Ask question

Ask question

All categories

2 years ago

7

A three-story school has interior column bays that are spaced 25 ft apart in both directions. If the loading on the flat roof is

estimated to be 20 lb/ft2, determine the reduced live load supported by a typical interior column at ground level.

1 answer:

Lady_Fox [76]2 years ago

7 0

Answer:

Explanation:

Floor Load:

Lo= 50psf

At= 25x25 = 625 square feet

$L= Lo(0.25 +15/\sqrt{KuAt)}$

$L=50(0.25+15/\sqrt{(4)(625)}$ = 13.1psf

%reduction= 13.1/50 = 26%

Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k

You might be interested in

Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e

puteri [66]

Given:

outer radius, R' = 10 m

inner diameter, d = 2 m

inner radius, R = $\frac{d}{2}$ = 1 m

surface temperature, T' = $20^{\circ}C$

Thermal conductivity of soil, K = 0.52 W/mK

Solution:

To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :

Q = $\frac{T - T'}{\frac{R' - R}{4\pi KRR'}}$

500 = $(T - 20)\frac{4\pi \times0.52\times 1\times 10}{10 - 1}$

T = 68.86 +20 = $88.865^{\circ}C$

Therefore, outside surface temperature is $88.865^{\circ}C$

4 0

2 years ago

The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate

Sonja [21]

Answer:

a) Mechanical efficiency ( $\varepsilon$ )=63.15% b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, $Wmec=0.95*6kW=5.7 kW$ .

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump $\Delta P$ . So $P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W$ .

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is $\varepsilon=3.6kW/5.7kW=0.6315=63.15\%$

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and $\rho$ the density of the water:

$[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T$ [/tex]

Finally, the temperature rise is:

$\Delta T=2100/75348 \ºC=0.028 \ºC$

7 0

2 years ago

A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.

Ivahew [28]

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

8 0

2 years ago

Universal Containers has a junction object called "Job Production Facility". With 2 master-detail relationships to the Job and P

alexdok [17]

Answer:

C. The user will not be able to see the junction object records or the field values.

Explanation:

For the profile of the user to give access permission such as create and read to the job without granting access permission to the production facility object, the value of the field or records of the junction object will not be seen by the user. This is one of the necessary criteria or principle for the universal container with a junction object.

3 0

2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

2 years ago