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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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3 years ago
After the elevator accelerates, then it moves at a constant speed of 6.0m/s, calculate the tension inthe cable.
Anon25 [30]

We know that

• The mass of the elevator is 5000 kg.

Let's draw a free-body diagram.

As you can observe, there are just two forces involved, the weight of the elevator and the tension force. Let's use Newton's Second Law.

\begin{gathered} \Sigma F_y=ma_y \\ T-W=ma_y \end{gathered}

But, W = mg = 5000kg*9.8m/s^2 = 49,000 N, and m = 5000 kg, a = 0 (because the speed is constant).

\begin{gathered} T-49,000N=5000\operatorname{kg}\cdot0 \\ T=49,000N \end{gathered}<h2>Therefore, the tension in the cable is 49,000 N.</h2>

3 0
2 years ago
A cylindrical rod with length (1.7 m) and radius (2 cm) is fixed from one end and a mass of (20 kg) is attached to the other fre
belka [17]

Answer:

0.44

Explanation:

Poisson's ratio is:

ν = (3K − E) / 6K

where K is the bulk modulus and E is Young's modulus.

Young's modulus is:

E = FL / (AΔL)

where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.

E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)

E = 0.530×10⁹ Pa

Bulk modulus is:

K = -ΔP / (ΔV/V)

where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.

K = -(180 atm × 101325 Pa/atm) / (-0.012)

K = 1.52×10⁹ Pa

Therefore, the Poisson's ratio is:

ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)

ν = (3(1.52) − 0.530) / 6(1.52)

ν = 0.442

Rounded to 2 significant figures, the Poisson's ratio is 0.44.

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3 years ago
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