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Anarel [89]
3 years ago
13

What are practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectio

nal area fixed?
Engineering
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

The answer is below

Explanation:

The practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectional area fixed are:

1. Shapes of moment of inertia: Engineers should consider or know the different shapes of moment of inertia for different shape

2. Understanding the orientation of the beam: this will allow engineers to either increase or decrease the moment of inertia of a beam without increasing its cross sectional area.

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Assignment 1: Structural Design of Rectangular Reinforced Concrete Beams for Bending
GREYUIT [131]

Answer:

Beam of 25" depth and 12" width is sufficient.

I've attached a detailed section of the beam.

Explanation:

We are given;

Beam Span; L = 20 ft

Dead load; DL = 0.50 k/ft

Live load; LL = 0.65 k/ft.

Beam width; b = 12 inches

From ACI code, ultimate load is given as;

W_u = 1.2DL + 1.6LL

Thus;

W_u = 1.2(0.5) + 1.6(0.65)

W_u = 1.64 k/ft

Now, ultimate moment is given by the formula;

M_u = (W_u × L²)/8

M_u = (1.64 × 20²)/8

M_u = 82 k-ft

Since span is 20 ft, it's a bit larger than the average span beams, thus, let's try a depth of d = 25 inches.

Effective depth of a beam is given by the formula;

d_eff = d - clear cover - stirrup diameter - ½Main bar diameter

Now, let's adopt the following;

Clear cover = 1.5"

Stirrup diameter = 0.5"

Main bar diameter = 1"

Thus;

d_eff = 25" - 1.5" - 0.5" - ½(1")

d_eff = 22.5"

Now, let's find steel ratio(ρ) ;

ρ = Total A_s/(b × d_eff)

Now, A_s = ½ × area of main diameter bar

Thus, A_s = ½ × π × 1² = 0.785 in²

Let's use Nominal number of 3 bars as our main diameter bars.

Thus, total A_s = 3 × 0.785

Total A_s = 2.355 in²

Hence;

ρ = 2.355/(22.5 × 12)

ρ = 0.008722

Design moment Capacity is given;

M_n = Φ * ρ * Fy * b * d²[1 – (0.59ρfy/fc’)]/12

Φ is 0.9

f’c = 4,000 psi = 4 kpsi

fy = 60,000 psi = 60 kpsi

M_n = 0.9 × 0.008722 × 60 × 12 × 22.5²[1 - (0.59 × 0.008722 × 60/4)]/12

M_n = 220.03 k-ft

Thus: M_n > M_u

Thus, the beam of 25" depth and 12" width is sufficient.

3 0
3 years ago
Question 3 (5 points)
Aleks [24]

Answer:

Technician A is corect

Explanation:

Because he said the correct reason

6 0
3 years ago
"It is better to be a human being dissatisfied than a pig satisfied; better to be Socrates dissatisfied than a fool satisfied. A
Flura [38]

Explanation:

It may be instructive to look at the opposite of the sentence here. Perhaps the smarter creature would be more unhappy when frustrated, recognizing how it gains from happiness relative to the creature with less experience and less knowledge of a situation that does not define it at the moment.  

Perhaps the argument is really about the fact that wisdom helps one to hypothetically live in multiple states and a lack of wisdom prevents or fails this possibility.

6 0
3 years ago
What are the 2 reasons an alignment should be done?
NikAS [45]

Answer:

because it will keep the tires in much better shape and it can improve the handling and keep your  car from pulling to one side

Explanation:

5 0
3 years ago
A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/(s^2). Determine the acceleration of an object o
storchak [24]

Answer:

See it in the pic

Explanation:

See it in the pic

5 0
3 years ago
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