ecture 8: Assembly of beam elements. 4. Example of Assemblage of Beam Stiffness Matrices. Place nodes at the load application points. Assembling the two sets of element equations (note the common elemental coordinate system)
ecture 8: Assembly of beam elements. Step 5: Assembly. f1y 1 6-1 6 d1y m EI 4 6 φ 1z 1z = f y 1 6 d y m SYM 4 φ z z () f y 1 6-1 6 d y m EI 4 6 φ () z z () = f y 1 6 dy () m SYM 4 φ z z Element #1 Element # f1y F1 y 1 6-1 6 d1y m m 1z 1z 6 4 6 φ1 () fy + f y F y EI 1 6 1 + 1 6+ 6 1 6 d y () = = m m 6 6 6 4 4 6 z + mz z + + φ () f F 1 6 1 6 y y d y () m m 6 z z 6 4 φ
ecture 8: Assembly of beam elements. Step 6: Apply boundary conditions & solve. φ = = = 1 d1y dy F1 y 1 6-1 6 d1 y m 1z 6 4 6 φ1 F y EI 1 6 1 + 1 6+ 6 1 6 d y = m 6 6 6 4 4 6 z + + φ F 1 6 1 6d y y m 6 6 4 z φ 1 lbf + 1 in lbf. ft lbf Fy 4 6 dy EI m z 8 = φ m z 6 4 φ
ecture 8: Assembly of beam elements. 4. Examples of Beam Analysis Using the Direct Stiffness Method. For each element, the state variables have been calculated in the solution step. We can deduce anything to deal with the loading, stress, and strain on the two beam elements. Global nodal forces (the net external forces). Shear and bending moment diagrams Recall from earlier classes (MECH 141 and MECH ) that a shear and bending moment diagram is a convenient way to locate a design point in a beam. The displacement function vx ˆ( ˆ) allows us to plot the V and M diagrams for any one element or a chain of elements. Continue using Figure 4-7 pg. #16 here. Examples 4.1 4.5 pg.#16-175 should also be reviewed.
ecture 8: Assembly of beam elements. Consider a case of: E I = = 8 in = 5. in 6 9 1 psi 4 Structural steel 1 4 6 d y EI 1 = 8 φ 6 4 φ 1 lbf.4. d 1 y.4 1 1 in lbf 1 5. = φ 4 1.5 1. in lbf. 5. 1. φ d 4 y.8 1 in 7 φ = 6 1 rad 6 φ 9.8 1 rad
ecture 8: Assembly of beam elements. We can now recover the global nodal forces from the original system of equations. F1 y.1..1.. m 1z. 1.. 5.. 1 4 F y.4 1.1..4..1..8 1 in = 4 1 m. 5.... 5. z 1.5 1 7 6 1 rad F.1..1. y. m. 5.. 1. 6 z 9.8 1 rad 8 lbf 1 in lbf -11 lbf = 1181 in lbf 181 lbf -77 in lbf A quick check to validate the solution process.
ecture 8: Assembly of beam elements. We can also recapture the nodal forces acting on each element. Element #1 f1y.1..1.. 818 lbf m1 z 8. 1.. 5.. 149 in lbf 1.9 1 = 4 = f y.1..1..8 1 in 818 lbf 7 m. 5.. 1. 6 1 rad 78 in l z bf Element # () f 4 y.1..1..8 1 in -181 lbf () 7 m 8. 1.. 5. 6 1 rad z -8955 in lbf 1.9 1 () = = f y.1..1.. in 181 lbf () 6 m. 5.. 1. 9.8 1 rad z -77 in lbf
ecture 8: Assembly of beam elements. From the free body diagrams of the two elements we can construct the shear and bending moment diagrams. ŷ 818 lbf 818 lbf 1 149 in lbf 1 ˆx 78 in lbf 181 lbf 181 lbf 8955 in lbf ˆx 77 in lbf
ecture 8: Assembly of beam elements. V and M diagrams are useful when we deal with rather simple beam assemblies. The calculation of maximum axial stresses (for beams these maximum stresses occur at the top and bottom beam surfaces) can be easily coded in advance. σ xmax dvˆ = Ec dx ˆ 1 1 1 1 = Ec x d + x + x+ d + x ( 1 ˆ 6 ) ˆ ( ˆ ) ˆ ( ˆ ) ˆ ( ˆ y 6 4 φ 1 6 y 6 ) 1 1 ˆ φ σ 6 ˆ ˆ 4 ˆ ˆ xmax xˆ Ec ( d y d1y ) 1 = = 6 ˆ ˆ ˆ 4 ˆ xmax xˆ Ec ( d 1y dy) 1 variation over the = = + + Node 1 value element is given by a linear interpolation of the two node values. Node value
ecture 8: Work Equivalence for Beams. 4.4 Distributed oading. Concept of work equivalence is the same as covered for bar elements. We are looking for four generalized nodal forces that are equivalent to the actual distributed load. This equivalence must hold for any set of generalized nodal displacements.
ecture 8: Work Equivalence for Beams. Case study: a rectangular (constant magnitude) distributed transverse load. w 1y 1 w 1z w y w z 4 fˆ mˆ fˆ mˆ w = N w dxˆ = w = N w dxˆ = 1 w = N w dxˆ = w = N w dxˆ = 1
ecture 8: Work Equivalence for Beams. Pg.# 178-179 emphasizes how these equivalent nodal loads must be superposed over other external loads. Example 4.6 looks specifically at how the equivalent point loads superpose over other external loads such as those created by beam supports. The revised form of the element equation becomes ˆ ˆw f + f = kˆ dˆ { ˆ ˆ } 1y φ1 y φ fˆ = fˆ fˆ { ˆ ˆ } 1y φ1 y φ fˆ = fˆ fˆ w w w w w T T
ecture 8: Work Equivalence for Beams. Appendix D of ogan (pg.# 75) gives the equivalent nodal loads for a variety of distributed load forms. These results can be applied in the first stages of any analysis. Once you have replaced a distributed load, it is removed from the FDB of the element/system. Note: these Appendix D results are specific to the beam element we have derived. If we were to change the element type it would affect the shape functions that are associated with each nodal load. If the shape functions change then the integration process tabulated in Appendix D is no longer valid.