Rolling Contact bearing
Problem-1 A bearing for axial flow compressor is used to support a radial load
of 2500N and thrust load of 1500N.The service required for bearing is 5 years at
the rate of 40 hrs per week. The speed of the shaft is 1000 rpm. Select suitable
ball bearing for the purpose. Diameter of the shaft is 50mm.
𝑮𝒊𝒗𝒆𝒏
𝑭𝒓 = 𝟐𝟓𝟎𝟎𝑵 𝒂𝒏𝒅 𝑭𝒂 = 𝟏𝟓𝟎𝟎𝑵
𝑳𝒊𝒇𝒆 = 𝟓 𝒚𝒆𝒂𝒓𝒔 @𝟒𝟎 𝒉𝒐𝒖𝒓𝒔 𝒑𝒆𝒓 𝒘𝒆𝒆𝒌
𝑺𝒑𝒆𝒆𝒅, 𝒏 = 𝟏𝟎𝟎𝟎 𝒓𝒑𝒎,
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 = 𝟓𝟎𝒎𝒎
Find Equivalent Load
Since bearing has to carry considerable axial/thrust load in addition to
radial load, Deep groove ball bearing has to be selected From PSG4.1
𝑷 = (𝑿 𝑭𝒓 + 𝒀 𝑭𝒂 )𝑺 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒 𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒓𝒂𝒕𝒊𝒐, 𝒆 𝒂𝒏𝒅 𝒕𝒚𝒑𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈,
𝑭𝒓
𝒈𝒆𝒕 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒓𝒂𝒅𝒊𝒂𝒍 𝒇𝒂𝒄𝒕𝒐𝒓 𝑿 𝒂𝒏𝒅 𝒂𝒏𝒅 𝒕𝒉𝒓𝒖𝒔𝒕 𝒍𝒐𝒂𝒅 𝒀
𝑭𝒂 𝟏𝟓𝟎𝟎
= = 𝟎. 𝟔;
𝑭𝒓 𝟐𝟓𝟎𝟎
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒, > 𝑒 ⇒ 𝑋 = 0.56 , 𝑌 = 2,
𝑭𝒓
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐, 𝑺𝒆𝒓𝒗𝒊𝒄𝒆 𝒇𝒂𝒄𝒕𝒐𝒓, 𝑺 = 𝟏. 𝟑
𝑷 = (𝟎. 𝟓𝟔 × 𝟐𝟓𝟎𝟎) + (𝟐 × 𝟏𝟓𝟎𝟎) × 𝟏. 𝟑 = 𝟓𝟕𝟐𝟎𝑵 = 𝟓𝟖𝟑. 𝟎𝟖𝑲𝒈𝒇
Find Dynamic load carrying capacity
𝟏
𝑳 𝑲
𝑪= 𝑷 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑳𝟏𝟎
𝑳 𝒊𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒊𝒏 𝒎𝒓 𝒐𝒓 (𝒐𝒏𝒆 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔)
𝑳 = 𝟓 𝒚𝒆𝒂𝒓𝒔 @𝟒𝟎 𝒉𝒐𝒖𝒓𝒔 𝒑𝒆𝒓 𝒘𝒆𝒆𝒌 = 𝟓 × 𝟓𝟐 × 𝟒𝟎 = 𝟏𝟎𝟒𝟎𝟎𝒉𝒓𝒔
⇒ 𝑳 = 𝟏𝟎𝟒𝟎𝟎 × 𝟔𝟎 𝒎𝒊𝒏𝒖𝒕𝒆𝒔 = 𝟔𝟐𝟒𝟎𝟎𝟎 𝒎𝒊𝒏 × 𝟏𝟎𝟎𝒓𝒑𝒎 = 𝟔𝟐𝟒 × 𝟏𝟎𝟔 𝒓𝒆𝒗
⇒ 𝑳 = 𝟔𝟐𝟒𝒎𝒓
𝑳𝟏𝟎 𝒊𝒔 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒇𝒐𝒓 𝟗𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒂𝒕𝟏𝒎𝒓
𝑳𝟏𝟎 = 𝟏𝒎𝒓
𝑲 = 𝟑 𝒇𝒐𝒓 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝟏
𝟔𝟐𝟒𝒎𝒓 𝟑
𝑪= × 𝟓𝟖𝟑𝑲𝒈𝒇 = 𝟒𝟗𝟖𝟐. 𝟔 𝑲𝒈𝒇
𝟏𝒎𝒓
Selection of bearing
𝑭𝒐𝒓 𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒐𝒓 𝒔𝒉𝒂𝒇𝒕 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 = 𝟓𝟎𝒎𝒎
𝒂𝒏𝒅 𝒅𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝟒𝟗𝟖𝟐. 𝟔𝑲𝒈𝒇 𝒔𝒆𝒍𝒆𝒄𝒕 𝒔𝒖𝒊𝒕𝒂𝒃𝒍𝒆
𝑫𝒆𝒆𝒑 𝒈𝒓𝒐𝒐𝒗𝒆 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟏𝟓 𝑫𝒆𝒆𝒑 𝒈𝒓𝒐𝒐𝒗𝒆 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝑺𝑲𝑭𝟔𝟒𝟏𝟎 𝒊𝒔 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟓𝟎𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟕𝟎𝟎𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟓𝟑𝟎𝟎𝑲𝒈𝒇
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒑𝒎 𝒊𝒔 𝟒𝟎𝟎𝟎
Reliability of selected bearing
𝟏 𝟏
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟑
′
𝑪= 𝑷 ⇒ 𝟕𝟎𝟎𝟎 = × 𝟓𝟖𝟑. 𝟎𝟖 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟏𝟕𝟑𝟎. 𝟑𝟒 𝒎𝒓
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟏𝟕𝟑𝟎. 𝟑𝟒𝒎𝒓
𝟏
𝟏 𝒃
𝑳 𝒍𝒏
𝒑
=
𝑳𝟏𝟎 ′ 𝟏
𝒍𝒏
𝒑𝟏𝟎
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟏𝟕𝟑𝟎. 𝟑𝟒𝒎𝒓
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳 = 𝟔𝟐𝟒 𝒎𝒓
𝒑 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒕𝒚 𝒐𝒇 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝒑𝟏𝟎 = 𝟎. 𝟗
𝟏
𝟏 𝟏.𝟏𝟕
𝟔𝟐𝟒 𝒍𝒏
𝒑
= ⇒ 𝒑 = 𝟎. 𝟗𝟔𝟖𝟓
𝟏𝟕𝟑𝟎. 𝟑𝟒 𝟏
𝒍𝒏
𝟎. 𝟗
Problem-2: Select suitable deep groove ball bearing for supporting radial
load is 10kN and axial load is 3kN for a life of 4000 hrs at 800 rpm. Select from
series 63. Calculate expected life of selected bearing and reliability of selected
bearing.
Find Equivalent Load
Since bearing has to carry considerable axial/thrust load in addition to
radial load, Deep groove ball bearing has to be selected From PSG4.1
𝑷 = (𝑿 𝑭𝒓 + 𝒀 𝑭𝒂 )𝑺 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒 𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒓𝒂𝒕𝒊𝒐, 𝒆 𝒂𝒏𝒅 𝒕𝒚𝒑𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈,
𝑭𝒓
𝒈𝒆𝒕 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒓𝒂𝒅𝒊𝒂𝒍 𝒇𝒂𝒄𝒕𝒐𝒓 𝑿 𝒂𝒏𝒅 𝒂𝒏𝒅 𝒕𝒉𝒓𝒖𝒔𝒕 𝒍𝒐𝒂𝒅 𝒀
𝑭𝒂 𝟑
= = 𝟎. 𝟑;
𝑭𝒓 𝟏𝟎
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒, 𝒇𝒐𝒓 𝒆 = 𝟎. 𝟐𝟕; > 𝑒 ⇒ 𝑋 = 0.56 , 𝑌 = 1.6,
𝑭𝒓
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐, 𝑺𝒆𝒓𝒗𝒊𝒄𝒆 𝒇𝒂𝒄𝒕𝒐𝒓, 𝑺 = 𝟏. 𝟑
𝑷 = (𝟎. 𝟓𝟔 × 𝟏𝟎𝒌𝑵) + (𝟏. 𝟔 × 𝟑𝒌𝑵) × 𝟏. 𝟑 = 𝟏𝟐𝟕𝟒𝟎𝑵 = 𝟏𝟐𝟗𝟖. 𝟔𝟕𝑲𝒈𝒇
Find Dynamic load carrying capacity
𝟏
𝑳 𝑲
𝑪= 𝑷 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑳𝟏𝟎
𝑳 𝒊𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒊𝒏 𝒎𝒓 𝒐𝒓 (𝒐𝒏𝒆 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔)
𝑳 = 𝟒𝟎𝟎𝟎𝒉𝒐𝒖𝒓𝒔 @𝟖𝟎𝟎𝒓𝒑𝒎 = 𝟒𝟎𝟎𝟎 × 𝟔𝟎 × 𝟖𝟎𝟎 = 𝟏𝟗𝟐𝒎𝒓
𝑳𝟏𝟎 𝒊𝒔 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒇𝒐𝒓 𝟗𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒂𝒕𝟏𝒎𝒓
𝑳𝟏𝟎 = 𝟏𝒎𝒓
𝑲 = 𝟑 𝒇𝒐𝒓 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝟏
𝟏𝟗𝟐𝒎𝒓 𝟑
𝑪= × 𝟏𝟐𝟗𝟖. 𝟔𝟕𝑲𝒈𝒇 = 𝟕𝟒𝟗𝟐. 𝟎𝟐 𝑲𝒈𝒇
𝟏𝒎𝒓
Selection of bearing
𝒇𝒐𝒓 𝒅𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝟕𝟗𝟒𝟐. 𝟎𝟐𝑲𝒈𝒇 𝒔𝒆𝒍𝒆𝒄𝒕 𝒔𝒖𝒊𝒕𝒂𝒃𝒍𝒆
𝑫𝒆𝒆𝒑 𝒈𝒓𝒐𝒐𝒗𝒆 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟏𝟒 𝑫𝒆𝒆𝒑 𝒈𝒓𝒐𝒐𝒗𝒆 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝑺𝑲𝑭𝟔𝟑𝟏𝟒 𝒊𝒔 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟕𝟎𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟖𝟏𝟓𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟔𝟑𝟎𝟎𝑲𝒈𝒇
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒑𝒎 𝒊𝒔 𝟓𝟎𝟎𝟎
Reliability of selected bearing
𝟏 𝟏
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟑
′
𝑪= 𝑷 ⇒ 𝟖𝟏𝟓𝟎 = × 𝟏𝟐𝟗𝟖. 𝟔𝟕 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟐𝟒𝟕. 𝟏𝟔 𝒎𝒓
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟐𝟒𝟕. 𝟏𝟔𝒎𝒓
𝟏
𝟏 𝒃
𝑳 𝒍𝒏
𝒑
=
𝑳𝟏𝟎 ′ 𝟏
𝒍𝒏
𝒑𝟏𝟎
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟐𝟒𝟕. 𝟏𝟔𝒎𝒓
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳 = 𝟏𝟗𝟐 𝒎𝒓
𝒑 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒕𝒚 𝒐𝒇 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝒑𝟏𝟎 = 𝟎. 𝟗
𝟏
𝟏 𝟏.𝟏𝟕
𝟏𝟗𝟐 𝒍𝒏
𝒑
= ⇒ 𝒑 = 𝟎. 𝟗𝟐𝟒𝟓
𝟐𝟒𝟕. 𝟏𝟔 𝟏
𝒍𝒏
𝟎. 𝟗
Problem-3
The radial reaction on the bearing is 9000N. It carries a thrust of 5000N.Speed of
the shaft is 1000 rpm. Outer ring is stationary. Expected average life of the
bearing is about 25000 hrs. Load on the bearing is smooth,service is 8 hrs per
day
1. Select a suitable roller bearing
2. Rated 90% life of selected bearing
3. Compute the probability of the selected bearing surviving 25000hrs.
Find Equivalent Load
Since bearing has to carry radial and axial /thrust load taper roller
bearing has to be selected From PSG4.1
𝑷 = (𝑿 𝑭𝒓 + 𝒀 𝑭𝒂 )𝑺 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒 𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒓𝒂𝒕𝒊𝒐, 𝒆 𝒂𝒏𝒅 𝒕𝒚𝒑𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈,
𝑭𝒓
𝒈𝒆𝒕 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒓𝒂𝒅𝒊𝒂𝒍 𝒇𝒂𝒄𝒕𝒐𝒓 𝑿 𝒂𝒏𝒅 𝒂𝒏𝒅 𝒕𝒉𝒓𝒖𝒔𝒕 𝒍𝒐𝒂𝒅 𝒀
𝑭𝒂 𝟓𝟎𝟎𝟎
= = 𝟎. 𝟓𝟔;
𝑭𝒓 𝟗𝟎𝟎𝟎
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒, 𝒇𝒐𝒓 𝒆 = 𝟎. 𝟑𝟕; > 𝑒 ⇒ 𝑋 = 0.4 , 𝑌 = 1.6,
𝑭𝒓
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐, 𝑺𝒆𝒓𝒗𝒊𝒄𝒆 𝒇𝒂𝒄𝒕𝒐𝒓, 𝑺 = 𝟏. 𝟑
𝑷 = (𝟎. 𝟒 × 𝟗𝟎𝟎𝟎𝑵) + (𝟏. 𝟔 × 𝟓𝟎𝟎𝟎𝑵) × 𝟏. 𝟑 = 𝟏𝟓𝟎𝟖𝟎𝑵 = 𝟏𝟓𝟑𝟕. 𝟐𝟎𝟔𝑲𝒈𝒇
Find Dynamic load carrying capacity
𝟏
𝑳 𝑲
𝑪= 𝑷 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑳𝟏𝟎
𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒓𝒂𝒕𝒆𝒅 𝒍𝒊𝒇𝒆 (𝟗𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍)
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑳𝒊𝒇𝒆 𝒊. 𝒆. 𝟓𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝟐𝟓𝟎𝟎𝟎𝒉𝒓𝒔
= = = 𝟓𝟎𝟎𝟎𝒉𝒓𝒔
𝟓 𝟓
𝑳 𝒊𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒊𝒏 𝒎𝒓 𝒐𝒓 (𝒐𝒏𝒆 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔)
𝑳 = 𝟓𝟎𝟎𝟎𝒉𝒐𝒖𝒓𝒔 @𝟏𝟎𝟎𝟎𝒓𝒑𝒎 = 𝟓𝟎𝟎𝟎 × 𝟔𝟎 × 𝟏𝟎𝟎𝟎 = 𝟑𝟎𝟎𝒎𝒓
𝑳𝟏𝟎 𝒊𝒔 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒇𝒐𝒓 𝟗𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒂𝒕𝟏𝒎𝒓
𝑳𝟏𝟎 = 𝟏𝒎𝒓
𝟏𝟎
𝑲= 𝒇𝒐𝒓 𝒓𝒐𝒍𝒍𝒆𝒓 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝟑
𝟏𝟎
𝟑𝟎𝟎𝒎𝒓 𝟑
𝑪= × 𝟏𝟓𝟑𝟕. 𝟐𝟔𝑲𝒈𝒇 = 𝟖𝟓𝟎𝟖. 𝟖 𝑲𝒈𝒇
𝟏𝒎𝒓
Selection of bearing
𝒇𝒐𝒓 𝒅𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝟖𝟓𝟎𝟖. 𝟖 𝑲𝒈𝒇 𝒔𝒆𝒍𝒆𝒄𝒕 𝒔𝒖𝒊𝒕𝒂𝒃𝒍𝒆
