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lesya692 [45]
3 years ago
8

A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l

b. Under this force, the bar elastically deforms so that the gauge length increases to 2.007 in. and the cross section decreases to 0.599 in. x 0.599 in. Determine the modulus of elasticity and Poisson’s ratio for this metal.
Engineering
1 answer:
Aleks [24]3 years ago
5 0

Answer:

modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

Explanation:

Modulus is the ratio of tensile stress to tensile strain

Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain within the direction of the stretching force

And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in

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Vlad1618 [11]

The function that you will use in order to test the reverse route back toward the original host is known as Extended ping. Thus, the correct option for this question is B.

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6 0
1 year ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 16.403
Dmitry [639]

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

8 0
4 years ago
Information such as tolerances and scale can be found in the _______________ of an engineering drawing.
nasty-shy [4]

Answer:

Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing

Explanation:

The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item

8 0
3 years ago
A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig
jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

             Section: 9-ft section of pipe.

                     Sum of forces perpendicular to member AC = 0

                     F_d - 0.8*W*L = 0

                     F_d = 0.8*10*9 = 72 lb

                     Sum of forces perpendicular to member BC = 0

                     F_e - 0.6*W*L = 0

                     F_e = 0.6*10*9 = 54 lb

              F_d = 72 lb ,  F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

              Section: Entire Frame.

                    Sum of moments about point B = 0

                    -A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

                    -A_y*(1.5625) + 15 + 39.375 = 0

                    A_y = 34.8 lb  

                   Sum of forces in vertical direction = 0

                     A_y + B_y - 0.8*F_d - 0.6*F_e = 0

                     B_y = 0.8*(72) + 0.6*(54) - 34.8

                     B_y = 55.2 lb  

                   Sum of forces in horizontal direction = 0

                     A_x + B_x - 0.6*F_d + 0.8*F_e = 0

                     A_x + B_x = 0

               Section: Member AC

                    Sum of moments about point C = 0

                     F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

                     72*2.5 - 34.8*12 - 9*A_x = 0

                     A_x = -237.6 / 9 = - 26.4 lb

                     B_x = - A_x = 26.4 lb

                     A_x = -26.4 lb  ,  B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

               Member AC:

                    From point A till just before point D

                     -0.6*A_x*x - A_y*0.8*x + M = 0

                     15.84*x - 27.84*x + M = 0

                      M = 12*x   ..... max value at D, x = 12.25 in

                      M_max = 12*12.25/12 = 12.25 lb-ft

               Member BC:

                    From point B till just before point E

                     -0.8*B_x*x + B_y*0.6*x + M = 0

                     -21.12*x + 33.12*x + M = 0

                      M = -12*x   ..... max value at E, x = 11.25 - 2.5 = 8.75 in

                      M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

                      AC: at D , M_max = 12.25 lb-ft

                      BC: at E , M_max = 8.75 lb-ft

5 0
3 years ago
A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum
Charra [1.4K]

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

7 0
3 years ago
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