hw21_Solutions - AAE 221 – Aerospace Structures II –...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AAE 221 – Aerospace Structures II – Spring 2004 Chapter 2.1 – Beam Bending and Extension Solutions Problem 0 1. (a) The Young’s modulus of the three sections are E1 = 3 ¥ 10 7 E 2 = E3 = 1 ¥ 10 7 First, the Young’s modulus of aluminum is chosen to be the reference modulus. The areas of each sub-component are A1 = 4at A2 = 8at A3 = 2at The modulus weighted area is ÊE A* =  Á ÁE Ëo ˆ ˜ Ai ˜ ¯i The distances from the origin of the coordinate axis to the m.w. centroid are found from ÊE A* z o* =  Á ÁE Ëo ÊE A* y o* =  Á ÁE Ëo ˆ ˜ ( z o ) i Ai ˜ ¯i ˆ ˜ ( y o ) i Ai ˜ ¯i where the quantity z o etc. are the distances from the origin to the centroid of each subcomponent. The location of the centroid is found to be z o* = 21a + 38t 248t = 22 22 * y o = 2a = 20t (due to symmetry) The moments and product of inertia can be found by first calculating the modulus weighted moments of inertia about our coordinate system origin, and then applying the parallel axis theorem to shift to the centroidal axis. The m.w. moments of inertia can be found from ÊEˆ 2 * I yo yo =  Á ˜ I yy + (z o ) A i ÁE ˜ Ë o¯ ÊEˆ 2 I z*o zo =  Á ˜ I zz + (y o ) A i ÁE ˜ Ë o¯ [ [ where the terms I yy , etc. are the moments of inertia of each sub-component about its own centroidal axis. For a symmetric rectangular piece, I yy = 13 bh 12 where the base b is parallel to the axis of the moment of inertia under consideration. For this problem, * I yo yo = I z*o zo 87400 4 t 3 344020 4 = t 3 From this, the m.w. moments of inertia are found by applying the parallel axis theorem. * * I yy = I yo yo - z o* A* * I zz = I z*o zo *2 o * () - (y ) A 2 Plugging in the appropriate values found previously, these were found to be * I yy = * I zz 427240 4 t 363 80020 4 = t 3 * The product of inertia, I yz = 0 due to symmetry. (b) For the various sub-components (aDT )1 = -1 ¥ 10 7 (aDT )2 = 0 (aDT )3 = 1 ¥ 10 7 The thermal axial load is ÊE N T = Eo  Á ÁE Ëo ˆ ˜[(aDT ) A]i ˜ ¯ The moments due to bending are ÊE M yT = - Eo  Á ÁE Ëo ÊE M zT = - Eo  Á ÁE Ëo ˆ ˜ z i [ DTA]i ˜a ¯i ˆ ˜ y i [ DTA]i ˜a ¯i Plugging in the appropriate values and centroidal distances, N T = -1 ¥ 10 6 t 2 M yT = 1.4 ¥ 10 7 t 3 M zT = 2.0 ¥ 10 7 t 3 Finally, the induced thermal stress may be found from s = EaDT Point A is in an aluminum sub-component, thus, E = 3 ¥ 10 7 psi. The thermal stress is ( s = ( ¥ 10 7 )1 ¥ 10 -3 )= 10000 psi 3 Point B is in the steel sub-component, thus E = 1 ¥ 10 7 psi. The thermal stress is ( s = ( ¥ 10 7 )- 1 ¥ 10 -3 )= -30000 psi 1 2. The area, as a function of x is : Ê 3 Ê x ˆˆ A( x) = a Á1 - Á ˜ ˜ Á 4L˜ Ë ¯¯ Ë 2 2 since the stress is s (x ) = du = dx P du =E A(x ) dx P E Ê 3 Ê x ˆˆ a Á1 - Á ˜ ˜ Á 4L˜ Ë ¯¯ Ë 2 2 du = dx P Ê 3 Ê x ˆˆ Ea Á1 - Á ˜ ˜ Á 4L˜ Ë ¯¯ Ë 2 2 Therefore, the displacement u(x) is P Ê 3 xˆ u (x ) = Á1 ˜ dx Ea 2 Ú Ë 4 L ¯ -2 integrating, u(x) becomes u ( x) = P Ea 2 Ê 4 L ˆÊ 3 x ˆ Á˜Á1 ˜ + C1 Ë 3 ¯Ë 4 L ¯ -1 We can apply the boundary condition of zero displacement at x=0 u (0) = 0 u (0) = P Ê 4L ˆ Á ˜ + C1 = 0 Ea 2 Ë 3 ¯ C1 = 4 PL 3Ea 2 The displacement u(x) is then Ê ˆ 4 PL Á 1˜ Á-1+ ˜ u ( x) = 3x˜ 3Ea 2 Á 1Á ˜ 4L¯ Ë To plot the displacement of the non-dimensional displacements divide u(x) by Ê ˆ ˜ u ( x) Ea 2 4 Á 1 ˜ = Á-1+ 3x˜ PL 3Á 1-1 Á ˜ 4L¯ Ë PL Ea 2 3. Cross section has width a and height 2a. The area of the cross-section is A = 2a3. Since the cross-section is symmetric, all the products of inertia (I*xy =0). I yy a ( 2a ) 3 2a 4 = = 12 3 Only u and w displacements occur since there is no bending in the y direction. The GDE for the problem are EAu ¢¢(x ) = - f x (x ) = -qo a EIyy w ¢¢( x ) = m¢ ( x ) + f z = 0 y since fz = 0 and my = qoa2 = const. † The boundary conditions are u(0) = 0 EAu¢( L) = 0 w(0) = 0 w( L) = 0 EIyy w ¢¢(0) = qo a 3 EIyy w ¢¢(0) = 0 (a) Starting from the GDE, † EAu ¢¢(x ) = -qo a EAu ¢(x ) = -qo ax + C1 Applying the boundary condition EAu ¢( L) = 0 , - qo aL + C1 = 0 C1 = qo aL Integrating once more, EAu ¢(x ) = -qo a x2 + qo aLx + C 2 2 Applying the second boundary condition, u(0)=0, u (0) = 0 C2 = 0 Therefore, Ê x2 ˆ EAu ( x) = qo aÁ Lx - ˜ Á 2˜ Ë ¯ 2 EAu ( x) 1 Ê x 1 Ê x ˆ ˆ = Á - Á ˜˜ 2Á L 2Ë L¯ ˜ qo L2 Ë ¯ (b) From the GDE for beam bending, EIyy w ¢¢¢¢( x ) = 0 EIyy w ( x ) = C1 x3 x2 + C2 + C3 x + C4 6 2 Applying the boundary condition w(0) = 0, w ¢¢(0) = 0 yields C4 = 0, C2 = 0 respectively. † † † The remaining boundary conditions furnish C1 and C3 yielding: C1 = q0 a 3 1 , C3 = - q0 a 3 L L 6 After substitution for †and simplification we get I 3 q0 L2 ÏÊ x ˆ Ê x ˆ¸ w( x ) = ÌÁ ˜ - Á ˜˝ 4 Ea ÓË L ¯ Ë L ¯˛ † 4. First consider the loading on the beam due to (a) steel and (b) aluminum. kg ˆ Ê (a) Á 7800 3 ˜ .05m 2 m¯ Ë ( Ê ) 10 m ˆ = 3900 N Á ˜ m Ë s2 ¯ Ê ) 10 m ˆ = 700 N Á ˜ m Ë s2 ¯ kg ˆ Ê (b) Á 2800 3 ˜ .025m 2 m¯ Ë ( To compute the moments of inertia I yy =  I zz =  I yz =  Ei 2 I yy + (z ) A i E0 [ Ei 2 I zz + (y ) A i E0 [ [ Ei I yz + (z )(y )A i E0 The location of the centroid can easily be determined from inspection and symmetry. I yy 2 È 210 Ê .25(.1)3 ˆ˘ Ê 2 Á (.025)˜˙ + Á .05(.5) = 2Í + .3 Á ˜ Á 12 Í 70 Ë 12 ¯˙ Ë Î ˚ ˆ ˜ = .0141458m 4 ˜ ¯ È 210 Ê .1(.25)3 ˆ˘ Ê .5(.05)2 Á I zz = 2 Í + .12 (.025)˜˙ + Á Á ˜ Á 12 Í 70 Ë 12 ¯˙ Ë Î ˚ ˆ ˜ = .00228646m 4 ˜ ¯ È 210 (0 + (- .3)(- .1)(.025))˘ + È 210 (0 + (- .3)(- .1)(.025))˘ = .0045m 4 I yy = Í ˙ Í 70 ˙ Î 70 ˚Î ˚ The GDE for this problem is * Eo I z*z v ¢¢¢¢ + Eo I yz w¢¢¢¢ = 0 * * Eo I yy w¢¢¢¢ + Eo I yz v ¢¢¢¢ = f z where fz=-4600 N/m The boundary conditions for this problem are w(0) = 0 w¢(0) = 0 * * Eo I yy w¢¢( L) + Eo I yz v ¢¢( L) = 0 v ( 0) = 0 v ¢(0) = 0 * * Eo I zz v ¢¢( L) + Eo I yz w¢¢( L) = 0 * * * * Eo I yy w¢¢¢( L) + Eo I yz v ¢¢¢( L) = 0 Eo I zz v ¢¢¢( L) + Eo I yz w¢¢¢( L) = 0 Solving the first equation by integration while noting the boundary conditions, * Eo I z*z v ¢¢¢ + Eo I yz w¢¢¢ = C1 * Eo I z*z v ¢¢ + Eo I yz w¢¢ = C 2 * Eo I z*z v ¢ + Eo I yz w¢ = C 3 * Eo I z*z v + Eo I yz w = C 4 Note that the constants of integration are 0 in each case, thus * Eo I z*z v + Eo I yz w = 0 (1) For the second equation, integrating, * * Eo I yy w¢¢¢ + Eo I yz v ¢¢¢ = f z x + C1 C1 = - f z L * * Eo I yy w¢¢ + Eo I yz v ¢¢ = f z x2 - f z Lx + C 2 2 C2 = f z L2 2 x3 x 2 f z L2 - fzL + x + C3 6 2 2 * * Eo I yy w¢ + Eo I yz v ¢ = f z C3 = 0 x4 x 3 f z L2 x 2 Eo I w + Eo I v = f z - fzL + + C4 24 6 22 C4 = 0 * yy * yz Ê x4 x 3 L2 x 2 ˆ ˜ E I w + E I v = fzÁ - L + Á 24 6 4˜ Ë ¯ * o yy * o yz Now, plug in the values for E and I and solve the system of equations for v and w .00228646v + .0045w = 0 (3.15 ¥ 10 )v + (9.90206 ¥ 10 )w = b 8 8 Ê x4 x 3 L2 x 2 ˆ ˜. Where b = f z Á - L + Á 24 6 4˜ Ë ¯ Solving for v and w, v = -.5315590062 ¥ 10 -8 b w = .2700863123 ¥ 10 -8 b Thus, Ê x4 ˆ v( x) = 2.445 ¥ 10 -5 Á - 3.33 x 3 + 100 x 2 ˜ Á 24 ˜ Ë ¯ 4 Êx ˆ w( x) = -1.2424 ¥ 10 -5 Á - 3.33 x 3 + 100 x 2 ˜ Á 24 ˜ Ë ¯ To plot the shear and bending moment diagrams, note the following: (insert diagram) ÂF