𝑻𝒂𝒑𝒆𝒓 𝒓𝒐𝒍𝒍𝒆𝒓 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐𝟓 & 𝑃𝑆𝐺 4.26 𝑇𝑎𝑝𝑒𝑟 𝑟𝑜𝑙𝑙𝒆𝒓 𝑩𝒆𝒂𝒓𝒊𝒏𝒈 𝟑𝟐𝟐𝟏𝟏𝑨 𝒐𝒓 𝟑𝟐𝟑𝟎𝟖𝑨 𝒎𝒆𝒆𝒕𝒔
𝒕𝒉𝒆 𝒅𝒆𝒔𝒊𝒈𝒏 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕𝒔
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟓𝟓𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟖𝟗𝟎𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟕𝟗𝟖𝟎𝑲𝒈𝒇
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒑𝒎 𝒊𝒔 𝟑𝟔𝟎𝟎
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟒𝟎𝟎𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟗𝟖𝟎𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟖𝟕𝟓𝟎𝑲𝒈𝒇
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒓𝒑𝒎 𝒊𝒔 𝟒𝟎𝟎𝟎
Reliability of selected bearing
𝟏
𝟏 𝒃
𝑳 𝒍𝒏
𝒑
=
𝑳𝟏𝟎 ′ 𝟏
𝒍𝒏
𝒑𝟏𝟎
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳 = 𝟑𝟎𝟎 𝒎𝒓
𝒑 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒕𝒚 𝒐𝒇 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝒑𝟏𝟎 = 𝟎. 𝟗
𝟏 𝟑
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟏𝟎
′
𝑪= 𝑷 ⇒ 𝟖𝟗𝟎𝟎 = × 𝟏𝟓𝟑𝟕. 𝟐𝟎𝟔 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟑𝟒𝟖. 𝟒𝟗
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟑𝟒𝟖. 𝟒𝟗𝒎𝒓
𝟏 𝟏
𝟏 𝒃 𝟏 𝟏.𝟏𝟕
𝑳 𝒍𝒏 𝟑𝟎𝟎 𝒍𝒏
𝒑 𝒑
= ⇒ = ⇒ 𝒑 = 𝟎. 𝟗𝟏𝟓
𝑳𝟏𝟎 ′ 𝟏 𝟑𝟒𝟖 𝟏
𝒍𝒏 𝒍𝒏
𝒑𝟏𝟎 𝟎. 𝟗
𝟏 𝟑
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟏𝟎
′
𝑪= 𝑷 ⇒ 𝟗𝟖𝟎𝟎 = × 𝟏𝟓𝟑𝟕. 𝟐𝟎𝟔 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟒𝟖𝟎. 𝟒𝒎𝒓
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟑𝟒𝟖. 𝟒𝟗𝒎𝒓
𝟏 𝟏
𝟏 𝒃 𝟏 𝟏.𝟏𝟕
𝑳 𝒍𝒏 𝟑𝟎𝟎 𝒍𝒏
𝒑 𝒑
= ⇒ = ⇒ 𝒑 = 𝟎. 𝟗𝟒
𝑳𝟏𝟎 ′ 𝟏 𝟒𝟖𝟎. 𝟒 𝟏
𝒍𝒏 𝒍𝒏
𝒑𝟏𝟎 𝟎. 𝟗
Problem-4
The operating schedule of ball bearing is as follows
Radial load of 1650N at 2000rpm for 5% of life time
Radial load of 1140N at 3300rpm for 15% of life time
Radial load of 560N at 1750rpm for 35% of life time
Radial load of 445N at 2200rpm for 45% of life time
The inner ring rotates and load is steady. Life is 10years for 2hrs per day of
operation. Select suitable ball bearing.
