LECTURE+35+COE-3001-A

# Tech aerospace ql x0 2 q x2 qlx 0 2 2 20 example

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Unformatted text preview: TION OF THE SHEAR FORCE AND LOAD EQUATIONS EIv 000 = V (8- 12b) EIv 0000 = q (8- 12c) •  Since loads are usually known quan??es, whereas bending moments must be determined from free- body diagrams and sta?cs, it is oZen desirable to work with the load equa?on ☞  Many computer programs for ﬁnding deﬂec?on are based on load equa?on ☞  Numerical integra?ons are then performed to obtain shear forces, bending moments, slopes, and deﬂec?ons ☞  Procedure for close form solu?ons (by hand) involves mul?ple integra?ons ☞  But we now get more arbitrary constants of integra?on… ☞  3 constants in the case of (8- 12b) and 4 constants in the case of (8- 12c) ☞ Constants are determined in the usual way from boundary, con?nuity, and symmetry 4/9/13 M. Mello/Georgia Tech Aerospace 25 EXAMPLE 8- 4 Problem: Can?lever beam suppor?ng a triangularly Distributed load of maximum intensity q. Beam has Length L and ﬂexural rigidity EI. Determine: 1.  equaBon of the deﬂecBon curve 2.  maximum deﬂecBon δB at the free end 3.  angle of rotaBon ΘB at the free end SOLUTION: solve by integraBng 4th order load equaBon 1.  First we need to describe q(x): q ( x) = q0 ( L x) L 2.  and so we can write: EIv 0000 = q= q0 ( L x ) L Deﬂec?ons of a can?lever beam With a triangular load 3.  and then begin integraBng away… unBl the cows come home 4/9/13 M. Mello/Georgia Tech Aerospace 26 EXAMPLE 8- 4 (cont.) EIv 0000 EIv 000 q0 ( L x ) L = q= = q0 ( L x) 2 + C1 2L ☞ Determine C1 ☞ Must now recognize that the right- hand- side is V(x)!! (equa?on 8- 12b) 000 EIv = V (8- 12b) ☞ Once this is established we look for a shear condi?on ☞ invoke V(L) = 0 (no shear force at the free end of the beam) 0 = q0 (L L)2 + C C1 = 0 1 2L 000 q0 ( L x ) 2 Hence, V = EIv = (shear force distribu?on in beam) 2L 4/9/13 M. Mello/Georgia Tech Aerospace 27 EXAMPLE 8- 4 (cont.) •  Next, begin to integrate again and con?nue the process V = EIv 00 EIv = 000 = q0 ( L x ) 2 2L q0 ( L x ) 3 + C2 6L ☞ Must now recognize that the right- hand- side is M(x)!! (equa?on 8- 12a) ☞ Once this is established we look for a bending- moment condi?on ☞ invoke M(L) = 0 (no bending moment at the free end of the beam) q0 ( L L ) 3 0= + C2 C2 = 0 6L 00 q0 ( L x ) 3 Hence, M = EIv = (bending- moment distribu?on in beam) 6L 4/9/13 M. Mello/Georgia Tech Aerospace 28 EXAMPLE 8- 4 (cont.) •  3rd and 4th integra?ons yield: EIv = q0 (L 24L EIv = q0 (L 120L 0 x) 4 + C3 x) 5 + C3 x + C 4 •  Apply boundary condi?ons at the ﬁxed support (x=0) where v’(0) = 0 and v(0) = 0 C3 = q0 L4 q0 L 3 and C4 = 120 24 Slope of the beam: 0 v= q0 x (4L3 24LEI 6L2 x + 4Lx2 x3 ) ✓B = q0 L 3 v ( L) = 24EI 0 deﬂec?on of the beam: v= 4/9/13 q0 x 2 (10L3 1204LEI 10L2 x + 5Lx2 x3 ) M. Mello/Georgia Tech Aerospace max = q0 L 4 v ( L) = 30EI 29...
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## 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.

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