Compute the contact stress number for the gear pair described in Example Problem 9_1.

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Accepted Answer

Data from Example Problem 9_1 are summarized as follows: [latex]\begin{aligned}N_{P} &=20 & N_{G} &=70, & F &=1.50 \mathrm{in} & W_{l} &=720 \mathrm{lb} & D_{P} &=2.500 \mathrm{in} \K_{c} &=1.50 & K_{\mathrm{x}} &=1.00 & K_{m}&=1.19 & K_{v} &=1.45\end{aligned}[/latex]

The gear teeth are 20°. full-depth, involute teeth. We also need the geometry factor for pitting resistance, I, From Figure 9_23(a), at a gear ratio of [latex]m_{G}[/latex]= [latex]N_{G}[/latex]/[latex]N_{p}[/latex]= 70/20 = 3.50 and for[latex]N_{p}[/latex]= 20, we read I = 0.108, approximately.The design analysis for bending strength indicated that two steel gears should be used. Then, from Table 9_9,

				Gear material and modulusof elasticity.[latex]E_{G},Ib/in^{2} [/latex] (MPa)
Pinion material	Modulus of elasticity.[latex]E_{p},Ib/in^{2} [/latex] (MPa)	Steel 30×[latex]10^{6}[/latex] (2×[latex]10^{5}[/latex])	Malleable iron 25×[latex]10^{6}[/latex] (1.7×[latex]10^{5}[/latex])	Nodular iron 24×[latex]10^{6}[/latex] (1.7×[latex]10^{5}[/latex])	Cast iron 22×[latex]10^{6}[/latex] (1.5×[latex]10^{5}[/latex])	Aluminum bronze 17.5×[latex]10^{6}[/latex] (1.2×[latex]10^{5}[/latex])	Tin bronze 16×[latex]10^{6}[/latex] (1.1×[latex]10^{5}[/latex])
Steel	30×[latex]10^{6}[/latex]	2300	2180	2160	2100	1950	1900
	(2×[latex]10^{5}[/latex])	(191)	(181)	(179)	(174)	(162)	(158)
Mall, iron	25×[latex]10^{6}[/latex]	2180	2090	2070	2020	1900	1850
	(1.7×[latex]10^{5}[/latex])	(181)	(174)	(172)	(168)	(158)	(154)
Nod. iron	24×[latex]10^{6}[/latex]	2160	2070	2050	2000	1880	1830
	(1.7×[latex]10^{5}[/latex])	(179)	(172)	(170)	(166)	(156)	(152)
Cast iron	22×[latex]10^{6}[/latex]	2100	2020	2000	1960	1850	1800
	(1.5×[latex]10^{5}[/latex])	(174)	(168)	(166)	(163)	(154)	(149)
Al. bronze	17.5×[latex]10^{6}[/latex]	1950	1900	1880	1850	1750	1700
	(1.2×[latex]10^{5}[/latex])	(162)	(158)	(156)	(154)	(145)	(141)
Tin bronze	16×[latex]10^{6}[/latex]	1900	1850	1830	1800	1700	1650
	(1.1×[latex]10^{5}[/latex])	(158)	(154)	(152)	(149)	(141)	(137)

we find that [latex]C_{p}[/latex]= 2300. Then the contact stress number is [latex]s_{c}=C_{p} \sqrt{\frac{W_{l} K_{o} K_{s} K_{m} K_{v}}{F D_{p} I}}=2300 \sqrt{\frac{(720)(1.50)(1.0)(1.19)(1.45)}{(1.50)(2.50)(0.108)}}[/latex]

[latex]s_{c}=156000 psi[/latex]

Question 9.EP.3: Compute the contact stress number for the gear pair describe...

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