hw3-2008sol

hw3-2008sol - Chapter 17, Solution 9 We have E = 210 GPa A...

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: Chapter 17, Solution 9 We have E = 210 GPa A = 5 10 -4 m2 Table P17.9 Data for the truss of Fig.P17.9 Element 1 2 Length 5 4 5313o . 90 o c s c2 cs s2 0.6 0.8 0.36 0.48 0.64 0 1 0 0 1 ( a ) Apply Eq.(17.14): u1 v1 u2 v2 0.48 - 0.36 - 0.48 u1 0.36 0.48 0.64 - 0.48 - 0.64 v1 AE [k ]1 = 0.36 0.48 u2 5 - 0.36 - 0.48 0.48 0.64 v 2 - 0.48 - 0.64 u1 v1 0 0 1 AE 0 [k ]2 = 0 4 0 0 - 1 u3 v3 0 0 u1 0 - 1 v1 0 0 u3 0 1 v3 ( b ) Global Stiffness Matrix u1 v1 u2 v2 u3 v3 1008 - 0.756 - 1008 0 0 . . 0.756 1008 3969 - 1008 - 1344 0 - 2.625 . . . . - 0.756 - 1008 0.756 1008 0 0 . . [ K ] = 10 7 1008 1344 0 0 . . . . - 1008 - 1344 0 0 0 0 0 0 0 0 0 2.625 - 2.625 0 (c) u1 v1 u2 v2 u3 v3 0.756 1008 - 0.025 . F1x 7 = 10 . . 1008 3969 v1 10000 . 100 103 = (1008 10 7 )( -0.025) + 3.696v1 or F1x = ( 0.756 10 7 )( -0.025) + 100887 10 7 v1 or v1 = 0.00887 m F1x = -99.6 kN ( d ) Support reactions: . F2 x - 0.756 - 1008 99.59 F - 1008 - 1344 - 0.025 . . 132.79 2y 7 = 10 kN = 0 0.00887 0 0 F3 x F3 y - 232.84 - 2.625 0 ( e ) Use Eq.(17.16): We have u2 = v 2 = 0, u3 = v3 = 0. - u1 105 10 6 0.025 AE F12 = [0.6 0.8]- v = 5 [0.6 0.8]- 0.00887 = 166 kN L1 1 - u1 105 10 6 0.025 AE F13 = [0 1]- v = 4 [0 1]- 0.00887 = -232.8 kN L2 1 (C) (T ) Chapter 17, Solution 11 We have EI = 216 10 6 lb in.2 , L = 5 12 = 60 in., ( a ) Use Eqs.(17.19): v1 1 v2 2 - 12 360 360 v1 12 360 14400 - 360 7200 1 EI [k ]1 = 3 12 - 360 v 2 L - 12 - 360 7200 - 360 14400 2 360 v2 2 360 12 360 14400 EI [ k ]2 = 3 L - 12 - 360 7200 360 v3 - 12 - 360 12 - 360 P = 10 kips 3 360 v2 7200 2 - 360 v3 14400 3 v3 3 0 0 v1 0 0 1 - 12 360 v 2 - 360 7200 2 12 - 360 v3 - 360 14400 3 ( b ) Global Stiffness Matrix v1 1 360 12 360 14400 EI - 12 - 360 [K ] = 3 7200 L 360 0 0 0 0 v2 2 - 12 360 - 360 7200 24 0 0 28800 - 12 - 360 360 7200 Use Eq.(17.20a): 24 F2 y EI M2 = 3 0 M L 360 3 0 28800 7200 360 v 2 7200 2 14400 3 { } = 1 AE [ K ]-1 { F } 0.072917 v 2 0.000521 - 0.002083 - 10 103 - 0.729 in. L3 0.000043 - 0.000035 0 = 0.0052 rad 2 = 0.000521 0.02083 rad EI - 0.002083 - 0.000035 0.000139 0 3 (c) F1 y - 12 EI M1 = 3 - 360 F L - 12 3y 360 7200 - 360 0 v 2 6.876 kips 0 2 = 225 kip in. - 360 3 3125 kips . From a free-body diagram of element 2: ( M 2 ) 2 = 187.5 kip in. and ( F2 ) 2 = -3125 kips. . (d) 1 1 V ( kips ) + 2 6.875 x 3.125 M ( kip in.) 187.5 + 225 x 10 kips 2 3 Analysis of a Statically Indeterminate Stepped Beam Given: A propped cantilever beam of flexural rigidity EI and 2EI for parts 1-2 and 2-3, respectively, is subjected to a concentrated load P at point 2 (Fig. 13.15a). Find: (a) Determine the nodal displacements. (b) Determine the nodal forces and moments. Assumption: The deflections are elastic . SOLUTION: We discretize the beam into elements 1 and 2 with nodes 1,2, and 3 as shown in Fig. 13.15a. There are a total of six displacement