Consider 2 hosts, Host A and Host B, transmitting a large file to a Server C over a bottleneck link with a rate of R kbps.

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 metal alloy has been tested in a tensile test with the following results for the flow curve

MA_775_DIABLO [31]

Answer:

dhjdjs

Explanation:

ehejsikajaudiieisisjsjsjsjsjjsjsjsjsjs

4 0

2 years ago

State three types of maintenance.

Nikitich [7]

Answer: Preventive Maintenance, Condition-Based Maintenance, Predictive Maintenance.

Explanation:

5 0

2 years ago

Read 2 more answers

Explain the following statement: In a windowless proportional counter, the output pulse height from an alpha particle source wil

-BARSIC- [3]

Answer / Explanation:

Regarding α and β Particles in Windowless Counter, the range of α particles is lower than β particles. Alpha particles typically have range less than the dimensions of the gas chamber so that proportional counters are able to easily record. Hence, with almost 100% efficiency, each particle which enters the so called active volume.

Then, since the pulse height spectra depends on the number of ion pairs which have formed, an aplha particle with higher energy creates more ion pairs in the chamber. However, beta particle range usually exceeds the dimensions of the chamber and therefore most of the betas hit the walls where they deposit energy. Then, fewer ion pairs are formed because very few β’s give their energy to the bulk gas.

5 0

2 years ago

A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that

Lina20 [59]

Answer:

The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )

Explanation:

width of channel = 3.0 m

Flow rate = 5 m^3/s

Normal depth = 0.50 m

Flow encounters a dam rise of 0.25 m

To know if the hydraulic jump will occur

we will Determine the new normal depth

Y2 = 3.77m

Yc ( critical depth )= 0.66m

Attached below is the detailed solution

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2 years ago