Ask question

Ask question

All categories

1 month ago

6

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

9 × 106 psi and αs = 6.5 × 10-6/°F). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 80°F.

1 answer:

Maru [3.3K]1 month ago

6 0

Answer:

the normal stress caused by the concrete post $\sigma_c = 67.26 psi$

the normal stress caused by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus of elasticity for the concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus of elasticity for the steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rod = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced from the temperature increase; we derive the expression:

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress caused by the concrete post $\sigma_c = 67.26 psi$

Also, the normal stress in the steel bars induced due to the temperature change is represented as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress caused by the steel $\sigma_s = - 1795.84 psi$

You might be interested in

In what processes do proteins in the electron transport chain participate?

kicyunya [3294]

Answer:

1. Reactions involving oxidation and reduction along with proton pumping

2. Reactions involving phosphorylation and proton pumping

Explanation:

During oxidative phosphorylation, there is a transfer of electrons from donors to acceptors, which constitutes a redox reaction.

These redox reactions liberate energy that is utilized to produce ATP. In eukaryotic cells, these reactions are performed by protein complexes found in the mitochondria, while in prokaryotic cells, the proteins are positioned in the intermembrane space of the cells. These interconnected protein complexes are referred to as electron transport chains.

8 0

1 month ago

Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How

Keith_Richards [3271]

The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.

6 0

1 month ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Yuliya22 [3333]

Answer:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

Explanation:

In this scenario, we determine the initial velocity as follows:

$v_i = 7 \frac{m}{s}$

The final velocity in this instance can be expressed as:

$v_f = 13 \frac{m}{s}$

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

$v_f = v_i +at$

Solved for acceleration, we find:

$a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}$

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

$v_f = v_ i +a t$

Substituting into the equation yields:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

5 0

3 months ago

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon

ValentinkaMS [3465]

When the boat submerges completely in the pond, the water level of the pond rises.

4 0

1 month ago