This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II SOLUTIONS FOR HOMEWORK NO. 4 Problem 1 (20 points) Part A: Because the beam is still in elastic region, the stress eld can be expressed as: ( y ) = yM y I (1) Since the most highly stressed region is at the verge of yielding, we have  ( y ) max  =  yM y I  y = h/ 2 = y M y = y I h/ 2 (2) For the diamondorientation beam, I is calculated as: h h 4 h 2 2 dA = 2 ( y ) 2 dy = I (3) = y 12 2 0 The area moment of inertia for this diamondorientation is the same as the crosssection appears as a square. Substitution of I into Eq. 2, we get that: M 2 y h 3 y ( diamond ) = (4) 12 Part B: h h 3 h 2 2 y M L ( diamond ) = y ( y ) dA = 2 y y ( y ) 2 dy = (5) 6 2 0 Part C: M L ( diamond ) 2 y h 3 6 = = 2 (6) 2 y h 3 M y ( diamond ) 12 When the crosssection appears as a square, M y is calculated as: y I y h 4 / 12 y h 3 M = = = y ( square ) y max h/ 2 6 1 (7) and M L is: h M h 3 L ( square ) = ( y ) dA = 2 y y hdy = (8) 2 y 4 0 So the ratio of M L /M y for squareorientation is: M L ( square ) y h 3 3 4 M = = (9) y ( square ) y h 3 2 6 Part D: The ratio of the rstyield bending moments for the two orientations is: M y ( diamond ) 2 y h 3 12 M = = 1 / 2 (10) y ( square ) y h 3 6 The rstyield bending moment is calculated by: y I y M y = (11) max Since y is a constant, and I is the same for these two orientations, the above ratio is only determined by the ratio of y max . The ratio of limit moment for the two orientations is: M L ( diamond ) 2 y h 3 2 6 M = = 2 < 1 . (12) L ( square ) y h 3 3 4 The limit moment for a symmetrical crosssection is calculated by: M L = 2 y ydA (13) A/ 2 where A/ 2 is the 1 / 2 area of the cross section in which y 0. For the diamondorientation, though the maximum value of y is 2 times larger than the squareorientation, the major part of its crosssection is located at the region with small y coordinate. Thus the ratio of limit moment calculated above is smaller than one. Since M y ( diamond ) < M y ( square ) and M L ( diamond ) < M L ( square ) , we should use the beam in the squareorientation....
View
Full
Document
This note was uploaded on 02/23/2012 for the course MECHANICAL 2.002 taught by Professor Davidparks during the Spring '04 term at MIT.
 Spring '04
 DavidParks
 Stress

Click to edit the document details