hw21_Solutions

# By u x ea 2 4 1 1 3x pl 3 1 1 4l pl ea 2 3

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Unformatted text preview: 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 = Â I zz = Â I yz = Â 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 noti...
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