All categories

2 years ago

Beam design what is the maximum code-allowable floor beam deflection under the given live load?

Physics

1 answer:

Irina18 [472]2 years ago

5 0

According to I.S. standards, the maximum deflection should typically not be greater than 1/325 of the span.

<h3>What is Beam deflection?</h3>

A beam is a long segment of a body that can support a weight while defying bending. Beam deflection is the term for the movement of a beam when a force is applied to it in a certain direction.

The beam has the ability to be bent or shifted from its initial position. The deflection is represented by this distance along each member point.

The amount of displacement that a certain structural member may experience with the aid of a sizable amount of load is known as deflection. It may also be known as the angle or the separation. The slope of the deflected form of the body under a load directly influences the distance of deflection of a member under that force.

To know more about Beam deflection:

brainly.com/question/15741463

#SPJ4

You might be interested in

The balls in the image above have different masses and speeds. Fill in the blanks in the following sentence: Ball ___ has the gr

s344n2d4d5 [400]

Yes I’m going home I was finna was your birthday and I got to get my kids ready

4 0

3 years ago

Based on the data collected, explain why a launch angle of 30 degrees (HIGH HORIZONTAL VELOCITY & SMALL TIME OF FLIGHT) will

Gennadij [26K]

Answer:

Explanation:

The horizontal distance traveled by the projectile is given by the formula

$R=\frac{u^{2}Sin2\theta }{g}$

The formula for the time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Case I: when the launch angle is 30°

So, $R_{1}=\frac{u^{2}Sin60 }{g}$

$R_{1}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 30 = 0.866 u

$T_{1} = \frac{2u Sin30 }{g}=\frac{u}{g}$

Case II: when the launch angle is 60°

$R_{2}=\frac{u^{2}Sin120}{g}$

$R_{2}=\frac{0.866u^{2}}{g}$

Horizontal velocity = u Cos 60 = 0.5 u

$T_{1} = \frac{2u Sin60 }{g}=\frac{1.73 u}{g}$

By observing the case I and case II, we conclude that

R1 = R2

Horizontal velocity 1 > Horizontal velocity 2

T1 < T2

5 0

3 years ago

A merry go round exerts a force of 1000 N on a rider on the

SVEN [57.7K]

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

$F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r$

Where;

m = The mass of the person

v = The velocity of the person

$F_c$ = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for the centripetal force, we have;

$F_c$ = m × ω² × r

1000 = 76.45 × (2·π/15)² × r

r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

3 0

3 years ago

What is the meaning of friend ?

MakcuM [25]

Answer:

person that you know and like (not a member of your family), and who likes you

6 0

3 years ago

Which description best describes a plateau?

Tamiku [17]

Answer:

The answer is "A raised, flat-surfaced area bound on one or more sides by cliffs or steep slopes."

Explanation:

5 0

2 years ago

Read 2 more answers