This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Then T= Example 4mm A stell bar G=200GPa of croos section shown is subject to a torque of 500 Nm Determine max shear stress? Solution J e= J e1+ J e2 = 1/3b 1 t 1 3 +1/3b 2 t 2 3 = 1/3*(100)*(10) 3 +1/3(125)*(4) 3 = 3,5787*10 4 mm 4 λmax= where t=t 1 = 500*(0,01)/3,578*10 4= 139,7 MPa θ= = 69,86*103 rad 10mm 100mm 125mm For this rectangular section force equation equals , Fz=0 λ 1 dzt 1 + λ 2 dzt 2 =0 λ 1 t 1= λ 2 t 2= q=shear flow and it is constant Area= the net force on the element is λ (ds.t)=qds the total torque T t 0 to (rcos )qds T=q 2dA T=2qA & λ = z Angle of Twist γL= θ r for ds element γL = θdsrcos = Θds=G Angle of Twist= θ = Θ= Θ= Θ= J e=...
View
Full
Document
This note was uploaded on 02/28/2011 for the course AEE 361 taught by Professor Daglas during the Spring '11 term at College of E&ME, NUST.
 Spring '11
 Daglas

Click to edit the document details