What is the design wind speed for average-risk building:____

A 1.7 cm thick bar of soap is floating in water, with 1.1 cm of the bar underwater. Bath oil with a density of 890.0 kg/m{eq}^3

PIT_PIT [208]

Answer:

The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm

Explanation:

The water level on the bar soap = 1.1 m mark

Therefore, the proportion of the bar soap that is under the water is given by the relation;

Volume of bar soap = LW1.7

Volume under water = LW1.1

Volume floating = LW0.6

The relative density of the bar soap = Density of bar soap/(Density of water)

= m/LW1.7/(m/LW1.1) = 1.1/1.7

Given that the oil density = 890 kg/m³

Relative density of the oil to water = Density of the oil/(Density of water)

Relative density of the oil to water = 890/1000 = 0.89

Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89

Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7

Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7

Where:

X = The height of the oil on the side of the bar when the soap is floating in only the oil

Therefore;

X = 1.236 cm.

3 0

4 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

a) Determine the yielding factor of safety

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

4 0

3 years ago

A student is working with three sealed containers filled with water. The first container is filled with ice. The second is fille

Temka [501]

Explanation:

The three containers each contains water in different states.

Solid state of matter is considered as not compressible because the molecules are already as closely packed as they can be.

The liquid sate of matter has a very minute to no compression ability at all as the molecules are relatively close to each other. Compression is difficult to achieve in the liquid state.

In the gaseous state of matter, the molecules have broken free of one another, and are fairly spaced one from another. This means that gases can be easily compressed.

Pressing down on the plunger, the container containing ice can't be compressed at all so it's volume stays the same.

For the container filled with water, only a minute compression can be achieved with great difficulty hence, the volume reduces by an insignificant amount.

For the container filled with vapour, compression can be easily achieved and the volume reduces significantly.

6 0

3 years ago

Select the type of job best demonstrated in each example. James is a Construction Manager who is paid the same amount every week

fredd [130]

Answer :

salaried

entry - level

benefits

full - time

Explanation:

6 0

4 years ago

Read 2 more answers

A stainless-steel wall (AISI 302) is 0.2 m thick with the top surface at a temperature 40°C and the bottom surface at a temperat

BaLLatris [955]

Answer/Explanation:

From the properties of stainless steel AISI 302.

At temperature 100°C, Thermal coefficient Ks = 16.2W/m.K.

Water at 40°C, Kf = 0.62W/m.K.

Using the rate equation, applying energy balance at x = 0,

Ks (T1 - T2)L = h(T1 - T[infinity])

h = (Ks/L) * (T1 - T2)/(T - T[infinity])

h is water flow convention coefficient, T1, T2, Ks, and L have their usual notations.

T1 = 100°C, T2 = 40°C,

T[infinity] = 25°C, L = 0.2m

h = (16.2W/m.K/0.2m) * (60/15)

h = 324W/m².K

Temperature gradient in the wall (dT/dx) = -(T1 - T2)/L = -60°C/0.2m = -300°C/m

In water at x = 0, the definition of h gives

(dT/dx)x=0 = -h/Kf(40°C - 25°C)

= (324W/m².K x 15°C)/0.62W/m.K = 7838.7°C/m

7 0

3 years ago

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What is the design wind speed for average-risk building:_____.