Ask question

Ask question

All categories

3 years ago

7

weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo

r each cable.

1 answer:

kari74 [83]3 years ago

7 0

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

Diameter = 2(0.325 in)

<u>Diameter = 0.65 in</u>

You might be interested in

Engineering Specificaitons is ...

kvv77 [185]

"Engineering specifications" means those records which provide detailed documentation of the construction, wiring, arrangement and related engineering details of the information processing equipment.

5 0

3 years ago

A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The

wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

$q=\frac{Ak}{L}dT$

$\frac{q}{A}=\frac{k}{L}dT$

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

${k_{fi}= 0.03 W/m.K$

${L_{fi}= 5 cm = 0.05 m$

${T_{fi}= 25 \°C$

${k_{cb} = 0.5 W/m.K$

${L_{cb}= 20 cm = 0.20 m$

${T_{cb}= 0 \°C$

${T_{m}= ? \°C =$ temperature at the interface

Solving for ${T_{m}$ will give the temperature at the interface:

$\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})$

$\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})$

$15 -0.6T_{m}=2.5T_{m}$

$3.1T_{m}=15$

$T_{m}=4.8$

3 0

3 years ago

A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is $\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}$

The shear strength= $0.58\sigma_y$ and in this case $\sigma_y=36 000 psi$

Shear strength= $\frac {Load}{Shear area}$ hence making load the subject then

Load=Shear area X Shear strength

Load= $\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi$

3 0

3 years ago

A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str

Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy = 370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = $ln\frac{1}{1-0.2}$ = 0.223

so

yield strength after cold is

yield strength = $K \varepsilon ^n$

yield strength = $600*0.223^{0.21)$

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = $\frac{\sigma u}{1-W}$

ultimate strength = $\frac{440}{1-0.2}$

ultimate strength after cold work = 550 MPa

8 0

3 years ago

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0

2 years ago