Wuhan University of Technology18.1 Beam flexure: elementary caseS3,4 =±aS1,2=±iaIncorporatingeachoftheserootsintoEq.(1811)separatelyandaddingtheresultingfourterms,oneobtainsthecompletesolutionΦ(r) =G1 exp(iar) + G2 exp(-iar) + G3 exp(ar) +G4 exp(-ar)o(r)=Aicosar+A2sinar+A3coshar+A4sinhaawhere A,Az,A3, and Ay are real constants which can be expressed in terms ofthe components of Gi, G2, G3, and G4. These real constants must be evaluatedsoastosatisfytheknownboundaryconditions(displacement,slope,momentorshear)attheendsofthebeam.18-6
18-6 Wuhan University of Technology 18.1 Beam flexure: elementary case Incorporating each of these roots into Eq. (1811) separately and adding the resulting four terms, one obtains the complete solution where A1, A 2, A 3, and A 4 are real constants which can be expressed in terms of the components of G1, G 2, G 3, and G 4. These real constants must be evaluated so as to satisfy the known boundary conditions (displacement, slope, moment, or shear) at the ends of the beam
Wuhan University of Technology18.1 Beam flexure: elementary caseto(x)xEI,m=constantsL(a)E(01=元mL401(t) =sin 元xLEI02=4元mL42元x02(x)=sinLEI03=9元mL3元(x)=sinL(b)18-7
18-7 Wuhan University of Technology 18.1 Beam flexure: elementary case
Wuhan University of Technology-18.1 Beam flexure: elementary caseExampleE181.SimpleBeamConsideringtheuniformsimplebeamshowninFigE181a,itsfourknownboundaryconditionsareb(0) = 0M(0) = EI @"(0) = 0Φ(L) = 0M(L)= EI"(L) =0Makinguseof Eq.(1815)and itssecondpartial derivativewith respecttox,Eqs.(a) can be written asΦ(O)=A1 cos0+A2 sin0+A3 cosh0+A4 sinh0= 0@"(0) = α2 (-A1 cos 0 - A2 sin0+ A3 cosh 0 + A4 sinh 0) = 018-8
18-8 Wuhan University of Technology Example E181. Simple Beam Considering the uniform simple beam shown in Fig. E181a, its four known boundary conditions are 18.1 Beam flexure: elementary case Making use of Eq. (1815) and its second partial derivative with respect to x, Eqs. (a) can be written as
Wuhan University of Technology-18.1 Beam flexure: elementary case(A1 +A3) = 0(-A1 + A3) = 0A1 = A3 = 0.Similarly, Eqs. (b) can be written in the formo(L)=A2 sinaL+A4 sinhaL=0d"(L)= a2 (-A2 sinaL + A4 sinhaL) = 02A4sinhaL=0(r) = A2 sinac18-9
18-9 Wuhan University of Technology 18.1 Beam flexure: elementary case Similarly, Eqs. (b) can be written in the form
Wuhan Universityof Technology18.1 Beam flexure: elementary caseA2 = 0. Φ(L) = 0sinaL=0which is the systemfrequency equation:it requiresthata=/Ln=0.1.2,..SubstitutingthisexpressionintoEg.(188)andtakingthesquarerootofbothsidesyieldthefrequencyexpressionEIWn=n2?㎡L4nTdn(α) = A2 sinn=1.2aL18-10
18-10 Wuhan University of Technology 18.1 Beam flexure: elementary case which is the system frequency equation: it requires that Substituting this expression into Eq. (188) and taking the square root of both sides yield the frequency expression