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blagie [28]
3 years ago
11

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

ains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point
Engineering
1 answer:
Murljashka [212]3 years ago
3 0

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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2 samples of water of equal volume are put into dishes and kept at room temp for several days. the water in the first dish is co
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Answer:

Vaporization is the process by which a substance changes from its solid or liquid state to a gaseous state.

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3 years ago
MODIFIED-BOTTOM-UP-CUT-ROD(p, n, c) to return not only the value but the actual solution, too. Hint: It is similar to how array
Vaselesa [24]

Answer:

b.

Matrix chain multiplication

M[i,j] = M[i,k] + M[(k+1),j] + p[i-1]*p[k]*p[j] i<=k<j

p[] = {5,10,3,12,5,50}

M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

M[4][5] = M[4][4]+M[5][5]+p[3]*p[4]*p[5] = 0+0+12*5*50 = 3000

M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

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M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

M[1][5] = min{M[1][1] +M[2][5]+p[0]*p[1]*p[5] , M[1][2] +M[3][5]+p[0]*p[2]*p[5],

M[1][3] +M[4][5]+p[0]*p[3]*p[5] , M[1][4] +M[5][5]+p[0]*p[4]*p[5]}

{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

8 0
3 years ago
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Answer:

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<u>kLa</u> = 0.17/s

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<u>The maximum specific oxygen uptake rate </u>= 4 mmol O2 / g h.

<u>Concentration of oxygen</u> =  0.5 × 10^-3 kg/ m^3

<u>**The maximum cell density</u> = 50 g/l

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The calculated maximum cell concentration:

xmax=  kLa · CAL*/ qo

CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate

Replacing the data given

xmax= ( 0.17/s ) ·   (8 × 10^-3 kg / m^3)  /  4 mmol O2 / g h

4 mmol O2 / g h  to kg O2/ g s

4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}

= 3.56 x 10^-3 kg O2/ g s

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xmax= ( 0.17/s ) ·   (8 × 10^-3 kg / m^3)  / 3.56 x 10^-3 o kg O2/ g s

xmax= 3. 8 x 10^4 g/ m^3   = 38 g/l

_____________________

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