Quiz4_2010 - EML 4507 Fall 2010 Quiz 4B-V1 A beam is...

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Unformatted text preview: EML 4507 Fall 2010 Quiz 4B-V1 A beam is clamped at the left end and rests on a spring at the right end. The right support is such that the beam cannot rotate at that end. Thus, the only active DOF is v2. A force F = 3,000 N acts downward at the right end as shown. The structure is modeled using two elements: one beam element and one spring element. The spring stiffness k = 1000 N/m. The beam properties are L = 1 m, E] = 1000 N—mz. (a) Write the element stiffness matrices of both elements and clearly show the row addresses. (b) Assemble the two element matrix equations and apply boundary conditions to obtain the global matrix equation. (c) Solve for unknown displacement 122. (d) Draw the BM and SF diagrams for the beam. 76) Solution (a) The stiffness matrix of the beam element is: 12 6 —12 6 '01 6 4 —6 2 91 — 12 —6 1 2 —6 122 6 2 —6 4 02 [km] = 1000 The stiffness matrix of the spring element is: ' 1 —1 v [162)] =1000 2 —1 1 v3 (b) Assembling we obtain the global " equations as 12 6 —12 6 0 v1 R1 6 4 —6 2 o 191 C1 1000 ~12 —6 12+1 —6 —1 v2 = —3000 6 2 —6 4 0 02 o 0 0 ~1 0 +1 v3 R3 (0) Deleting all zero DOFs we obtain the global equation as 13,0001)2 = —3,000 :> v2=—0.231 m (d) Use the equation 12 6L ~12 6L {,1 4x,”1 k E] 6L 4L2 —6L 2L2 01 —M1 [Md—F42 —6L 12 —6L 112 _ +Vy2 6L 2L2 —6L 4L2 62 +M2 Substituting for E1, L and {41(1)} —12 2, 772 —Vy1 ~6 1,386 —M1 1000 {v2 = —0.23 1} = = 12 —2, 772 +Vy2 —6 1,386 +M2 SF 1 —2,772 N 2:17.210 . .3000 N g N5358é New ‘585 NT“ T225110 1 i 1 l 1 l 1 EML 4507 Fall 2010 Quiz 4B-V2 A beam is clamped at the left end and rests on a spring at the right end. The right support is such that the beam cannot rotate at that end. Thus, the only active DOF is 122. A force F = 1,200 N acts downward at the right end as shown. The structure is modeled using two elements: one beam element and one spring element. The spring stiffness k = 500 N/m. The beam properties are L = l m, E] = 500 N—mz. (a) Write the element stiffness matrices of both elements and clearly show the row addresses. (b) Assemble the two element matrix equations and apply boundary conditions to obtain the global matrix equation. (0) Solve for unknown displacement v2. (d) Draw the BM and SF diagrams for the beam. 2@ Solution (a) The stiffness matrix of the beam ' element is: 12 6 —12 6 '01 6 4 ~6 2 61 —12 —6 12 —6 112 6 2 —6 4 92 [1(0)] = 500 The stiffness matrix of the spring element is: 1 ——1 v [k‘2)]=500[1 1L: (b) Assembling we obtain the global equations as 12 6 —12 6 0 v1 R1 6 4 —6 2 ’0 91 C1 500 —12 —6 12+1 —6 —1 v2 = —1200 6 2 —6 4 0 92 o 0 0 —1 0 +1 v3 R3 (c) Deleting all zero DOFs we obtain the global equation as 6,5001}2 = ~1200 2) v2=-0.185 m ((1) Use the equation 12 6L —12 6L {,1 _Vy1 A E] 6L 4L2 —6L 2L2 01 _ —M1 [HQ—F42 —6L 12 —6L 112 _ +Vy2 6L 2L2 ~6L 4L2 62 +M2 Substituting for E], L and {gm} —12 1,110 —Vyl —6 555 —M1 500 {v2 = —O.185} = = 12 —1,110 +Vy2 —6 555 +M2 SF _ -1,110N ' +555 N—In BM ~555 N—m ES: Al“) N WC ’1‘ mo N y ESSN’Y“ E i5 1 EML 4507 Fall 2010 A beam is clamped at the left end and rests on a spring at the right end. The right support is such that the beam cannot rotate at that end. Thus, the only active DOF is v2. A force F = 2,400 N acts-downward at the right end as shown. The ‘ structure is modeled using two elements: one beam element and one spring element. The spring stiffness k = 1000 N/m. The beam properties are L = 1 m, E] = 1000 N-m2. (a) Write the element stiffness matrices of both elements and clearly show the row addresses. (b) Assemble the two element matrix equations and apply boundary conditions to obtain the global matrix equation. (c) Solve for unknown displacement v2. (d) Draw the BM and SF diagrams for the beam. 7 Solution (a) The stiffness matrix of the beam _ element is: ' 12 6 —12 6 121 6 4 —6 2 61 —12 —6 12 —6 v2 6 2 —6 4 6 [km] = 1000 The stiffness matrix of the spring element is: 1 —1v [1<<2>]=1000[_1 1}): (b) Assembling we obtain the global equations as 12 6 —12 6 0 v1 R1 6 4 ~6 2 0 .91 C1 1000 —12 —6 12+1 —6 —1 v2 = —2400 6 2 ~6 4 o 62 0 0 0 —1 0 +1 v3 R3 (0) Deleting all zero DOFs we obtain the global equation as 13,0001)2 = —2400 :> v2=—0.185 m ((1) Use the equation 12 6L —12 6L 1,1 44,1 k E] 6L 4L2 —6L 21;2 01 —M1 [HQ—E42 —6L 12 —6L 112 _ +Vy2 6L 2L2 —6L 4L2 62 +M2 Substituting for E1, L and {qm} —12 2,215 —Vyl —6 1,108 4141 1000 {v2 = —0.185} = = 12 —2,215 +Vy2 —6 1,108 +M2 SF _ -2,215 N +1,108 N—m —1,108N—m 15"I ' 9-9— 9.qu N g = 1108 N“ HOS—NH“ /Yj \"b\r“\l\m i5 ...
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This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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Quiz4_2010 - EML 4507 Fall 2010 Quiz 4B-V1 A beam is...

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