Answer:
Throughout the clarification section elsewhere here, the definition of the concern is mentioned.
Explanation:
The stress-strain curve provides designers with a long list of critical parameters that are needed for utility development. Including capacity, longevity, elasticity, apparent viscosity, tension electricity, resilience, as well as flexural strength, a load-pressure assignment gave us several mechanical households at a certain point of operation. It also assists in manufacturing.During which the overarching force can inform us about the maximum energy either workload the substance will experience, which could also be inferred within the action of the stress-strain. The dynamic properties can be seen by pre maximum activity because it will be before even the maximum yield intensity as well as the posted maximum would display plastic behavior however after the peak becomes achieved, the natural frequencies continue to decline.A universal shift register can shift in both the left-to-right and right-to-left directions, and it has parallel-load capability. Draw a circuit for such a shift register.
Answer:
Explanation:
A unidirectional shift register allows for the capability of shifting in one direction as the name unidirectional implies.
A bidirectional shift register has the capabilities of shifting in both the left to right and right to left directions.
For a Universal shift register, there are possibilities of bidirectional shifts and parallel-load capabilities that have the following properties.
There is the existence of a clear control input whose main FUNCTION is to set all the register to 0.'
A shift control to both the right and left direction to enable both the shift right operation and shift left operation.
Finally, a parallel-load control whose function is to activate a parallel transfer from input to output.
The diagram for the circuit for such a shift register can be seen in the diagram attached below.
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P
Answer:
hello a diagram attached to your question is missing attached below is the missing diagram
The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m
Answer :
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa
Explanation:
Given data:
Type of steel = A-36
cross-sectional area = 500 mm^2
Calculate the average normal stress in each bar
we have to make some assumptions
assume forces in AB, CD, EF to be p1,p2,p3 respectively
∑ Fy = 0 ; p1 + p2 + p3 = 70kN ---------- ( 1 )
∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0
where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )
Take ; Tan∅
Tan∅ = MN / 2d = OP/d
i.e. s1 - 2s2 - s3 = 0
[tex]\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0[/tex]
L , E and A are the same hence
P1 - 2p2 + p3 = 0 ----- ( 3 )
Next resolve the following equations
p1 = 40.03 kN, p2 = 23.33 kN, p3 = 5.33 kN
Stress = force / area
for Bar 1 = (40.03 * 1000) / 500 = 80.06 MPa
Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa
for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa