Direct Stiffness - Beam Example

Transcription

1 Direct Stiffness - Beam Example Equations we will use from the beam notes k EI EI () a x y y x x y y x (4) x y y xy x y y 3 xy xy y x 3 x xy y x x y x y x 2EI x y y 3 xy x y y xy xy y x x xy y x 3 x -- y x 2EI y x , Ke (2) of 7 Direct Stiffness - Beam Example

2 Example: Analyze the following frame structure for displacements, reactions and member forces. I 6 in 4 A 2 in 2 E 29 ksi 5 ft 2 ft 2 2 ft Stiffness matrix for elements: ote - All units converted to kips and inches as we go... Element #: inding x and y and applying Eq.. : x change in x / (2 - )/25.8 y change in y / (5 - )/25.6 near end D.O.. ( 4, 5, 6) far end D.O.. (, 2, 3) Ke e e3 symmetric e3 Element #2: inding x and y and applying Eq.. : x change in x / y change in x / near end D.O.. (, 2, 3) far end D.O.. ( 7, 8, 9) Ke e e3 symmetric e3 2 of 7 Direct Stiffness - Beam Example

3 Assembly of Global Stiffness matrix: KG Ke + Ke2 After assembly we have KG e e e e e e e3 Define known loads: Using the fixed end moment table and some statics to reduce the uniform load to equivalent nodal loads (E...) > 2 k*in reactions 3 k 2 k*in 3 k > 3 k 2 k*in E... 3 k 2 k*in released D.O.. fixed D.O.. R > R k R 2 R not included 3 of 7 Direct Stiffness - Beam Example

4 Solve the system for r u and R u The symbolic system is : Rk K K2 Ru K2 K22 and this system with the numbers looks like: ru rk e e e e e e e3 Displacements: now expanding we get... R k K*r u + K2*r k or with r k : R k K*r u with the numbers: > e3.247 in.954 in.27 rad Reactions: expanding the bottom half... R u K2*r u + K22*r k or with r k : R u K2*r u with the numbers: e e3.247 in.954 in.27 rad > k k k k*in 4 of 7 Direct Stiffness - Beam Example

5 Reactions continued: The reactions on the previous page are reactions to the equivalent nodal loads at the and degrees of freedom. The E... at the fixed degrees of freedom and were not included. ote that the vertical reactions ( and ) only add up to 3 kips, even though the distributed load totals 6 kips. Remember that only half of that 6 kips made it into the load vector. The complete reactions at and (complete meaning it accounts for the original distributed load) will have the reactions to the fixed E...s added back in. What we add is the opposite of the missing load (its reaction), thus we add + 3 kips to, and subtract 2 k*in from. 2 3 k 2 k*in E... 3 k 2 k*in k k k + 3 k k*in - 2 k*in orce Recovery: ow force recovery is given by applying Eq. 8 from the beam note packet S k*a*r for each element, where k is equation () and a is equation (4) from the beam packet. (Eq. is OT the same as Eq. 2) Element : S4 S5 S e3 6 6e e e S4 S5 S e3 6 6e e e k.8 k 398 k*in 5 of 7 Direct Stiffness - Beam Example

6 ow using the results of force recovery to draw the axial, shear and moment at each end. S4 S5 S6.8 k.8 k 398 k*in 46 k*in.8 k 398 k*in.8 k Element 2: ote that eq() eq(2) when member is horizontal S7 S8 S e e e e3 S7 S8 S e e e e3 82 k*in 488 k*in ow using the results of force recovery to draw the axial, shear and moment at each end. S7 S8 S9 82 k*in 488 k*in 82 k*in 488 k*in 6 of 7 Direct Stiffness - Beam Example

7 ow What?? These are the internal forces on member two due to the equivalent nodal loads as determined earlier. The final step is for us to reconcile the equivalent nodal loads back to the original distributed load. We will call this load replacement. oad Replacement: We were given the problem on the left to solve, and we approximated it as the problem on the right. 3 k 2 k*in We now want to adjust the answers we found so that they reflect the original loads rather than the E...s. We have done this for the reactions already, by adding in the effects of the E... at the right wall. We still need to do this for the horizontal member itself. ortunately this is easily done. We take the internal forces we solved for on the previous page, and add to that the original distributed load and the reactions to it: member 2 solution to E forces 82 k*in in equilibrium 488 k*in + original load and reactions 2 k*in 2 k*in 3 k 3 k in equilibrium member 2 final solution to original load 688 k*in 398 k*in 24.6 k 35.4 k in equilibrium This now is the final free body diagram with the correct loads and internal forces for member two. ote that the shear (35.4 k) and moment (688 k*in) at the right end now match the reactions at the right wall (see page 5). The end. 7 of 7 Direct Stiffness - Beam Example