Question

Lifetime of the bearings. A shaft is supported by a pair of bearing 30307. The shaft works under the load shown in figure, $F_r$=2500N, $F_a$=1000N. $F_p$=1.1. Please determine : Which bearing have the short lifetime? For bearing 30307 , $C_r$=39800N, $C_e$=35200N, $F_d = \frac{F_r}{2Y}$; e $\frac{F_a}{F_r}$ ? e $\frac{F_a}{F_r}$ > e 0.32 X Y X Y 1 0 0.4 1.9

          Lifetime of the bearings.
A shaft is supported by a pair of bearing 30307. The
shaft works under the load shown in figure, $F_r$=2500N,
$F_a$=1000N. $F_p$=1.1. Please determine : Which bearing
have the short lifetime?
For bearing 30307 , $C_r$=39800N, $C_e$=35200N,
$F_d = \frac{F_r}{2Y}$;
e $\frac{F_a}{F_r}$ ? e $\frac{F_a}{F_r}$ > e
0.32 X Y X Y
1 0 0.4 1.9

Lifetime of the bearings.
A shaft is supported by a pair of bearing 30307. The
shaft works under the load shown in figure, Fr=2500N,
Fa=1000N. Fp=1.1. Please determine : Which bearing
have the short lifetime?
For bearing 30307 , Cr=39800N, Ce=35200N,
Fd = (Fr)/(2Y);
e (Fa)/(Fr) ? e (Fa)/(Fr) > e
0.32 X Y X Y
1 0 0.4 1.9

Added by Lydia L.

Mechanics of Materials

R. C. Hibbeler 10th Edition

Chapter 5

Instant Answer

Solved by Expert Narayan Hari

Step 1

The equivalent load is given by the formula: P = (X + Y) * F where X and Y are factors that depend on the load distribution on the bearings. In this case, X = 0.32 and Y = 0.4. For bearing 30307, the equivalent load is: P1 = (0.32 + 0.4) * 1000 = 720 N Show more…

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Please write down the answer on paper. Lifetime of the bearings: A shaft is supported by a pair of bearings 30307. The shaft works under the load shown in the figure, E = 2500 N, F = 1000 N. F = 1.1. Please determine: Which bearing has the shorter lifetime? For bearing 30307, c = 39800 N, c = 35200 N. 2Y F 0.32 X X 0.4 1.9