z = V - 4600(20 - x) = 0 A ÂM Ê 20 - x ˆ = M + (4600(20 - x ))Á ˜=0 Ë2¯ Thus, the shear V, and bending moments are V = 4600(20 - x) N M = -2300(20 - x) 2 N-m 5. The GDE for this problem is EIw¢¢¢¢ = f z ( x) 3L ˆ 2q o x Ê L ˆˆ Ê Ê where the term fz is f z = qo Ld Á x Á1 - stpÁ x - ˜ ˜ ˜+ Á 4¯ LË 2 ¯˜ Ë Ë ¯ The boundary conditions for this problem are w(0) = w( L) = 0 w¢( L) = w¢¢(0) = 0 Integrating this function while noting that ˆ Ú d (x - x )dx = stp(x - x ) o o ˆ ˆ ˆ ˆ Ú stp(x - x )dx = stp(x - x )Ú dx o o The non-dimensional displacement w is EIw(x ) 9 1 1 Ê x 3ˆ 1 Ê x ˆ Ê x 1ˆ Ê x 1ˆ =stpÁ - ˜ + Á ˜ - stpÁ - ˜ x 5 stpÁ - ˜ + 4 128 Ë L 4 ¯ 60 Ë L ¯ 60 Ë L 2 ¯ 480 Ë L 2 ¯ qo L 5 9 1 3 Ê x 3ˆ 1 Ê x 3ˆ Ê x 1ˆ Ê x 1ˆ stpÁ - ˜ x + stpÁ - ˜ x - stpÁ - ˜ x 2 stpÁ - ˜ x 2 32 Ë L 4 ¯ 64 Ë L 2 ¯ 8 Ë L 4¯ 24 Ë L 2 ¯ 3 x3 77 x 139 Ê x ˆ Ê x 3ˆ x Ê x 1ˆ + stpÁ - ˜ + stpÁ - ˜ + Á˜ 6 Ë L 4 ¯ 24 Ë L 2 ¯ 3840 L 3870 Ë L ¯ 3 6. The GDE for this problem is EIv ¢¢¢¢( x) = f y ( x) = The boundary conditions are a2 rg 2 EIv ¢¢(0) = + Kv ¢(0 ) EIv ¢ ¢ ¢(L ) = + kv(L ) v(0) = 0 EIv ¢¢( L) = 0 Integrating, we have EIv ¢¢¢ = f y x + C1 x2 + C1 x + C 2 2 x3 x2 EIv ¢ = f y + C1 + C 2 x + C3 6 2 x4 x3 x2 EIv = f y + C1 + C2 + C3 x + C 4 24 6 2 EIv ¢¢ = f y From v(0) = 0, C4 = 0 From the other B.C., EIv ¢¢(0) = EIC 2 = KC 3 Ê L2 ˆ EIv ¢¢(L ) = EI Á f y + C1 L + C 2 ˜ = 0 Á ˜ 2 Ë ¯ EIv ¢¢¢(L ) = f y L + C1 = ˆ k Ê x4 x3 x2 Áy + C1 + C2 + C3 x + C 4 ˜ Á 24 ˜ EI Ë 6 2 ¯ Solving this system of three equations for the constants of integration, È 0 Í Í L Í EI L3 Í+ 6 Îk the constants of integration are: EI 1 L2 2 È ˘ ˘ Í ˙ 0 - K ˙ È C1 ˘ Í ˙ 2 L ˙ 0 ˙ ÍC 2 ˙ = Í - f y ˙Í ˙ Í ˙ 2 L ˙ ÍC 3 ˙ Í EIf y L f y L4 ˙ Î˚ ˚ Í ˙ 24 ˚ Îk 2 3 1 Lf y 24 KEI + 12 L kEI = 5 KL k C1 = 8 3KEI + 3L2 kEI + kL3 k 2 3 1 KL f y EI 12 EI + L k 8 3KEI + 3L2 kEI + kL3 k ( ) C2 = ( ) 2 3 1 L f y EI 12 EI + L k C3 = 8 3KEI + 3L2 kEI + kL3 k ( ) Alternately, using the non dimensional parameters a and b characterizing the stiffness of the rotational and linear springs respectively, a= KL EI kL2 b= EI C1 = C2 = C3 = f y 24a + 12 b + 5ab 8 3a + 3b + ab 12a + ab 8 3a + 3b + ab fy 12 + b 8 3a + 3b + ab fy Thus, 2 EIv( x) 1 Êxˆ 1 24a + 12 b + 5ab Ê x ˆ 1 12a + ab Ê x ˆ 1 12 + b Êxˆ =- Á ˜ + Á ˜Á ˜Á˜ 2 4 24 Ë L ¯ 48 3a + 3b + ab Ë L ¯ 16 3a + 3b + ab Ë L ¯ 8 3a + 3b + ab Ë L ¯ a rgL 4 3 2 The first plot was done with (a,b)=(5,10), the second with (a,b)= (•,0 ) As a approached 0, the problem becomes ill-defined. As a approaches infinity the problem is equivalent to a rigidly supported cantilever beam for which the deflection of the beam as the end is - f y L4 8 EI and is the solution’s limit. Since the cross-section is known, the moment of inertia can be found. A strip of material parallel to the base has the area dA = ydz = (a - z )dz By definition, a 2 a I y = Ú z dA = Ú z 0 0 2 (a - z ) a4 dz = 12 Using the parallel axis theorem the moment of inertia I = I - Ad 2 = The final answer is therefore Ea 2 v( x) 1 Êxˆ 1 24a + 12 b + 5ab =- Á ˜ + 4 24 Ë L ¯ 48 3a + 3b + ab 18 rgL 4 a4 a2 12 2 a4 Êaˆ Á˜= 36 Ë3¯ 2 1 12a + ab Êxˆ Á ˜Ë L ¯ 16 3a + 3b + ab 3 12 + b Êxˆ 1 Á ˜Ë L ¯ 8 3a + 3b + ab 2 Êxˆ Á˜ ËL¯ For b =0, Ea 2 v(a ) 11 =- 4 8 2a 18 rgL 7. The boundary conditions for this problem are w(0) = w( L) = 0 EIw¢¢(0) = EIw¢¢( L) = 0 The governing differential equation is then EIw¢¢¢¢( x) = f ( x) = q ( x) - kw( x) 8. The GDE for this problem is EAu ¢¢( x) = - f x (x ) where the distributed force term is 3L ˘ È Ê 2x ˆ Ê L ˆ f x ( x) = po