Shear and Moment Diagram on a Simply Supported Flexural member

Transcription

1 Shear and Moment Diagram on a Simply Supported Fleural member Eercise announcement: Construct a shear and moment diagram to analyze the element indicated below. Data Preparation: It is noted that a non uniformly distributed load is applied along the a certain portion of the fleural member. A mathematical epression of the loading pattern that will allow the process of integration, and the definition of limits, needs to be developed. The uniformly distributed load and the point loads are easier to epress. Point Loads: P 1 := P := 5 Locations of Pt.Loads: p1 := p := 1 Uniformly distributed Load w1: Locations and length of load w1: w 1uni := 1. Lw1 := Rw1 := 8 l U1 := Rw1 Lw1 l min1 := Lw1 l ma1 := Rw1 The first uniformly distributed load is described in the above epressions as W1 in terms of load per unit length for the distributed load, distance Lw1 is determining the left end, Rw1 the right end, and length lu1 gives the total length along which the aforementioned load is distributed along the beam. It is however necessary to establish a mathematical relation that can be integrated to the epression of the load and that will limit its boundaries. An epression of w=1. does not create limits for the graph of the shear or the moment diagram. A simple mathematical relation that will yield a value of for locations beyond the length of the uniformly distributed load has been composed as seen below: l min1 l ma1 limits w1 := w 1 := w 1uni ( l min1 ) ( l ma1 ) The resultant of the above load will occur at its center: limits w1 Rw1 Lw1 Res w1 := Lw1 + Res w1 = 4 Load tot := w 1uni l U1 = 9.6 For the second load, a similar set of equations had to be developed, but in this case the triangular form oc the loading pattern had to be addressed. w ini := w fin := Lw := Rw := 7 l U := Rw Lw = 5 l min := Lw l ma := Rw w := w ini w fin limits w := ( ) limits w l min l min.5 l ma l ma limits w Limits w Limits w l U ma w ini, w fin Load tot := = 5 limits w Rw Lw Res w := Lw + Res 3 w =

2 Solution Process: Solving for reactions: w ini w fin Rw Lw Load tot := w 1uni ( Rw1 Lw1) + + P 1 + P Load tot = 1.6 Beam's Length L beam := 16 Support Locations: RA := 4 RB := 15 U1 := Lw1 U := Lw w 1uni Rw1 Lw1 Res w1 RA R B := Generating Diagrams: + R A V P := P 1 p1 P p ( Rw Lw ) w ini w fin + ( Res w RA) + P 1 ( p1 RA) + P p RA RB RA R B = 3.11 R A := Load tot + R B RA R B RB R A = Th equation above is used to calculate the shear that is produced by the point loads, including the support reactions as indicated in the graph on the right. Each point load is adding shear to the relation as the analysis is conducted from left to right. V P V w1 = w 1 d In this equation the relation of the uniformly distributed load 1 is applied in an integral from to any value of "" which represents the length. A typically linear slope is generated to represent constant the value of 1. kips per foot added to the beam V w V w = w d Last and most complicated is the unevenly distributed load. The continuous decrease of the load along the length of the beam in a linear fashion yields an integrated result of a curvilinear distribution of the shear V w

3 V := V P + V w1 + V w U1 := Lw1 l U1 := Rw1 Lw1 U := Lw l U := Rw Lw 1 All three of the above epressions can be added to generate the complete shear diagram. 8 4 V V u := if V min V ma, V min, V ma V u = 1 M := V P d + V w1 d + V w d Finally, the Moment diagram is the integral of the Shear diagram! M M u := if M min M ma, M min, M ma M u = 15.41

