Question

145. The cross section of a diameter of the ring is 15.6 in. Determine the value of P that will cause a maximum ring is the T section shown in Fig. P-1345. The inside stress of 18 ksi. P = 22.3 kips Ans P 6 in.- 1 in. 115.6 in. 1 in 4 in. A Section AB Figure P-1345 4

Please solve this by using the formulas of curved beams.

Answer 1

Ans) To determine load, first calculate the distance of centroid from outer edge of web,

$\dpi{100} \bar{y}$ = A1y1 + A2y2 / (A1 + A2)

= (6 x 1 x 3) + (4 x 1 x 6.5) / ( 6 + 4)

= 4.40 in

$\dpi{100} \bar{y}$₂ (from left end) = 7 - 4.40 = 2.60 in

Diameter of ring = 15.6 in (given)  

Radius of ring = 15.6 / 2 = 7.8 in

Radius upto inner surface (R1) = 7.8 - 2.60 = 5.2 in

Radius upto outer edge of flange (R2) = 5.2 + 1 = 6.2 in

Radius upto outer edge of web (R3) = R1 + 7 = 5.2 + 7 = 12.2 in

Width of flange (B) = 4 in (from figure)

Width of web (b) = 1 in (from figure)

Using equation of curves,

h²/R² = R/A [B ln (R2/R1) + b ln (R3/R2)] - 1

Putting values,

= 7.8/10 [ 4 ln (6.2/5.2) + ln(12.2/6.2)] - 1

=> h²/R² = 0.076

=> R²/h² = 13.14  

Maximum stress = (M / AR) [ y₂ (R²/h²) / (R - y₂) - 1]

=> 18 = M / 78 [ 2.6 (13.14)/(7.8 - 2.6) - 1]

=> 18 = (M / 78) x 5.57

=> M = 252 .1 k-in

Also, M = P x (R₃ -1)

=> P = M/R = 252.1/11.2

= 22.5 kip