Ld Í x - ˙ + po Á1 - ˜ stpÁ - x ˜ 4˚ L ¯ Ë2 Î Ë ¯ This may be rewritten as po - 2 po 2p Ê Lˆ Ê Lˆ 3L ˆ Ê x + o Á x - ˜ stpÁ x - ˜ + po Ld Á x ˜ L LË 2¯ Ë 2¯ 4¯ Ë Therefore, the GDE becomes EAu ¢¢( x) 2 2Ê Lˆ Ê Lˆ 3L ˆ Ê = -1 + x - Á x - ˜ stpÁ x - ˜ - Ld Á x ˜ po L LË 2¯ Ë 2¯ 4¯ Ë The boundary conditions are u (0) = 0 ~ EAu ' ( L) = N ( L) Integrating, EAu ¢( x) x2 1Ê Lˆ Lˆ 3L ˆ Ê Ê = -x + x - Á x - ˜ stpÁ x - ˜ - LstpÁ x ˜ + C1 po L LË 2¯ 2¯ 4¯ Ë Ë EAu ( x) x2 x3 1Ê Lˆ Lˆ 3L ˆ Ê 3L ˆ Ê Ê =+ xÁ x - ˜ stpÁ x - ˜ - LÁ x ˜ stpÁ x ˜ + C1 x + C 2 po 2 3L 3L Ë 2¯ 2¯ 4¯ Ë 4¯ Ë Ë 3 2 From the boundary condition u(0) = 0, C2 = 0 Using the second B.C., ˆ Ê L2 ˆ k Ê L2 L2 1 Ê L3 ˆ ÊLˆ ÁÁ ˜ - Lp o + C1 po = Á ˜ - LÁ ˜ + C1 L ˜ po + Á4˜ Á8˜ ˜ EA Á 2 3 3L Ë ¯ Ë4¯ ˯ Ë ¯ Solving for the constant of integration, EAu ¢( L) = - Lp o + po L po L 5 kL2 11 LÊ 5 + kL 11 ˆ Á ˜ L+ 4 EA 24 = Ë 4 EA 24 ¯ C1 = kL kL 1+ 1+ EA EA Therefore, the u displacement is Ê 5 kL 11 ˆ LÁ + ˜ EAu ( x) x x 1Ê Lˆ Lˆ 3L ˆ Ê 3L ˆ Ê Ê Ë 4 EA 24 ¯ x =+ xÁ x - ˜ stpÁ x - ˜ - LÁ x ˜ stpÁ x ˜+ kL po 2 3L 3L Ë 2¯ 2¯ 4¯ Ë 4¯ Ë Ë 1+ EA kL Dividing both sides by L2 and letting a = , EA 2 3 3 Ê 5 kL 11 ˆ Á+ ˜ EAu ( x) 1Ê xˆ 1Ê x ˆ 1Ê x 1ˆ Ê x 1ˆ Ê x 3 ˆ Ê x 3 ˆ Ë 4 EA 24 ¯ Ê x ˆ = - Á ˜ + Á ˜ - Á - ˜ stpÁ - ˜ - LÁ - ˜ stpÁ - ˜ + Á˜ kL 2Ë L¯ 3Ë L¯ 3Ë L 2¯ po L2 Ë L 2¯ Ë L 4¯ Ë L 4¯ ËL¯ 1+ EA 2 3 3 9. The governing differential equation for this problem is ÊL ˆ EI yy w¢¢¢¢ = f z ( x) = - f o stpÁ - x ˜ Ë3 ¯ with the boundary conditions w(0) = w( L) = 0 EIw¢¢(0) = EIw¢¢( L) = 0 Solve by successively integrating the GDE Ê L ˆˆ ÊL ˆ Ê EI yy w¢¢¢¢ = f z ( x) = - f o stpÁ - x ˜ = - f z Á1 - stpÁ x - ˜ ˜ Á 3 ¯˜ Ë3 ¯ Ë Ë ¯ Ê Lˆ Ê L ˆˆ Ê EI yy w¢¢¢ = - f o Á x - Á x - ˜ stpÁ x - ˜ ˜ + C1 Á 3¯ Ë 3 ¯˜ Ë Ë ¯ 2 Ê x2 1 Ê Lˆ L ˆˆ Ê EI yy w¢¢ = - f o Á - Á x - ˜ stpÁ x - ˜ ˜ + C1 x + C 2 Á 2 2Ë 3¯ 3 ¯˜ Ë Ë ¯ Applying the second set of boundary conditions on w’’, EIw¢¢(0) = C 2 = 0 EIw¢¢( L) = 0 fi C1 = fo L Ê L2 1 4 L2 ÁÁ2 2 9 Ë ˆ 5 ˜ = fo L ˜ 18 ¯ Returning to the GDE, 3 Ê x3 1 Ê Lˆ L ˆˆ x2 Ê EI yy w¢ = - fo Á - Á x - ˜ stpÁ x - ˜ ˜ + C1 + C3 Á 6 6Ë 3¯ 3 ¯˜ 2 Ë Ë ¯ 4 3 Ê x4 1 Ê ˆ Á - Á x - L ˆ stpÊ x - L ˆ ˜ + C1 x + C3 x + C4 EI yy w = - fo ˜ Á ˜˜ Á 24 24 Ë 3¯ 3 ¯¯ 6 Ë Ë From the first set of boundary equations, w(0) = 0 fi C 4 = 0 w( L) = 0 fi fo L Ê L4 1 Ê 2 L ˆ 4 ˆ C1 L3 25 Á - Á ˜ ˜=f o L3 Á 24 24 Ë 3 ¯ ˜ 6 L 1944 Ë ¯ Therefore, the non-dimensional deflection is EI yy w( x) f o L4 1 Êxˆ 1 Ê x 1ˆ 25 Ê x ˆ Ê x 1ˆ 5 Ê x ˆ = - Á ˜ + Á - ˜ stpÁ - ˜ + Á ˜Á˜ 24 Ë L ¯ 24 Ë L 3 ¯ Ë L 3 ¯ 108 Ë L ¯ 1944 Ë L ¯ 4 4 3 The non-dimensional bending moment EI yy w¢¢( x) f o L2 5 Ê x ˆ 1 Ê x 1ˆ Ê x 1ˆ 1 Ê x ˆ = Á ˜ + Á - ˜ stpÁ - ˜ + Á ˜ 18 Ë L ¯ 2 Ë L 3 ¯ Ë L 3¯ 2 Ë L ¯ 2 2 10. (a) deflection: Lˆ 3L ˆ Ê Ê EI yy w¢¢¢¢ = - Fod Á x - ˜ - M od Á x ˜ 4¯ 4¯ Ë Ë The boundary conditions are w(0) = w¢(0) = 0 EIw¢¢¢( L) - kw( L) = 0 EIw¢¢( L) - kq w¢( L) = 0 Integrating four times, Fo Ê Lˆ L ˆ M o Ê 3L ˆ x3 x2 Ê Ê 3L ˆ EIw( x) = - Á x - ˜ stpÁ x - ˜ Á x - ˜ stpÁ x - ˜ + C1 + C 2 + C 3 + C 4 6Ë 4¯ 4¯ 2 Ë 4¯ 4¯ 6 2 Ë Ë Applying the first two boundary conditions, w(0) = 0 fi C 4 = 0 w¢(0) = 0 fi C 3 = 0 3 2 Applying the next two B.C., EIw¢¢¢( L) - kw( L) = 0 fi Fo + C1 + Fo k 27 L3 M o k L2 C1 kL3 C 2 kL2 + =0 6 EI 64 2 EI 16 6 EI 2 EI EIw¢¢( L) - kq w¢( L) = 0 F k 9 L2 M o k L kq C1 L2 kq C 2 L 3Fo L - M o + C1 L + C 2 - o q + + =0 4 2 EI 16 EI 4 EI 2 EI fi- This produces a 2x2 linear system È kL3 1Í 6 EI Í 2 Í L + kq L Í 2 EI Î È ˘ Ê 9kL3 ˆ kL2 M o kL2 ˘ Fo Á1 Í ˙ Á 128 EI ˜ - 32 EI ˜ ˙ Ë ¯ 2 EI ˙ È C1 ˘ = Í ˙ kq L ˙ ÍC 2 ˙ Í Ê 3L kq 9 L2 ˆ k L ˆ˙ Ê Î˚ ˜ + M o Á1 + q ˜ ˙ 1+ + Í Fo Á Á EI ˙ ˚ 32 EI ˜ Ë 4 EI ¯˙ Í Ë4 ¯ Î ˚ (b) Bending moment distribution Lˆ Ê Lˆ 3L ˆ Ê Ê M y = EIw¢¢ = - Fo Á x - ˜ stpÁ x - ˜ - M o stpÁ x ˜ + C1 x + C 2 4¯ Ë 4¯ 4¯ Ë Ë (c) if M o = Fo L Fo 2L FL kq = o 2 k= then, setting a = Fo L2 , We can substitute these values into our 2x2 linear system EI 13a ˘ a 2˘ È È Í 1 - 256 ˙ Í 1 - 12 ˙ È C1 ˘ 4L = Fo Í Í Ê aˆ a ˙ ÍC 2 ˙ 7 17 a ˆ˙ Î˚ Í LÊ + Í LÁ1 + ˜ 1 + ˙ Á ˜˙ 2˙ Í Ë 4¯ Í Ë 4 64 ¯˙ Î ˚ Î ˚ Solving for the two constants of integration, 3Fo 512 + 454a + 21a 2 C1 = 32 48 + 32a + a 2 ( ( ) ) ) C2 = Fo L 2304 - 244a + 29a 2 64 48 + 32a + a 2 ( ( ) Therefore, the displacement (non-dimensional) is EI 1Ê x 1ˆ Lˆ 1Ê x 3ˆ 3L ˆ x3 x2 Ê Ê w( x) = - Á - ˜ stpÁ x - ˜ - Á - ˜ stpÁ x + C2 ˜ + C1 6Ë L 4¯ 4 ¯ 2Ë L 4¯ 4¯ Fo L3 6 Fo L3 2 Fo L3 Ë Ë 3 2 11. (i) A* =  (ii) Ei Ai = 2 + 3 * 4 + 1.5 * 2 = 17 in 2 Eo z* = 0 y o* = (iii) ÊE * I yy =  Á ÁE Ëo ˆ Ê1 ˆ Ê1 ˆ Ê1 ˆ ˜ I yy + z 2 A i = 1 * Á * 1 * 2 3 ˜ + 3Á * 2 * 2 3 ˜ + 1.5Á * 1 * 2 3 ˜ = 5.66 in 4 ˜ Ë 12 ¯ Ë 12 ¯ Ë 12 ¯ ¯i 1 A* ÊEˆ  Á E ˜ (y ) A Á˜ Ë o ¯i oi i = 1 (1 * 1.5 * 2 + 1.5 * -1.5 * 2) = -.088235 17 [ ÊE * I zz =  Á ÁE Ëo ˆ Ê1 ˆ ˜ I zz + y 2 A i = 1 * Á * 2 * 13 + 1.588235 2 * 2 ˜ + ˜ Ë 12 ¯ ¯i [ Ê1 ˆ Ê1 ˆ 3Á * 2 * 2 3 + .088235 2 * 4 ˜ + 1.5Á * 2 * 13 + 1.4112 * 2 ˜ = 15.5343in 4 Ë 12 ¯ Ë 12 ¯ (iv) Axial deflection Eo A*u ¢¢(x ) = 0 Eo A*u ¢(x ) = C1 Eo A*u (x ) = C1 x + C 2 Using the following boundary conditions, u (0) = 0 fi C 2 = 0 EA* u ¢( L) = 100 fi C1 = 100 Therefore, Eo A* u ( x) = 100 x u ( x) = (v) 100 x = 5.882 ¥ 10 -7 x 6 10 ¥ 10 ( ) 17 Eo I z*z v ¢¢¢¢( x) = -10 integrating four times, and applying the following boundary conditions v ¢¢¢( L) = 0 v ¢¢( L) = -100 ⋅ 2.088235 = -208.8235 v ¢(0) = 0 v(0) = 0 The displacement is found to be v( x) = 1 * Eo I zz Ê 5 4 500 3 ˆ x - 25104.41175 x 2 ˜ Á- x + 3 Ë 12 ¯ v( x) = -2.68223 ¥ 10 -4 x 4 + 1.07289 ¥ 10 -6 x 3 - 1.61606 ¥ 10 -4 x 2 (vi) The w displacement is * Eo I yy w¢¢¢¢( x) = 0 and contains the following boundary conditions w¢¢¢( L) = 0 w¢¢( L) = -100 w¢(0) = 0 w(0) = 0 integrating and applying the boundary conditions results in: w( x) = -8.8235 ¥ 10 -7 x 2 (b) The stress is found from the equation E = Eo ÏNc Ê M cI* - M cI* Ô y yz Á z yy Ì * - yÁ * * *2 ÔA Ë I yy I zz - I yz Ó ˆ Ê M cI* - M cI* z yz ˜ - z Á y zz ˜ Á I* I* - I* 2 yz ¯ Ë yy zz ˆ¸ ˜Ô ˜˝ ¯Ô ˛ s xx () () * * Simplifying, with M z = Eo I zz v ¢¢ and M y = Eo I yy w¢¢ s xx = Ê - 5 x 2 + 1000 x - 50208.8235 ˆ Ê - 100 ˆ¸ E ÏNc ˜ - zÁ - yÁ ˜˝ Ì Á ˜ Eo Ó A* 15.5343 Ë ¯ Ë 5.8 ¯˛ With the coordinates of a, b, and c being (50, 2.088235, 1), (50, -1.911765, 1), and (0, 2.088235, 1), plugging (x, y, z) into the stress equation produces a: s xx = 1732 psi b: s xx = -2311 psi c: s xx = 6773 psi 12. (a) The governing differential equation is of the form EI yy w¢¢¢¢ = f ( x) However, since both distributed forces are acting downward f ( x) = - f o - kw( x) Thus the GDE may be written as EI yy w¢¢¢¢ + kw = - f o with the following boundary conditions w(0) = w¢(0) = w( L) = w¢( L) = 0 (b) This is a non-homogenous differential equation, and the solution may be broken into a homogenous solution and a particular solution, i.e., w( x) = wh ( x) + w p ( x) By inspection, w p ( x) = fo k The homogenous solution may be found by solving the characteristic equation associated with the homogenous equation. wh ( x) = e bx (A1 cos bx + A2 sin bx )+ e - bx (A3 cos bx + A4 sin bx ) where b = k 4 EI Using Mathematica to solve for the four unknown coefficients by applying the boundary conditions, f o Ï - 1 + e bL cos bL + e bL sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ f Ï1 - e bL cos bL + e bL sin bL ¸ A2 = o Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ A1 = A3 = e bL A4 = e bL f o Ï e bL - cos bL + sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ f o Ï e bL - cos bL - sin bL ¸ Ì ˝ k Ó - 1 + e 2 bL + 2e bL sin bL ˛ To plot, w(x), make the assumptions fo =L k bL = 1 13. The GDE for this problem is EIw¢¢¢¢( x) = -q By integrating this DE successively four times EIw = 1 1 1 qx 4 + C1 x 3 + C 2 x 2 + C 3 x + C 4 24 6 2 The following boundary conditions are applied w(0) = w¢(0) = w( L) = EIw¢¢( L) = 0 From the first two boundary conditions C3 = C 4 = 0 From the last two boundary conditions, a system of equations can be set up È1 3 Í6 L ÍL Î C1 = È1 ˘ 1 2 ˘ È C ˘ Í qL4 ˙ L 1 2 ˙ 2 ˙Í ˙ = Í 1 1 ˙ ÎC 2 ˚ Í qL2 ˙ ˚ Î2 ˚ 5 1 qL C 2 = - qL2 8 8 Thus, the deflection is w( x) = q - 3L2 x 2 + 5 Lx 2 - 2 x 4 48 EI [ Non dimensionalizing, 2 2 4 EI 1 È Êxˆ Êxˆ Êxˆ ˘ w( x) 4 = Í- 3Á ˜ + 5Á ˜ - 2Á ˜ ˙ 48 Í Ë L ¯ qL ËL¯ ËL¯ ˙ Î ˚ The moment is 2 EI 1 È Êxˆ Êxˆ ˘ w( x)¢¢ 4 = Í- 1 + 5Á ˜ - 4Á ˜ ˙ 8Í qL ËL¯ ËL¯ ˙ Î ˚ 14. (a) Since s ( x) = P du =E A( x) dx E du P 1 = 2 dx Ao Ê xˆ Á1 - e ˜ L¯ Ë PÊ xˆ u (x ) = Ú Á1 - e L ˜ dx EAo Ë ¯ -2 integrating this expression gives u ( x) = - P EAo Ê L ˆ -1 1 Á ˜( ) + C Ëe ¯ Applying the boundary condition of u(o)=0, The constant of integration, C is found to be C= Thus, the axial deflection u(x) is PL 1 EAo e È ˘ Í P ÊLˆ 1˙ u ( x) = Á ˜ Í- 1 + x˙ EAo Ë e ¯ Í 1- e ˙ Í L˙ Î ˚ È ˘ EAo Ê 1 ˆ Í 1˙ (b) u ( x) = Á ˜ Í- 1 + x˙ PL Ë e ¯ Í 1- e ˙ Í L˙ Î ˚ (c) The stress is then s ( x) = P Ao xˆ Ê Á1 - e ˜ L¯ Ë -2 or non-dimensionally, AÊ xˆ s ( x ) = o = Á1 - e ˜ PË L¯ -2 15. The GDE for this problem is Lˆ 3L ˆ Ê Ê EIv ¢¢¢¢( x) = - f o + f o stpÁ x - ˜ - f o Ld Á x ˜ 4¯ 4¯ Ë Ë The boundary conditions are v(0) = v( L) = 0 v ¢(0) = 0 EIv ¢¢( L) = 0 Integrating four times with respect to x 1 1Ê Lˆ Lˆ 1 3L ˆ 3L ˆ 1 1 Ê Ê Ê 3 2 EIv( x) = fo x4 + f o Á x - ˜ stpÁ x - ˜ - f o LÁ x ˜ stpÁ x ˜ + C1 x + C 2 x + C 3 x + C 4 24 24 Ë 4¯ 4¯ 6 4¯ 4¯ 6 2 Ë Ë Ë 4 3 Applying the boundary conditions C 4 = C3 = 0 1249 fo L 2048 - 289 C2 = f o L2 2048 C1 = Thus the non dimensional deflection is 4 4 3 EIv( x) 1 È Ê x ˆ Ê x 1 ˆ Ê x 1 ˆ˘ 1 Ê x 3 ˆ Ê x 3 ˆ 1249 x 289 = - Á ˜ + Á - ˜ stpÁ - ˜˙ - Á - ˜ stpÁ - ˜ + Í 4 24 Í Ë L ¯ Ë L 4 ¯ fo L Ë L 4 ¯˙ 6 Ë L 4 ¯ Ë L 4 ¯ 2048 L 2048 Î ˚ The moment is 2 2 EIv ¢¢( x) 1 È Ê x ˆ Ê x 1 ˆ Ê x 1 ˆ˘ Ê x 3 ˆ Ê x 3 ˆ 1249 x 289 = Í- Á ˜ + Á - ˜ stpÁ - ˜˙ - Á - ˜ stpÁ - ˜ + 4 2 Í Ë L¯ Ë L 4¯ fo L Ë L 4 ¯˙ Ë L 4 ¯ Ë L 4 ¯ 2048 L 2048 Î ˚ 16. Boundary conditions: w(0 ) = 0 w' (0 ) = 0 w' ' (L ) = 0 EIw' ' ' (L ) = kw(L ) Governing differential equation: Lˆ Ê EIw' ' ' ' = - M 0 d' Á x - ˜ 2¯ Ë Lˆ Ê EIw' ' ' = - M 0 dÁ x - ˜ + C1 2¯ Ë Lˆ Ê EIw' ' = - M 0 stpÁ x - ˜ + C1 x + C 2 2¯ Ë Lˆ Ê Lˆ 1 Ê EIw' = - M 0 Á x - ˜ stpÁ x - ˜ + C1 x 2 + C 2 x + C 3 2¯ Ë 2¯ 2 Ë Lˆ Ê Áx - ˜ Lˆ 1 1 2¯ Ê EIw = - M 0 Ë stpÁ x - ˜ + C1 x 3 + C 2 x 2 + C 3 x + C 4 2 2¯ 6 2 Ë fi C1 = kw(L ), C 2 = M 0 - Lkw(L ), C 3 = C 4 = 0 2 Therefore, we have EIw(x ) = - M0 Ê Lˆ Lˆ 1 12 Ê 3 Á x - ˜ stpÁ x - ˜ + kw(L )x + x (M 0 - Lkw(L )) 2Ë 2¯ 2¯ 6 2 Ë 2 We need the expression for w(L ) Ê L2 ˆ 1 1 Á ˜ + kw(L )L3 + L2 (M 0 - Lkw(L )) Á4˜ 6 2 ˯ 2 (3 8)M 0 L EI fi w(L ) = 3 , where a = 3 kL (a + 1 3) kL EIw(L ) = M0 2 