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Papessa [141]
2 years ago
7

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

tic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Engineering
1 answer:
zlopas [31]2 years ago
3 0

Answer:

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation )

Explanation:

Yield strength = 275 Mpa

Tensile strength = 380 Mpa

elastic modulus = 103 GPa

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation ) .

Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given

strain = yield strength / elastic modulus

          = 0.0027

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Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

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non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

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2 years ago
A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob
scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

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0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

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8 0
2 years ago
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

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б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

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attached below is the remaining solution

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Measures the power output of the machine
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Answer:

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7 0
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