Ask question

Ask question

All categories

kakasveta [241]

4 years ago

13

Electrical strain gauges were applied to a notched specimen to determine the stresses in the notch. The results were εx=0.0019 a

nd εy=0.00072. Find σx and σy if the material is made of carbon steel where E=200GPa, Sy=360 MPa.

1 answer:

pickupchik [31]4 years ago

6 0

Answer:

$\sigma_x=451.8MPa$

$\sigma_y=265.99MPa$

Explanation:

$\varepsilon _x=0.0019$

$\varepsilon _y=0.00072$

Poission's ratio for steel =0.28

We know that

$\varepsilon _x=\dfrac{\sigma _x}{E}-\mu\dfrac{\sigma _x}{E}$

$\varepsilon _y=\dfrac{\sigma _y}{E}-\mu\dfrac{\sigma _x}{E}$

Now by putting the values

$0.0019=\dfrac{\sigma _x}{200\times 1000}-0.28\dfrac{\sigma _y}{200\times 1000}$

$0.00072=\dfrac{\sigma _y}{200\times 1000}-0.28\dfrac{\sigma _x}{200\times 1000}$

So $\sigma_x=451.8MPa$

$\sigma_y=265.99MPa$

You might be interested in

Examples of reciprocating motion in daily life

bonufazy [111]

Answer:

Examples of reciprocating motion in daily life are;

1) The needles of a sewing machine

2) Electric powered reciprocating saw blade

3) The motion of a manual tire pump

Explanation:

A reciprocating motion is a motion that consists of motion of a part in an upward and downwards $(\updownarrow)$ or in a backward and forward (↔) direction repetitively

Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump

In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.

8 0

3 years ago

To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01

liq [111]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

3 0

3 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

3 years ago

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

6 0

3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

1)

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

2)

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

3)

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

4)

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

5)

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

6)

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

7)

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

8)

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago