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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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Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

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Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

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Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

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Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

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