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Gnom [1K]
3 years ago
9

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

designated by size, with the size being the diameter in eights of an inch. For example, a #4 bar is 4/8 inch in diameter. A 11 inch square concrete column is constructed with eight #4 bars. A 50000 pound load is applied to this column. The modulus of elasticity of the reinforcing steel is 29x106 psi, while the modulus of elasticity of the concrete is 4.1x106 psi. (area of the concrete= 119.4 in^2)
a. What is the stress in the steel?
b. What is the stress in the concrete?
Physics
1 answer:
Sergio039 [100]3 years ago
4 0

Answer:

21678.47223\ lbf-in^2

383.1109\ lbf-in^2

Explanation:

d = Diameter of column = 0.5 inch

A_c = Area of concrete = 119.4\ in^2

The strain in the system is conserved

\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s

Now

F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf

F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf

Stress is given by

\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2

The stress in the steel is 21678.47223\ lbf-in^2

\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2

The stress in the steel is 383.1109\ lbf-in^2

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8 0
1 year ago
Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as
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<u>We are given:</u>

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

__________________________________________________________

<u>Momentum of Tarzan before swinging:</u>

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

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__________________________________________________________

<u>Speed of Tarzan after grabbing the bananas:</u>

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v                           [where v is the velocity of Tarzan]

v = 5.9 m/s

<u>__________________________________________________________</u>

<h3><u>Finding the Initial and Final KE and PE:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

<u>Initial and Final KE:</u>

We know that KE = 1/2*(mv²)

<u>Initial KE:</u>

Initial KE = 1/2*(mv²)          [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

<u>Final KE:</u>

Final KE = 1/2*(mv²)            

<em>[where v is the velocity at the maximum point of the swing]</em>

Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

<u>Initial and Final PE:</u>

We know that:

PE = mgh                

<em>[where g is the acceleration due to gravity and h is the height]</em>

<u>Initial PE:</u>

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

<u>Final PE:</u>

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3><u>Finding the maximum height Tarzan will reach:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

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1653.5 = 950h

h = 1653.5/950         [dividing both sides by 950]

h = 1.74 m

Therefore, the maximum height reached by Tarzan is 1.74 m

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Answer:

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Doubling the starting speed gives

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Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

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