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Troyanec [42]
3 years ago
11

Compute the estimated actual endurance limit for SAE 4130 WQT 1300 steel bar with a rectangular cross section of 20.0 mm by 60 m

m. It is to be machined and subjected to repeated and reversed bending stress. A reliability of 99% is desired
Engineering
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

estimated actual endurance strength = 183.22 MPa

Explanation:

given data

actual endurance limit for SAE 4130 WQT 1300

cross section = 20.0 mm by 60 mm

reliability = 99%

solution

we use here table for get some value for AISI 4130 WQT 1300 steel

Sut = 676 MPa

Sn = 260 MPa

and here

stress factor Cst = 1

and wrought steel Cm = 1

and reliability factor Cr is = 0.81

so here equivalent diameter will be

equivalent diameter = 0.808 \sqrt{bh}    ...........1

equivalent diameter = 0. 808 \sqrt{20*60}  

equivalent diameter = 28 mm

so here size factor will be = 0.87 by the table

so now we can get estimated actual endurance strength will be

actual endurance strength = Sn × Cm × Cst × Cr × Cs ...............2

put here value

actual endurance strength = 260 × 1 × 1 × 0.81 × 0.87

actual endurance strength = 183.22 MPa

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Read 2 more answers
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
GarryVolchara [31]

Answer:

a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}

b) U_A = 3.66 m/s

V_B = 4.32 m/s

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let U_A be the initial velocity of rod A

Let U_B be the initial velocity of rod B

Let V_A be the final velocity of rod A

Let V_B be the final velocity of rod B

Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

Substitute equation (2) into equation (1)

M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

Solving for V_B in equation (3) above:

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

c) Impulse, I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)

I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

I = 0 kg m/s^2

d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

7 0
3 years ago
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