Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve. does not hold for compression test
Question Completion Status:
The point of application of the total pressure on surface is
O a. None of the above
O b. Centroid of the surface
O c. Centre of pressure
Answer: c. Centre of pressure
Explanation:
Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.
The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.
The answer is all income (&) expenses
Answer:
NICE
Explanation:
What is "Engineering"?
what is an unit resistance
Answer: Ohm Ω
Explanation:
The resistance can be defined as the hindrance offered to the flow of current. Lower the current the higher is the resistance. The SI unit of resistance is ohm Ω. When one ampere current flows from an object or wire the application of one volt of potential difference on it the resistance so produce is one ohm. Resistance can be achieved by the application of the certain gadgets in the circuit which can prevent the flow of excess current from devices and can prevent shock circuit.
Draw a FBD of the beam with reactions at A & B. A is a pin, B is a roller. Try to guess intuitively which way the vertical components of A & B are pointing. Do not show the 6 kN forces in your FBD. Only show the couple moment, or pure moment
Answer:
kindly check the drawing of the FBD of the beam with reactions at A & B. A is a pin, B is a roller in the attached picture.
Explanation:
Without further ado, let's dive straight into the solution to the question above. From the diagram of the FBD of the beam with reactions at A & B it can be shown that the reaction moment is anticlockwise while the moment is clockwise.
The system is at equilibrium and the it does not matter where you place the couple (pure) moment.
The distance from A to C can either be equal or not. If AY = 2.15 kN and M = 25.8. Then, the distance between A and B = 25.8/2.15 = 12m.
Convert the following measurement to their unit equivalents.
14.5 ft =___inch
15. 60 inch=___feet
16. 10 inch = __centimeter
17. 5 meter =__decimeter
18. 8 meter = ___feet
Explanation:
(1) 174 inch
(2)1 foot 3.6 inches
(3) 39.649 centimetre
(4)175 decimeter
(5)26.25 feet
The point of contact of two pitch circles of mating gears is called?
Answer:
the pitch point im pritty sure what is is
The point of contact of two pitch circles of mating gears is called the pitch point.
What is the pitch point?
A pitch point is where two pitch circles frequently come into touch. A common point of contact between two pitch circles of mating gears is called a pitch point. The pitch circle and is centered on the top of the teeth. Dedendum circle, also known as the root circle, is the circle traced through the base of the tooth.
The pressure angle, often known as the “tooth shape,” is the angle at which pressure from one gear's tooth is transferred to another gear's tooth. Pure rolling motion will occur at pitch point.
Therefore, it is pressure point.
Learn more about the pressure point, refer to:
https://brainly.com/question/2510654
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Rigid bar ABC is supported by three symmetrically-positioned vertical rods, which are initially unstrained. After load P is applied, the normal strain in rods (2) is 0.0010 mm/mm. Determine the normal strain in rod (1) if there is a gap of 1.0 mm in the connection between rod (1) and the rigid bar at B. Report the strain in mm/mm
Answer:
The answer is below
Explanation:
The lengths of the rods are not given.
Let us assume the length of rod 1 = 1500 mm and the length of rod 2 = 800 mm
Solution:
The normal strain is defined as the change in member length δ divided by the initial member length L. The normal strain (ε) is:
ε = δ / L
δ = εL
For rod 1:
[tex]\delta_2=\epsilon_2 L_2\\\\\delta_2=0.0010\ mm/mm*1500\ mm=1.5\ mm[/tex]
The axial elongation of rod 2 is 1.5 mm. Since rigid bar ABC is attached to rod 2, the rigid bar move down by same amount.
The rigid bar moves down 1.8 mm but rods 1 will not be stretched by this amount. Because there is a gap between rod (1) and the rigid bar at B, the first deflection of 1 mm would not cause an elongation in rod 1. Therefore, the elongation in rods (1) is:
[tex]\delta_1=1.5\ mm-1\ mm=0.5\ mm[/tex]
The normal strain in rod 1 is:
[tex]\epsilon_1=\frac{\delta_1}{L_1} =\frac{0.5\ mm}{800\ mm} \\\\\epsilon_1=0.000625\ mm/mm[/tex]