces are zero
b. Equilibrium exists
c. The resultant is reduced
d. The resultant is increased
e. The equilibrium is greater than the resultant
b. Equilibrium exists
- For a lever to be in equilibrium the:
a. Clockwise moments must equal zero
b. Anti-clockwise moments must equal zero
c. Lever must have a torquing effect
d. Clockwise moments must equal the anti-clockwise moments
e. Clockwise moments must be greater than the anti-clockwise moments
d. Clockwise moments must equal the anti-clockwise moments
- The moment of a force is the turning effect:
a. Opposite to the perpendicular force
b. About a given point
c. Of the resistance to movement
d. Of a short period of time
e. On an object due to gravity
b. About a given point
- A beam supported at both ends has a concentrated load of 70 kg at a distance of 4 m from one end and 8 m from the other end. This load produces a stress of the type classed as:
1. Shearing 2. Tensile 3. Torsional 4. Bending
a. 2, 3
b. 1, 3
c. 3, 4
d. 1, 4
e. 1, 2, 3, 4
d. 1, 4
- The magnitude of the moment of a force is equal to:
a. Force times the area
b. Force times the time
c. Force times the pressure
d. Force times the perpendicular distance
e. Force times the circular distance
d. Force times the perpendicular distance
- A 6 m long cantilever beam carries a concentrated load of 45 kN at its free end. What is the shear force at the wall?
a. 270 kNm
b. 270 kN
c. 45 kN
d. 7.5 kN/m
e. 0.133 kN
b. 270 kN
- Reaction forces are those:
a. Those which support the beam
b. Clockwise movements
c. Internal forces in the beam
d. The sum of the clockwise movements
a. Those which support the beam
- A simply supported beam is 8 m long and has a concentrated load of 40 kN at 3 m from the left end support and a concentrated load of 30 kN at 2 m from the right hand end. Find the support reactions. (Neglect the mass of the beam).
a. R1 = 37.5 kN, R2 = 32.5 kN
b. R1 = 32.5 kN, R2 = 37.5 kN
c. R1 = 35 kN, R2 = 35 kN
d. R1 = 30.5 kN, R2 = 39.5 kN
e. R1 = 40 kN, R2 = 30 kN
b. R1 = 32.5 kN, R2 = 37.5 kN
- When a simply supported, horizontal beam has a load of 19 620 kN hanging from the centre of the beam, the supports on each end of the beam carry an equivalent mass of:
a. 900 kg
b. 1200 kg
c. 9810 kg
d. 100 t
e. 1000 t
e. 1000 t
- A cantilever pivoted at one end is 6 m long. The load due to the lever’s weight is 9 N, and it acts through its mid-point. At 1 m from the pivoted end a force of 298 N acts in an upward direction. To keep the system in balance, a load must be placed at the extreme end of the cantilever. What is the magnitude of that load?
a. 243 N
b. 149.79 N
c. 45.17 kg
d. 45.17 N
e. 41.75 kg
d. 45.17 N