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Elena-2011 [213]
3 years ago
15

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

ed at the centers of the ends of the bar. Knowing that a = 30 mm and σall = 135 MPa, determine the smallest allowable depth d of the milled portion of the bar.
Engineering
1 answer:
monitta3 years ago
3 0

The smallest allowable depth is d=16.04 \mathrm{mm} for the milled portion of bar.

<u>Explanation:</u>

Given,

Magnitude of force,\mathbf{p}=18 \mathrm{kN}

a=30 \mathrm{mm}

=0.03 \mathrm{m}

Allowable stress,\sigma_{a l l}=135 \mathrm{MPa}

cross sectional area of bar,

A=a \times d

A=a d

e - eccentricity

e=\frac{a}{2}-\frac{d}{2}

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

M=P e

=P\left(\frac{a}{2}-\frac{d}{2}\right)

=\frac{P(a-d)}{2}

Allowable stress

\sigma=\frac{P}{A}+\frac{M c}{I}

c=\frac{d}{2}

Moment of Inertia,

I=\frac{b d^{3}}{12}

=\frac{a d^{3}}{12}

\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}

\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\

\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)

\sigma\left(a d^{2}\right)=P d+3 P a-3 P d

\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a

\left(\sigma a d^{2}\right)=-2 P d+3 P a

\sigma d^{2}=-\frac{2 P}{a} d+3 P

By substituting values we get,

\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0

\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0

On solving above equation we get,d=0.01604 \mathrm{m}\\

d=16.04 \mathrm{mm}

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The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

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Explanation:

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rows is initialized to 1 to ensure that row starts at 1 or you can remove it then seat number will start from 0.

The first loop is used for digits starting from 1 to number of rows

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Please refer to the attached image.

Output:

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