This preview shows page 1. Sign up to view the full content.
Unformatted text preview: is 2 2, then ⇡ 2 EI2
Pcr =
(n = 1) (L/2)2 ☞ Subs>tute numerical values for E=200 GPa, I2 = 163 cm4, and (L/2) = 4m to obtain: Pcr = 200kN ☞ If column buckles out of plane (about strong axis 1 1, then ⇡ 2 EI1 ‘
Pcr = L2 ☞ Subs>tute numerical values for E=200GPa, I1 = 3060 cm4, and (L) = 8m to obtain: Pcr = 943.8kN ☞ Cri>cal load is smaller of the two values… Pcr = 200kN 4/17/13 x x x 1 1 (strong axis) out of plane buckling 2 2 (weak axis) In plane buckling M. Mello/Georgia Tech Aerospace EXAMPLE 9 1 (ﬁnish) x Pcr ☞ CRITICAL STRESSES: cr =
A ☞ check to see if both cri>cal load es>mates are s>ll below the propor>onal limit of the material pl = 300M P a 943.8kN =
= 238.9M P a cr
39.5cm2 ☞ conclusion: both cri>cal load es>mates are sa>sfactory Pcr
☞ allowable load: Pallow = where n= 2.5 n Pallow = 79.9kN 4/17/13 x x 1 1 (strong axis) out of plane buckling 2 2 (weak axis) In plane buckling M. Mello/Georgia Tech Aerospace Columns with other support condiFons Case 1: Case 2: column with both ends pinned column ﬁxed at base and free at top Case 3: column with both ends ﬁxed against rota>on Case 4: column ﬁxed at the base and pinned at the top ☞ SO WHAT IS Pcr FOR THESE OTHER 3 CASES?? Pcr ⇡ 2 EI
=
☜ so far we have solved this case only L2
4/17/13 M. Mello/Georgia Tech Aerospace Columns with other support condiFons ☞ In each case a unique 2nd order diﬀeren>al equa>on arises as a consequence of the bending moment rela>onship (which is uniquely deﬁned for each case) ☞ The ODE’s which result in cases 2 4 are only slightly more complicated than the one we encountered for the column with 2 pinned ends. ☞ Each of the the ode’s contains a combina>on of a homogeneous and a par>cular solu>on which lead to a unique Pcr once the boundary condi>ons are subs>tuted in order to determine the arbitrary constants ☞ In each successive case Pcr is seen to increase in response to greater and greater constraint. 4/17/13 M. Mello/Georgia Tech Aerospace Summary of results for all support condiFons ☞ cri>cal loads, cri>cal lengths, and cri>cal length factors ☞ cri>cal length is the length of the equivalent pinned end column 4/17/13 M. Mello/Georgia Tech Aerospace EXAMPLE 9 2 GIVEN: A viewing planorm is supported by a row of Aluminum pipe columns having length L = 3.25m and outside diameter d = 100 mm. The columns are designed to support compressive loads of P=100kN. Also E= 72GPa and σpl = 480MPa. DETERMINE: Minimum required thickness (t) of the columns if a safety factor of n=3 is required with respect to Euler buckling. SOLUTION: Model each column as a ﬁxed pinned column Pcr 4/17/13 2.046⇡ 2 EI
=
L2 ⇡⇥ 2
A=
d
(d
4
A = 1999mm2 ☜ Case (d) in table of previous slide M. Mello/Georgia Tech Aerospace 2t) 2 ⇤ EXAMPLE 9 2 ⇡⇥ ⇤ ☞ moment of iner>a: I =
d4 d4
i
64 o ⇤
⇡⇥ 4 I=
d
( d 2 t) 4
64 ⇡⇥
I=
(0.1m)4 (0.1m 64 ☞ Invoke safety factor: Pcr = nPallow 2t)4 ⇤ Pcr = 3(100kN ) = 300kN
2.046⇡ 2 EI
= 2 and solve for thickness (t). L ☞ Now subs>tute into Pcr t = 0.006825m = 6.83mm ☜ ANSWER ☞ Check cri>cal stress in column against propor>onal limit σpl = 480MPa cr = 4/17/13 Pcr
A cr = 300kN
1999mm2 cr = 150M P a M. Mello/Georgia Tech Aerospace () cr < pl )...
View
Full
Document
This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Institute of Technology.
 Spring '09
 ZHU

Click to edit the document details