“All dogs bark. Fido barks. Thus, Fido is a dog,” is an example of which of the following?

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and

Lelu [443]

Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD, $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D $_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r $_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D $_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r $_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However $\phi_r = \phi_{C/D} - \phi_{B/A}$

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$ = Torque on shaft AB and CD respectively

$T_{AB}$ = Required

$T_{CD}$ = 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$ = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$ = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

= $\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T $_{AB}$ = 79.54 N.m

Where T = $T_{AB}$ + T $_{CD}$ =

Therefore T $_{CD total }$ = 500 - 79.54 = 420.46 N·m

τ $_{max}$ = $\frac{T\times R}{J}$

$\tau_{ABmax}$ = $\frac{T_{AB}\times R_{AB}}{J_{AB}}$ = $\frac{79.54\times 0.015}{7.95\times 10^{-8}}$ = 15 MPa

$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa

7 0

3 years ago

The Kelvin scale is also know as the absolute zero<br> TRUE OR FALSE

Bond [772]

The Kelvin scale is also called absolute zero scale

Explanation:

true

6 0

3 years ago

Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer

Katarina [22]

Let's start by differentiating the terms distance and displacement. They both refer to the length of paths. Distance only accounts for the total length regardless of the path taken. Displacement measures the linear path from the starting point to the end point. So, it does not necessarily follow the actual path. However, for this problem, assuming that the path is just in one direction, displacement and distance would just be equal. The equation would be:

Distance = Displacement = v₀t + 0.5at² = 0(10 s) + 0.5(+1.2 m/s²)(10 s)²
Distance = Displacement = 60 meters

6 0

3 years ago

How much heat is needed to raise the temperature of 100 g of water by 50 C, if the specific heat of water is 4,184 J/kg.C?

Yuliya22 [10]

Heat required to raise the temperature of water is given as

$Q = ms\Delta T$