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miss Akunina [59]
2 years ago
13

Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with

a factor of safety of 3, determine the maximum allowable loadthat will satisfy geotechnical ULS requirements.
Engineering
1 answer:
ehidna [41]2 years ago
7 0

Answer:

Um...

Explanation:

This is what I like to see teachers giving out.

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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
SSSSS [86.1K]

Answer:

The solution is attached in the attachment.

3 0
2 years ago
Read 2 more answers
A cylindrical container is 150mm in diameter and weights 2.25n when empty
stellarik [79]

Answer:

.

Explanation:

.

3 0
2 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     <em>=  191 mPa </em>

3 0
2 years ago
An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.
saul85 [17]

Answer:

a. L_o  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, L_o  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)

For the school, K_{LL} = 2

Therefore, we have;

L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, W_d = b × W_{D + L} =

∴  W_d =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, W_d = 812.8 ft./lb

d. For the uniformly distributed load, we have;

V_{max} = 812.8 × 26/2 = 10566.4 lbs

M_{max} =  812.8 × 26²/8 = 68,681.6 ft-lbs

v_{max} = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

P_L = 1000 lb

W_{D} = 20 × 16 = 320 lb/ft.

V_{max} = 1,000 + 320 × 26/2 = 5,160 lbs

M_{max} =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

v_{max} = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0
2 years ago
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface
Svet_ta [14]

Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

= 4.52W/m

The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

= 0.13/10

= 0.013m

= 13mm

5 0
2 years ago
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