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fenix001 [56]
2 years ago
10

A steel bar with a diameter of .875 inches and a length of 15.0 ft is axially loaded with a force of 21.6 kip. The modulus of el

asticity of the steel is 29 *106 psi. Determine
Engineering
1 answer:
svetoff [14.1K]2 years ago
3 0

Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

\sigma  = Load, P/ Area, A

\sigma  = 4P/\pi d^2

substitute the value

\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi

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A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
Another name for your computer, running the web browser program is: Web user The client The mainframe Browsing agent
timofeeve [1]

Answer:

Browsing agent

Explanation:

Hope this helps!

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2 years ago
In several sentences, please discuss:
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Answer:

Everything is explained below in the Explanation section.

Explanation:

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Whereas, the formula to calculate TMDL is as follows:

TMDL =  Sum of waste load allocations (point source) + Sum of load allocations (non point sources and background) + Margin of Safety

 

<em>b) Dead Zones :</em>

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8 0
3 years ago
A packet weighs 40kg in air but when it is totally submerged into a 1mx1m square tank the weight of the packet is only 18kg. How
Irina18 [472]

Answer:

water  rise = 22 mm

Explanation:

weight of packet IN AIR = 40 *9.81 =392.4 N

weight of packet  IN WATER= 18 *9.81 =176.58 N

by Archimedi's principle

difference in weight = weight of displaced water

w_a - w_w = \rho_w v_d g

392.4 - 176.58 = 1000* v_d* 9.81

v_d = 0.022 m^3

v_d = A*H_rise

0.022 =1*H_rise

H_rise = 0.022 m = 22 mm

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QUESTIONS
igomit [66]

Answer:

According to many of the states' driving rules.

All vehicles of 3,000 pounds or more are required to have a brake system that makes them break as a response to the breaking of the vehicle's tow.

Explanation:

The reason behind this answer is that vehicles of more than 3,000 pounds are extremely dangerous and difficult to control. Therefore, when the tow breaks the automatic brake system is required. Because they are too big to be controlled, and if they are left without a brake system to reduce their damage they can destroy entire houses or other cars if this mechanism is not implemented.

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