Ask question

Ask question

All categories

Artyom0805 [142]

2 years ago

9

A plate of width 84mm and thickness 10mm is to be welded to another plate by means of double parallel fillets. The plates are su

bjected to a static load of 86kN and the maximum permissible shear stress is 3.5MPa.. 1. Find the length of the weld in millimeters(mm).

1 answer:

alexgriva [62]2 years ago

6 0

Answer:

d

Explanation:

i did the test

You might be interested in

Three spheres are subjected to a hydraulic stress. The pressure on spheres 1 and 2 is the same, and they are made of the same ma

r-ruslan [8.4K]

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒ $P_1=P_2$

⇒ $\frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}$

and the bulk modulus be,

⇒ $B_1=B_2$

Sphere 3:

⇒ $\frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1$

then,

⇒ $P_3=B\times \frac{\Delta V_3}{V_3}$

⇒ $=B\times 1$

⇒ $=150000\times 1$

⇒ $=150000 \ N/m^2$

5 0

3 years ago

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6 g, respectively.

PilotLPTM [1.2K]

Answer:

N=945.76 RPM

Explanation:

Given that

A= 0.15 m

Acceleration = 0.6 g

a=0.6 x 9.81 m/s²

a= 5.886 m/s²

We know that acceleration a given as

a = ω² A

ω=Angular speed

$\omega=\sqrt{\dfrac{0.6\times 9.81}{0.6\times 10^{-3}}}$

ω=99.04 rad/s

We know that

$\omega=\dfrac{2\pi N}{60}\\\\N=\dfrac{60\times \omega}{2\pi }$

$N=\dfrac{60\times 99.04}{2\pi }$

N=945.76 RPM

Therefore the speed of the motor will be 945.76 RPM.

6 0

4 years ago

10. To cut 1/4" (6 mm) thick mild steel at a rate of 40 inches per minute, the current would be set to

pishuonlain [190]

Answer: Idk but try downloading more apps it’s a lot easier to have more than one

Explanation:

5 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$

6 0

4 years ago

It was found experimentally that a certain material does not change in volume when subjected to an elastic state of stress. Calc

lora16 [44]

How are you? because i’m great

5 0

3 years ago