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Strength of Materials, Summaries of Mathematics

Strength of Materials: solutions to torsion problems

Typology: Summaries

2023/2024

Uploaded on 03/05/2025

josel-mer-florida
josel-mer-florida 🇵🇭

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@ the gears attached tothe fixed-end steei snaft are subjected to the torques. if the Snean modulus of eloshcity is 80 GPa, ana the Shaft has a diameter of |4mm, derermin€ the displacement of tooth Pon gear A. the Shafi turns freely within the bearing ai B- Enter the amount in mm. . lanm 220nm 4onm To _ A ey ts is Tane Y ten Tpe (50 YY Y, 10 Hh 'To fo find Baye = TL G = 80 GPa JG find J first : Sat el. . I= F (0) 5 (14mm) J= 3771. 481981 @ a oe as LI AC = (150x107) Nm (0-4 x10?) o (- 20x10?) umm (0.3x10?) ® t (71'- 48/981 ) (€0,c00) © T (= ttoxt0? Ynmm) (0-9x10?) (271. 481981) (80, 00 ) 8 Ale = -0.212 rad = 0-212 rad G\ — convert io degrees 8 wfe = 0.2)ar( 10) 2 ip ype ( Trad ) es Displacement q= cama“? = 0.2/2( comm) = | 21-2 mm | A tubular qluminum bar (G+ 4xlo psi) of Square Cross sechON with Outer dimensions 2in 2in must resist a torque T= 3000 bin. Calculate the minimum requireq wail thickness t mintf the allowabie Shear stress js 4500 psi qnd the allowable rare of twist 15 0.01 rad/ft - In put the numerical answer only in inchess. civen : +. G= 4x10% psi #in G = Din b = 2in U = 4o00 psi 2in T = 3000 Ib'in = outer — thickness) * & = O01 rad/ rt leh c= thickness 6-0.01 rad (4 ) Siac) alae)? el =) it) ® £9032 x14 @ 6 = 445 ARCT + w] respect to shear stress * Sa toe eee 2Ab ZAU to find § bs e008 Ib “Ih 2( 2-4)? (40) tigate t= See tarde) 2 (4500) tA -4t tt) = _2000_ 2 (4500) 4t- 462442 = 3000 2( 4500) C2771. 401981) (90,000) + = 0.0915 (7 —+ w/ resped i Snear Sress £ with eespect fo tate of mul st te Te = t(b-t = b= Tt al oe eniane 66 t(2-tP =__so0 (4x10*) ( 9.28ax1074) 3 t(2- t)= 9/0 t = 0-140 in Always choose the larger t .