exam 2 sp 10 solution

exam 2 sp 10 solution - with contents, of q . What is the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
P L/2 L/2 L/3 Find the reaction and the bending moment at the interior pier EI EI Name______________________ 2L/3 P/2 “P”= P “P” = P/2 “P” = R v P v P/2 v R x x x a=1.5L a=1.667L a=L b=.5L b=L b=.333L “L” = 2L
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
P v R = R (2L)3/(48 EI) = RL3/6EI v P/2 = (P/2)(L/3)L[(2L)2 –(L/3)2 –L2]/[6(2L)E = .0401 PL3/EI v P = (P (L/2) L [(2L)2 –(L/2)2 – L2] / [6 (2L) EI = .1146 PL3/EI R P/2 + P = R R = .928P Mint pier = -.132 PL P/12 .75P .5R V P/2 5P/12 .25P .5R M .368P -.632P .298P -.202P .184PL .067PL -.132PL
Background image of page 2
Name______________________ d t L L The cylindrical pressure vessel shown has an internal pressure of “p” and has a total weight per unit length,
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: with contents, of q . What is the state of stress at point A , at the top of the vessel, over the interior support ? A M-qL2/8 R = 1.25 qL .375 qL .375 qL I 2245 pd3t/8 bend = M (-d/2) / I = (-qL2/8)(-d/2) / I hoop = p(d/2)/t long = p(d/2)/2t A hoop long + bend v R = R (2L)3/(48 EI) = RL3/6EI v q = (5/384)q (2L)4/ EI = .2083qL4/EI R = 1.25 qL...
View Full Document

This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

Page1 / 3

exam 2 sp 10 solution - with contents, of q . What is the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online