Substituting, we get the deflection 2 2 ÈÊ x ˆ 3 Ê x ˆ 2 ˘ EIw(x ) 1Ê x 1ˆ 1 Ê x 1ˆ 1Ê x ˆ = - Á - ˜ stpÁ - ˜ + Á ˜ + ÍÁ ˜ - 3Á ˜ ˙ 2Ë L 2¯ M 0 L2 Ë L 2 ¯ 2 Ë L ¯ 16(a + 1 3) ÍË L ¯ ËL¯ ˙ Î ˚ For bending moment, Lˆ Ê EIw' ' (x ) = - M 0 stpÁ x - ˜ + kw(L )x + M 0 - Lkw(L ) 2¯ Ë EIw' ' (x ) 3 Ê x 1ˆ Êx ˆ fi = - stpÁ - ˜ + 1 + Á - 1˜ M0 8(a + 1 3)Ë L ¯ Ë L 2¯ The axial stress along the bottom, s max (x ) = ˘ M 0b È Lˆ 3Ê 1 ˆ Ê Í1 - stpÁ x - ˜ + Á Á a + 1 3 ˜(x - L )˙ ˜ 2I Î 2¯ 8Ë Ë ¯ ˚ 17. By inspection, y * = y, z * = z fi (0,0 ) By symmetry, I *z = 0 y * I zz =  Ei I zz + y 2 A i = 13.33 in 4 E0 [ ( )] * GDE: E 0 I zz v' ' ' ' (x ) = -10 BC: v(0 ) = v' (0 ) = v' ' ( ) = v' ' ' ( ) = 0 100 100 * E 0 I zz v' ' ' (x ) = -10 x + C1 * E 0 I zz v' ' (x ) = -5 x 2 + C1 x + C 2 C 5 * E 0 I zz v' (x ) = - x 3 + 1 x 2 + C 2 x + C 3 3 2 C C 5 * E 0 I zz v(x ) = - x 4 + 1 x 3 + 2 x 2 + C 3 x + C 4 12 6 2 fi C1 = 1000, C 2 = -50000, C 3 = C 4 = 0 Therefore, we have * E 0 I zz v(x ) = - 5 4 1000 3 50000 2 x+ xx 12 6 2 Deflection: v(x ) = 1 Ê - 5 4 500 3 ˆ x+ x - 25000 x 2 ˜ *Á 3 E 0 I zz Ë 12 ¯ Moment: * E 0 I zz v' ' (x ) = -5 x 2 + 1000 x - 50000 Stress: s xx ÈNc Ê M cI* - M cI* z yy y yz Í - yÁ * Á I* I* - I* 2 ÍA yz Ë yy zz Î Mc ˆ EÊ Á - y *z ˜ = E0 Á I zz ˜ Ë ¯ E = E0 () ˆ Ê M cI* - M cI* z yz ˜ - z Á y zz ˜ Á I* I* - I* 2 yz ¯ Ë yy zz () ˆ˘ ˜˙ ˜˙ ¯˚ At point A, v( 00 ) = -0.9377 in 1 s xx = 0 since M zc = 0 at x = 100 At point C, v(0 ) = 0 by the first boundary condition lb È (- 50000 )˘ s xx = 1Í- 2 ˙ = 7502 in 2 13.33 ˚ Î 18a) Boundary conditions: w(0 ) = w' (0 ) = w(L ) = w' ' (L ) = 0 GDE: (EIw' ')' ' = f z (x ) = q0 Ê1 - x ˆ fi EIw' ' ' ' (x ) = q0 Ê1 - x ˆ Á ˜ Á ˜ Ë L¯ Ë L¯ b) q0 x 2 EIw' ' ' = q 0 x + C1 2L q x2 q x3 EIw' ' = 0 - 0 + C1 x + C 2 2 6L 3 q0 x q 0 x 4 C1 2 EIw' = + x + C 2 x + C3 6 24 L 2 q x4 q x5 C C EIw = 0 - 0 + 1 x 3 + 2 x 2 + C 3 x + C 4 24 120 L 6 2 fi C1 = - 2q 0 L q L2 , C 2 = 0 , C3 = C 4 = 0 5 15 q 0 x 4 q 0 x 5 q 0 Lx 3 q 0 L2 x 2 EIw(x ) = + 24 120 L 15 30 Nondimensionalizing, EIw(x ) - 1 Ê x ˆ 1 Êxˆ 1 Êxˆ 1 Êxˆ = Á ˜+ Á ˜- Á ˜+ Á ˜ 4 120 Ë L ¯ 24 Ë L ¯ 15 Ë L ¯ 30 Ë L ¯ q0 L 5 4 3 2 c) M y (x ) = EIw' ' (x ) M y (x ) = q 0 L2 Ê x ˆ q L2 Ê x ˆ 2q L2 Ê x ˆ q L2 - 0 Á ˜ - 0 Á ˜+ 0 Á˜ 2 ËL¯ 6 ËL¯ 5 Ë L ¯ 15 2 3 Nondimensionalizing, M y (x ) q 0 L2 -1Ê x ˆ 1 Ê x ˆ 2Ê x ˆ 1 = Á ˜ + Á ˜ - Á ˜+ 6 ËL¯ 2Ë L¯ 5ËL¯ 5 3 2 19a) Ê 3x ˆ A(x ) = a(x )b = abÁ1 ˜ Ë 4L ¯ ba 3 (x ) a 3 b Ê 3 x ˆ I yy (x ) = = Á1 ˜ 12 12 Ë 4 L ¯ 3 The applied load will have two effects - extension: force p0b - bending: moment –p0ba/8 No distributed load: the only loading is applied at x = L which shows up in the boundary conditions. GDE: (EA(x )u ' (x ))' = - f x = 0 (EI yy (x )w' ' (x ))' ' = f z = 0 Boundary conditions: u (0 ) = w(0 ) = w' (0 ) = ( yy w' ')L = 0 EI ' EA(L )u ' (L ) = p 0 b (EI yy w' ') = - p 0 L ab 8 Note that the applied moment at x = L is negative, according to our “happy face convention.” b) (EA(x )u ' (x ))' = 0 with u (0) = 0 and EA(L )u ' (L ) = p0 b Integrating, EA(x )u ' (x ) = C1 fi EA(L )u ' (L ) = C1 = p 0 b p0 b Ê 3x ˆ EabÁ1 ˜ Ë 4L ¯ 4 Lp 0 Ê 3 x ˆ u (x ) = lnÁ1 ˜ 3EA Ë 4 L ¯ u ' (x ) = Nondimensionalizing, Eau (x ) 4 Ê 3x ˆ = - lnÁ1 ˜ p0 L 3 Ë 4L ¯ 20. Let E1 be the reference modulus, i.e. E0 = E1. Also, symmetry implies I *z = 0 . y ÊE A* =  Á i ÁE Ë0 ˆ E ˜ Ai = 2bt + at 2 ˜ E1 ¯ Since the location of the centroid is obvious, we only need to compute I *y y È bt 3 Ê a + t ˆ 2 ˘ a 3 t E 2 I * = 2Í +Á ˜ bt ˙ + yy Í 12 Ë 2 ¯ ˙ 12 E1 Î ˚ We expect only u and