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3 years ago

The principal value of a Pareto diagram is as a

Engineering

1 answer:

vlada-n [284]3 years ago

4 0

The Pareto principle is that most things in our life are not commonly distributed.

<u>Explanation:</u>

Pareto chart shows that most of the things which we have in our life and the resources in our life are not equally distributed. The ratio is not always 50:50 according to this principle.

The most important use of a Pareto diagram is to show the most important factor among the set of factors that have been shown. Along with that it also shows the sources which lead to the common defects in the system and tries to solve those defects which occur most often.

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3 years ago

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as

katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:

A = as

$=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}$

Calculating the kinematics equation:

$\to v^2 = v^2_{o} + 2as\\\\$

$=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}$

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$\to a= \frac{dv}{dt}$

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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

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2 years ago