𝑷𝟏 𝟑 𝒏𝟏 + 𝑷𝟐 𝟑 𝒏𝟐 + 𝑷𝟑 𝟑 𝒏𝟑 + 𝑷𝟒 𝟑 𝒏𝟒
𝑷𝒎 =
∑𝒏
𝟓
𝒏𝟏 = 𝟐𝟎𝟎𝟎 × = 𝟏𝟎𝟎𝒓𝒑𝒎
𝟏𝟎𝟎
𝟏𝟓
𝒏𝟐 = 𝟑𝟑𝟎𝟎 × = 𝟒𝟗𝟓𝒓𝒑𝒎
𝟏𝟎𝟎
𝟑𝟓
𝒏𝟑 = 𝟏𝟕𝟓𝟎 × = 𝟔𝟏𝟐. 𝟓𝒓𝒑𝒎
𝟏𝟎𝟎
𝟒𝟓
𝒏𝟒 = 𝟐𝟐𝟎𝟎 × = 𝟗𝟗𝟎𝒓𝒑𝒎
𝟏𝟎𝟎
(𝟏𝟔𝟓𝟎𝟑 × 𝟏𝟎𝟎) + (𝟏𝟏𝟒𝟎𝟑 × 𝟒𝟗𝟓) + (𝟓𝟔𝟎𝟑 × 𝟔𝟏𝟐. 𝟓) + (𝟒𝟒𝟓𝟑 × 𝟗𝟗𝟎)
𝑷𝒎 =
𝟏𝟎𝟎 + 𝟒𝟗𝟓 + 𝟔𝟏𝟐. 𝟓 + 𝟗𝟗𝟎
𝒏= 𝒏 = 𝟏𝟎𝟎 + 𝟒𝟗𝟓 + 𝟔𝟏𝟐. 𝟓 + 𝟗𝟗𝟎 ≅ 𝟐𝟐𝟎𝟎 𝒓𝒑𝒎
𝒎
Find Equivalent Load
Since bearing has to carry radial load deep groove ball bearing has to
be selected From PSG4.1
𝑷 = (𝑿 𝑭𝒓 + 𝒀 𝑭𝒂 )𝑺 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒 𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒓𝒂𝒕𝒊𝒐, 𝒆 𝒂𝒏𝒅 𝒕𝒚𝒑𝒆 𝒐𝒇 𝒃𝒆𝒂𝒓𝒊𝒏𝒈,
𝑭𝒓
𝒈𝒆𝒕 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒐𝒇 𝒓𝒂𝒅𝒊𝒂𝒍 𝒇𝒂𝒄𝒕𝒐𝒓 𝑿 𝒂𝒏𝒅 𝒂𝒏𝒅 𝒕𝒉𝒓𝒖𝒔𝒕 𝒍𝒐𝒂𝒅 𝒀
𝑭𝒂
= 𝟎;
𝑭𝒓
𝑭𝒂
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟒, 𝒇𝒐𝒓 𝒆 = 𝟎. 𝟑𝟕; > 𝑒 ⇒ 𝑋 = 1 , 𝑌 = 0,
𝑭𝒓
𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐, 𝑺𝒆𝒓𝒗𝒊𝒄𝒆 𝒇𝒂𝒄𝒕𝒐𝒓, 𝑺 = 𝟏. 𝟓
𝑷 = (𝟏 × 𝟖𝟓𝟔𝑵) + (𝟎 × 𝟎) × 𝟏. 𝟓 = 𝟏𝟐𝟖𝟒𝑵 = 𝟏𝟑𝟎. 𝟖𝟖𝑲𝒈𝒇
Find Dynamic load carrying capacity
𝟏
𝑳 𝑲
𝑪= 𝑷 … 𝑭𝒓𝒐𝒎 𝑷𝑺𝑮 𝟒. 𝟐
𝑳𝟏𝟎
𝑳 𝒊𝒔 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒊𝒏 𝒎𝒓 𝒐𝒓 (𝒐𝒏𝒆 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒓𝒆𝒗𝒐𝒍𝒖𝒕𝒊𝒐𝒏𝒔)
𝑳 = 𝟕𝟗𝟏. 𝟏𝒎𝒓
𝑳𝟏𝟎 𝒊𝒔 𝒍𝒊𝒇𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒇𝒐𝒓 𝟗𝟎% 𝒔𝒖𝒓𝒗𝒊𝒗𝒂𝒍 𝒂𝒕𝟏𝒎𝒓
𝑳𝟏𝟎 = 𝟏𝒎𝒓
𝑲 = 𝟑 𝒇𝒐𝒓 𝒃𝒂𝒍𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈
𝟏
𝟕𝟗𝟏. 𝟏𝒎𝒓 𝟑
𝑪= × 𝟏𝟑𝟎. 𝟖𝟖𝑲𝒈𝒇 = 𝟏𝟐𝟏𝟎. 𝟒𝟓𝑲𝒈𝒇
𝟏𝒎𝒓
Selection of bearing
You can Deep groove ball bearing or self aligning bearing as there is no axial
load on bearing. Only radial loads are acting.