w displacements because there is no bending effect in the y-direction. GDE: EA*u ' ' (x ) = - f x (x ) = - p 0 b EI *y w' ' (x ) = f z (x )+ m 'y (x ) y f z (x ) = -(btr1 + atr 2 + btr 3 ) = constant with Êa ˆ m y (x ) = p 0 Á + t ˜ = constant fi m 'y = 0 Ë2 ¯ Boundary conditions: u (0 ) = EA*u ' (L ) = w(0 ) = w' (0 ) = 0 ~ Êa ˆ E 0 I *y w' ' ' (L ) = V z (L )+ m y (L ) = Q + p 0 Á + t ˜ y Ë2 ¯ ~ * E 0 I yy w' ' (L ) = M y (L ) = 0 21. GDE: 2L ˆ Ê EIw' ' ' ' (x ) = - f 0 stpÁ x ˜ 3¯ Ë Boundary conditions: w(0 ) = w(L ) = w' (L ) = 0 EIw' ' (0 ) = Kw' (0 ) 2L ˆ Ê 2L ˆ Ê EIw' ' ' (x ) = - f 0 Á x ˜ stpÁ x ˜ + C1 3¯ Ë 3¯ Ë 1Ê 2L ˆ 2L ˆ Ê EIw' ' (x ) = - Á x ˜ stpÁ x ˜ + C1 x + C 2 2Ë 3¯ 3¯ Ë 1Ê 2L ˆ 2 L ˆ C1 2 Ê EIw' (x ) = - Á x x + C 2 x + C3 ˜ stpÁ x ˜+ 6Ë 3¯ 3¯ 2 Ë 1Ê 2L ˆ 2 L ˆ C1 3 C 2 2 Ê EIw(x ) = - Á x x+ x + C3 x + C 4 ˜ stpÁ x ˜+ 24 Ë 3¯ 3¯ 6 2 Ë 4 3 2 fi C1 = 2 L2 È f 0 L3 Ê f 0 L3 K f 0 L3 EI KL ˆ˘ - Á1 + , C3 = , C4 = 0 ˜˙, C 2 = Í EI ¯˚ 108(4 EI + KL ) 108(4 EI + KL ) Î 162 Ë Defining a = EI KL and b = , we get 4 EI + KL 4 EI + KL 2 2 È 11 Ê x ˆ 3 È5 EIw(x ) 1 Ê x 2ˆ 1 x˘ 1 Êxˆ ˘ Ê x 2ˆ = - Á - ˜ stpÁ - ˜ + a Í Á ˜Á ˜˙ ˙ + bÍ 24 Ë L 3 ¯ f 0 L4 Ë L 3¯ Í 972 Ë L ¯ 108 L ˙ Í 972 216 Ë L ¯ ˙ Î ˚ Î ˚ For K = 0 (simply supported), a = 0.25, b = 0. For K = ∞ (doubly clamped), a = 0, b = 1. For K = _EI/L, a = 1/(4+_), b = _/(4+_). 22. GDE: Lˆ Ê EAu ' ' (x ) = - f x (x ) = - p 0 dÁ x - ˜ 2¯ Ë Ê L ˆˆ Lˆ Ê Ê EIv' ' ' ' (x ) = f y (x )- m (x ) = 0 - Á - M 0 dÁ x - ˜ ˜ = M 0 d ' Á x - ˜ Á ˜ 4 ¯¯ 4¯ Ë Ë Ë ' z ' Boundary conditions: EAu ' (0 ) = u (L ) = v(L ) = EIv' ' (L ) = 0 EIv' ' (0 ) = Kv' (0 ) EIv' ' ' (0 ) = -kv(0 ) For axial displacement, Lˆ Ê EAu ' (x ) = - p 0 stpÁ x - ˜ + C1 2¯ Ë Lˆ Ê Lˆ Ê EAu (x ) = - p 0 Á x - ˜ stpÁ x - ˜ + C1 x + C 2 2¯ Ë 2¯ Ë L fi C1 = 0, C 2 = p 0 2 Therefore, EAu (x ) 1 Ê x 1 ˆ Ê x 1 ˆ = - Á - ˜ stpÁ - ˜ p0 L 2 Ë L 2¯ Ë L 2¯ For vertical deflection, Lˆ Ê EIv' ' ' = M 0 dÁ x - ˜ + C1 4¯ Ë Lˆ Ê EIv' ' = M 0 stpÁ x - ˜ + C1 x + C 2 4¯ Ë Lˆ Ê Lˆ C Ê EIv' = M 0 Á x - ˜ stpÁ x - ˜ + 1 x 2 + C 2 x + C 3 4¯ Ë 4¯ 2 Ë MÊ C Lˆ Lˆ C Ê EIv = 0 Á x - ˜ stpÁ x - ˜ + 1 x 3 + 2 x 2 + C 3 x + C 4 2Ë 4¯ 4¯ 6 2 Ë fi 3L(32 EIkM 0 + 7 kKLM 0 ) C1 = 32 3EIK + 3EIkL2 + kKL3 2 ( ) C2 = C3 ) EI ( EIM + 11kL M ) 96 =32( EIK + 3EIkL + kKL ) 3 3 0 0 2 3 96 EIKM 0 + 11kKL3 M 0 32 3EIK + 3EIkL2 + kKL3 ( C4 = 3EIL(32 EIM 0 + 7 KLM 0 ) 32 3EIK + 3EIkL2 + kKL3 ( ) 3 2 Substituting and then nondimensionalizing, we get EIv(x ) 1 Ê x 1 ˆ 1 96a + 11ab Ê x ˆ Ê x 1 ˆ 1 32b + 7ab Ê x ˆ = Á - ˜ stpÁ - ˜ Á ˜Á˜ 2 2Ë L 4¯ M0L Ë L 4 ¯ 64 3a + 3b + ab Ë L ¯ 64 3a + 3b + ab Ë L ¯ 1 96 + 11b Ê x ˆ 3 32 + 7a Á ˜+ 32 3a + 3b + ab Ë L ¯ 32 3a + 3b + ab 2 where a = nondimensionalized number characterizing the stiffness of the rotational spring = kL3 b = nondimensionalized number characterizing the stiffness of the linear spring = EI KL EI Note that if _ = _ = 0, the problem is ill-defined. 23. Boundary conditions: v(0 ) = v' (0 ) = EI zz v' ' ' (L ) = 0, EI zz v' ' (L ) = M 0 A* =  Ai Ei = at + 3at = 40t 2 E0 By symmetry, z * = 0 y* = 1 40t 2 È Ê1 ˘ 33 1ˆ Í3Á 2 a + 2 t ˜at + 1(0 )˙ = 8 t ¯ ÎË ˚ 2 * fi I *z = 0 , I zz =  y i =1 Ei 1435 4 I zz + y 2 A i = t E0 9 [ GDE: E 0 I zz v' ' ' ' (x ) = -10tp - g (r1 A1 + r 2 A2 ) E 0 I zz v' ' ' ' (x ) = -10tp - 30t 2 rg ...
View Full Document

This note was uploaded on 01/18/2010 for the course AE 322 taught by Professor Lambros during the Spring '04 term at University of Illinois at Urbana–Champaign.

Ask a homework question - tutors are online