Respuesta :
Answer / Explanation:
Given the data :
Length of the rod (L) = 0.6m
Load; F = ± 2 KN
Ultimate strength; S u t = 770 M P a = 770 / 6.89 kpsi = 112 kpsi
Yield strength; S y = 420 M P a
Design factor = 1.5
Surface factor; k a = 0.488
Size factor; k b = 0.85
Now by using mechanical engineering design data book,
We calculate for fatigue strength
So, Se¹ = 0.5 × 770
Se¹ = 385 M P a
K a = 57.7 × 770⁻⁰⁷¹⁸
Kₐ = 0.488
Assume Kb = 0.85 (check later)
Se = Kₐ Kb Se¹
Se = 0.488 × 0.85 × 385
S e = 160 M P a
By using design data book,
a = ( f Sut )² / Se
b = - 1 / 3 log f Sut / Se
S f = ₐ N ᵇ
Sf = 2553 (10⁴) ⁻ ⁰.²⁰⁰⁵
S f = 403 M P a
The expression for stress,
M max = 2000 × 0.6
M max = 1200 Newton per meter
= M ( b/2) / I = M ( b/2) / bb³/12
= 6M / b³ = 6 x 1200 / b³
= 7200 / b³ Pa
Now by comparing strength and stress,
S f = n × σ a
= 403 × 10 ⁶
= 1.5 × 7200/ b³
where b = 0.0299 m
b = 30 m m
So, we assume the size factor, so the expression for checking the size factor is,
By using design data book,
de = 0.808 (hb)∧ 1/2
de = 0.808 b
de = 0.808 × 30
de = 24.2
Kb = ( 24.2 / 7.62) ∧-0.107
Kb = 0.88
We have assume 0.85 which is nearer to the 0.88 and it is acceptable.
So the dimension of the square is 30 mm