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6 Wood Beam Design Southern Pine is selected for a fleural member to carry the loads descrbed. The beam shall be adequately braced to prevent Lateral Torsional Buckling. Assume normal temperature conditions: Data: Beam's Length L beam := 15ft Try a standard 8 Support Locations: RA := 3ft RB := 14ft b w := 1.5in Spcwt := 36 lbf ft 3 h := 7.5in w beam := b w h Spcwt w beam =.7 lbf Other UDLs: w LL := 5 lbf ft ft w DL1 4 lbf b w h := w ft DL := w beam S := S 6 = in 3 w 1 := 1. w DL w LL w 1 = 44.8 lbf w ft := 1. w DL w = 3.6 lbf ft A:= b w h A = 1.87 in Beginning and end locations of uniformly distributed loads: Point Loads (Dead loads already considered factorized): Location Pt.Loads: Solution: Using the LRFD method: Lw1 := 1ft Rw1 := 14ft Lw := ft Rw := 15ft P 1 :=.kip P :=.4kip P 3 :=.kip p1 := ft p := 8ft p3 := 15ft Resultant locations of uniformly distributed loads: Res w1 := Lw1 + Rw1 Lw1 Res w1 = 7.5ft Res w := Lw + Rw Lw Res w = 7.5ft w Rw Lw Solving for reactions using sum of moments and sum of forces: ( ) ( Res w1 ) RA + w ( Rw ) Lw ( Res w RA) + P 1 ( p1 RA) + P ( p RA) + P 3 ( p3 RA) Load tot := w 1 Rw1 Lw1 + + P 1 + P + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 Load tot = kip w 1 Rw1 Lw1 R B := Generating Diagrams: RB RA R B = lbf R A := Load tot + R B R A = lbf V lbf Shear Diagram ft V u =.53 kip

7 1.6 Moment Diagram M k' ft M u = 1.34 k' M ma = 16.3 kip in M min =. k'

8 Wood Grade/Type: Resistance factors: Reference values: Ref. Modului Elast: Southern Pine No. 3 Time effect factor: λ :=.8 Φ bend :=.85 Φ tension :=.8 Φ v :=.75 Φ comp :=.9 Φ stability :=.85 Φ z_connect :=.65 F b := 875psi F t := 4psi F v := 135psi E min := 6ksi E := 17ksi From Table 4A of NDS supplement Format conv. Factors:.16 K Fb := K Φ Fb =.54 bend Table 4A of NDS supplement K Ft := K Φ Ft =.7 K Fv := K tension Φ Fv =.88 K F_Emin := K v Φ F_Emin = 1.76 stability F C := 65psi Ref. Des. Compr. From Table 4A of NDS supplement Parellel to grain: Element size factors CF (Tables 4A from NDS Supplement) and other adjustment factors Size factor for sawn lumber in bending: C Fb := 1. Wet service factor for sawn lumber in bending: C Mb := if F b C Fb 1.15ksi, 1,.85 C Mb = 1 Wet service factor for sawn lumber in tension: C Ft := 1. Wet service factor for sawn lumber in shear: Beam stability factor: C Mv :=.97 C L := 1 Wet service tensile factor: Temperature factor: C Mt := 1. C t := 1 Wet service Modulus of Elasticity factor: Load duration factor: C ME :=.9 C D := K Fb Φ bend λ C D = 1.78 The adjusted Bending Design value is: F' b_lrfd := F b C D C Mb C t C L C Fb F' b_lrfd = psi The adjusted Tensile Design value is: F' t_lrfd := F t K Ft Φ tension λ C Mt C t C Ft F' t_lrfd = psi The adjusted Shear Design value is: F' v_lrfd := F v K Fv Φ v λ C Mv C t F' v_lrfd = 6.8 psi The adjusted Modulus of Elasticity is: E' LRFD := EC ME C t E' LRFD = 153 ksi

9 The adjusted Modulus of Elasticity for beam stability is: E' min_lrfd := E min K F_Emin Φ stability C ME C t E' min_lrfd = 837 ksi The Moment Resistance is: ΦM n := F' b_lrfd S ΦM n = k' M u = 1.34 k' Moment_Condition := if ΦM n M u, "Approved", "Not Approved" Moment_Condition = "Approved" The Shear Resistance is: ΦV n :=.667 F' v_lrfd A ΦV n = kip V u =.53 kip Shear_Condition := if ΦV n V u, "Approved", "Not Approved" Shear_Condition = "Approved"