Upcoming SlideShare
×

# H#8

5,344 views

Published on

Published in: Health & Medicine, Business
3 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
5,344
On SlideShare
0
From Embeds
0
Number of Embeds
7
Actions
Shares
0
57
0
Likes
3
Embeds 0
No embeds

No notes for slide

### H#8

1. 1. Problem Number (3.136)<br />A 3-m-long steel angle has an L203*152*12.7 cross section. Where the thickness of the section is 12.7 mm and that its area is 4350 mm2. Knowing that ιall = 50 MPa and that G = 77.2 GPa and ignoring the effect of stress concentration, determine (a) the largest torque T that can be applied. (b) the corresponding angle of twist.<br />Solution<br />τ= T tJ , T= τ Jt , J= b t33 , T= τ b t23<br />J= 0.203+0.152×(0.0127)33=2.424 ×10-7<br />T= 50 ×106 ×2.424 × 10-70.0127=954.3 N.m<br />ϕ= T LJ G= 954.3 ×32.424 × 10-7 ×77.2 ×109=0.153rad=8.77o <br />Problem Number (3.138)<br />A 3-m-long steel member has a W250*58 cross section. Knowing that G = 77.2 GPa and that the allowable shearing stress is 35 MPa, determine (a) the largest torque T that can be applied. (b) the corresponding angle of twist. Knowing that the dimensions of flange is 203*13.5 and the dimensions of web is 252*8. <br />Solution<br />τ= T tJ , T= τ Jt , J= b t33 , T= τ b t23<br />J1= 0.203 ×(0.0135)33+ 0.203 ×(0.0135)33=3.33 ×10-7<br />J2= 0.252 ×(0.008)33=4.3008 × 10-8<br />Jt= J2+ J1=3.76 × 10-7<br />T1= 35 × 106 ×3.33 × 10-70.0135=863.26 N.m <br />T2= 35 × 106 × 4.3008 × 10-80.008=188.16 N.m<br />Tt= T1+ T2=863.26+188.16=1051.42 N.m<br />ϕ= T LJ G= 1051.42 ×33.76 × 10-7 ×77.2 ×109=0.1087rad=6.23o<br />Problem Number (3.139)<br />A torque T = 5 KN.m is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b.<br />Solution<br />area=115×69= 7935 mm2<br />τa= T2ta= 50002 ×0.006 × 7935 × 10-6=52.51 MPa<br /> τb= T2ta= 50002 ×0.01 × 7935 × 10-6=31.51 MPa<br />Problem Number (3.142)<br />A 90-N.m torque is applied to a hollow shaft having the cross section shown. Neglecting the effect of stress concentrations, determine the shearing stress at points a and b <br />Solution <br />area=52 ×13+ 39 ×13+ 3.14 ×39×394= 2376.985 mm2<br />τa= T2ta= 902 ×0.004 × 2.376985× 10-3=4.73 MPa<br />τb= T2ta= 902 ×0.002 × 2.376985× 10-3=9.46 MPa<br />Problem Number (3.148)<br />A cooling tube having the cross section shown is formed from a sheet of stainless steel of 3-mm thickness. The radii c1 = 150 mm and c2 = 100 mm are measured to the center line of the sheet metal. Knowing that a torque of magnitude T = 3 KN.m is applied to the tube, determine (a) the maximum shearing stress in the tube. (b) the magnitude of the torque carried by the outer circular sell. Neglect the dimension of the small opening where the outer and inner shells are connected.<br />Solution<br />area= 3.14 0.152- 0.12- (0.003×0.05) = 0.0391 m2<br />τ= T2ta= 30002 ×0.003 × 0.0391=12.79 MPa<br />T=2τta=2 ×12.79 ×106 ×0.003 ×4.239 ×10-4=5.4 KN.m<br />Problem Number (3.149)<br />Equal torques are applied to thin-walled tubes of the same length L same thickness t, and same radius c. one of the tubes has been slit length-wise as shown. Determine (a) the ratio τbτa of the maximum shearing stresses in the tubes, (b) the ratio ϕbϕa of the angles of twist of the shafts.<br />Solution<br />for a : Ta=2τatA , ϕa= TL2AGt2<br />for b : Tb= τb Jt , ϕb= TLJG<br />Ta= Tb<br />2τatA= τb Jt , J= Lt33<br />6τaA= τb L t<br />τbτa= 6ALt , A=3.14 c2<br />τbτa= 18.84 c2Lt<br />ϕbϕa= TLJG × 2AGt2TL= 2At2J= 6ALt= 18.84 c2Lt<br />ϕbϕa= 18.84 c2Lt<br />