components for the beam before the boundary conditions are applied. So, the order of the system stiffness matrix must be 6 x 6. Through the use of Eq. (13.22), the stiffness matrix for element 1, with (EI/L 3 ) I = EI/L 3 and L 1 = L, ( .I R] M IL ( L (a) -t 19P123 2 R3 (b) - 4P/23 8PLl23 D&\IWsJ -IIPLl23 3 x (c ) FIGURE 13.15 (a) Stepped beam with a load; (b) shear diagram; (c) bending-moment diagram. may be expressed as VI 81 V2 82 E/ [k]1 = 0 12 6L 6L 4L2 -12 -6L 6L 2L2 --- - 10 0 10 0 L - 12 6L -6L 2L2 1 1 12 -6L 1 -6L 4L 2 : -- --- . 0 0 0 0 ---1 1 0 0I v3 83 VI 0 0 0 1 0 1 81 0 1 V2 01 1 82 0 0 0 1 v3 0 1 83 ~ Similarly, with (El/eh rearrangement: = VI El/4L 3 and ~ 81 V2 = 2L, for eleme nt 2 we obtain after v3 83 1-0E/ [k]2 = 0 0 - - 0 - - 0 - - - 0 - - 0- 1 82 0 0 0 0 0: r-- - - - - ------ 0 1 3 3L - 3 3L 0 1 3L 4L 2 -3L 2L2 0 1- 3 - 3L 3 -3L ~O_ .9 _: 3L -2L2 -3L 4L4 :0 10 :0 10 Observe that in the forego ing, the nodal displacements are shown to indicate the associativity of the rows and columns of the member stiffness matrices; accordingly, rows and columns of zero s are added (boxed in by the dashed lines). (a) The system stiffness mat r ix of the beam can now be superimposed: [K] = [k] I + [kh. The governing equation s for the beam , with F 2y = - P (Fig. 13.ISb), are then Rl Ml -p 0 R3 / 0 0 12 6L -12 6L 0 0 6L 4L2 -6L 2L2 0 0 -12 -6L IS - 3L - 3 3L 6L 2L2 - 3L 8L2 - 3L 2L2 0 0 3 -3L 3 -3L 0 0 3L 2L2 -3L 4L2 VI 81 V2 82 V3 83 The boundary conditions are VI = 0, 8 1 = 0, and V3 = O. Multiplyingout these equations associated with the unknown displacements, we get {-P -EI [- 3L -3 3L ] L o } 15 8L 2L o -0 3L 2 2 2 2 r} 82 83 2L 4L and r} MI R3 { = El 0 [-12 - 6L - 3 6L 2L2 -3L jJ {~ } -P} = - - {-2L} PL -188 0 2 Solving Eq. (d) by the method of matrix inversion (see Section C.3), we obtain the deflection and slopes as V2 } 82 = - L2 - [ 83 276/ 28L 18 -30 18 SI /L -39/L -30 ] { -39/L III /L 0 276/ 30 The minus sign indicates a downward deflection at node 2 and clockwise rotation of left end 1; the positive sign means a counterclockwise rotation at right end 3 (Fig . 13.15), as appreciated intuitively. ) Introducing the displacements obtained into Eq. (e), after multiplying and simplifying, nodal forces and moments are obtained as R { MJ R3 I} = 276L P [ -12 -6L -3 6L 2 2L -3L o ] { - 28L } o - 18 -3L 30 = 23 I{19P } IIPL 4P In general, it is nece ssary to ascertain the nodal force s and ioments associated with each element to analyze the whole structure. For the ase treated, it may readily be seen from a free-body diagram of element 2 that 12 = R 3 (2L) = 8PLl23 . Hence, we obtain the shear and moment diagrams for ie beam as illustrated in Figs . 13.l5b and c, respe ctively. 'omments: ...
View Full Document

This note was uploaded on 05/19/2008 for the course MAE 314 taught by Professor Rabiei,heeter during the Spring '08 term at N.C. State.

Ask a homework question - tutors are online