Medium series Deep groove ball bearing SKF6305 is selected
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟐𝟓𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟏𝟔𝟔𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟏𝟒𝟒𝟎𝑲𝒈𝒇
From Self aligning ball bearing SKF2207K is selected
𝒃𝒐𝒓𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓, 𝒅 = 𝟑𝟓𝒎𝒎,
𝑫𝒚𝒏𝒂𝒎𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 , 𝑪 = 𝟏𝟕𝟎𝟎𝑲𝒈𝒇 ,
𝑺𝒕𝒂𝒕𝒊𝒄 𝒍𝒐𝒂𝒅 𝒄𝒂𝒓𝒓𝒚𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚, 𝑪𝟎 = 𝟖𝟎𝟎𝑲𝒈𝒇
Reliability of selected bearing
𝟏 𝟏
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟑
′
𝑪= 𝑷 ⇒ 𝟏𝟔𝟔𝟎 = × 𝟏𝟑𝟎. 𝟖𝟖 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟐𝟎𝟒𝟎. 𝟑𝟒 𝒎𝒓
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟑𝟒𝟖. 𝟒𝟗𝒎𝒓
𝟏 𝟏
𝟏 𝒃 𝟏 𝟏.𝟏𝟕
𝑳 𝒍𝒏 𝟕𝟗𝟏. 𝟏 𝒍𝒏
𝒑 𝒑
= ⇒ = ⇒ 𝒑 = 𝟎. 𝟗𝟔𝟔
𝑳𝟏𝟎 ′ 𝟏 𝟐𝟎𝟒𝟎. 𝟑𝟒 𝟏
𝒍𝒏 𝒍𝒏
𝒑𝟏𝟎 𝟎. 𝟗
𝟏 𝟏
𝑳𝟏𝟎 ′ 𝑲 𝑳𝟏𝟎 ′ 𝟑
′
𝑪= 𝑷 ⇒ 𝟏𝟕𝟎𝟎 = × 𝟏𝟑𝟎. 𝟖𝟖 ⇒ 𝑳 = 𝑳𝟏𝟎 = 𝟐𝟏𝟗𝟏. 𝟒𝟐 𝒎𝒓
𝑳𝟏𝟎 𝟏𝒎𝒓
𝑳𝒊𝒇𝒆 𝒐𝒇 𝒔𝒆𝒍𝒆𝒄𝒕𝒆𝒅 𝒃𝒆𝒂𝒓𝒊𝒏𝒈, 𝑳𝟏𝟎 = 𝟑𝟒𝟖. 𝟒𝟗𝒎𝒓
𝟏 𝟏
𝟏 𝒃 𝟏 𝟏.𝟏𝟕
𝑳 𝒍𝒏 𝟕𝟗𝟏. 𝟏 𝒍𝒏
𝒑 𝒑
= ⇒ = ⇒ 𝒑 = 𝟎. 𝟗𝟔𝟖
𝑳𝟏𝟎 ′ 𝟏 𝟐𝟏𝟗𝟏. 𝟒𝟐 𝟏
𝒍𝒏 𝒍𝒏
𝒑𝟏𝟎 𝟎. 𝟗
