Design of a Shaft Key A shaft of diameter D rotates at 600 rpm and transmits 100 hp through a gear. A square key of width w is to be used (Figure 9.10(a)) as a mounting part. Determine the required length of the key. Given: D=50 mm, w=12 mm. Design Decisions: The shaft and key will be made of AISI

Question

Design of a Shaft Key

A shaft of diameter [latex] D [/latex] rotates at 600 rpm and transmits 100 hp through a gear. A square key of width [latex] w [/latex] is to be used (Figure 9.10(a)) as a mounting part. Determine the required length of the key.

Given: [latex] D [/latex]=50 mm, [latex] w [/latex]=12 mm.

Design Decisions: The shaft and key will be made of AISI 1035 cold-drawn steel having the same tensile and compressive yield strength and that yield strength in shear is [latex] S_{y s^{\prime}}=S_y / 2 [/latex]. The transmitted power produces intermittent minor shocks and a factor of safety of [latex] n [/latex]=2.5 is used.

Accepted Answer

From Table B.3, for AISI 1035 CD steel, we find [latex] S_y [/latex]=460 MPa. Through the use of Equation (1.16),

[latex] hp =\frac{F V}{745.7}=\frac{T n}{7121} [/latex] (1.16)

[latex] T=\frac{7121(100)}{600}=1.187 kN \cdot m [/latex]

The force [latex] F [/latex] at the surface of the shaft (Figure 9.13) is

[latex] F=\frac{T}{r}=\frac{1.187}{0.025}=47.48 kN [/latex]

On the basis of shear stress in the key,

[latex] \frac{S_y}{2 n}=\frac{F}{w L} \quad ext { or } \quad L=\frac{2 F n}{S_y w} [/latex] (9.23)

Substitution of the given numerical values yields

[latex] L=\frac{2(47,480)(2.5)}{460\left(10^6 ight)(0.012)}=43 mm [/latex]

Based on compression or bearing on the key or shaft (Figure 9.10(a)),

[latex] \frac{S_{y c}}{n}=\frac{F}{(w / 2)} \quad ext { or } \quad L=\frac{2 F n}{S_{y c} w} [/latex] (9.24)

Inasmuch as [latex] S_y=S_{y c^{\prime}} [/latex] this also results in [latex] L [/latex]=43 mm.

TABLE B.3 Mechanical Properties of Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels
UNS Number	AISI/ SAE Number	Processing	Ultimate $\text {Strength}^a$ $S_u$ (MPa)	Yield $\text {Strength}^a$ $S_y$ (MPa)	Elongation in 50 mm (%)	Reduction in Area (%)	Brinell Hardness ( $H_B$ )
G10060	1006	HR	300	170	30	55	86
		CD	330	280	20	45	95
G10100	1010	HR	320	180	28	50	95
		CD	370	300	20	40	105
G10150	1015	HR	340	190	28	50	101
		CD	390	320	18	40	111
G10200	1020	HR	380	210	25	50	111
		CD	470	390	15	40	131
G10300	1030	HR	470	260	20	42	137
		CD	520	440	12	35	149
G10350	1035	HR	500	270	18	40	143
		CD	550	460	12	35	163
G10400	1040	HR	520	290	18	40	149
		CD	590	490	12	35	170
G10450	1045	HR	570	310	16	40	163
		CD	630	530	12	35	179
G10500	1050	HR	620	340	15	35	179
		CD	690	580	10	30	197
G10600	1060	HR	680	370	12	30	201
G10800	1080	HR	770	420	10	25	229
G10950	1095	HR	830	460	10	25	248
Source: ASM Handbook, vol. 1, ASM International, Materials Park, OH, 2020.
Note: To convert from MPa to ksi, divide given values by 6.895.
${ } ^ {a}$ Values listed are estimated ASTM minimum values in the size range 18–32 mm.

Question 9.7: Design of a Shaft Key A shaft of diameter D rotates at 600 r......

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