Question

Problem 4 (30 pts): Deflection A diving board is simply supported with an overhang portion from C to B. A diverstanding at the free end B has weight W = 205 lb. Assume the diving board has constant properties El with length L=8 fi and E = 10,000 ksi. Use these bending moment equations to answer the questions below: M(x) = -2Wx (0 SX S1/3) M2(x) = Wx - WL, (L/3 SX SL) W ΕΙ h-2.5 in. B kb 17 in. L/3 (a) Use the method of integration to determine the slope v '(x) and deflection v(x) equations for the entire beam. Clearly state and box the conditions used to solve for the constants of integration. (b) If the maximum allowable deflection of the beam at free end B is Smar= 0.250 in., then what is the largest weight Wmax [16] of a diver that the overhang portion of the diving board can sustain? Use the moment-area method to address this problem.

Answer 1

The bending moment equations for whole beam is given

using these equations and as stated we will find the slope and deflection curves using integration method and also rest of the calculation.

see below the complete solution.

solutions 22:5in Guiven, WS 205 eb izin I bh = 22-135 in 12 EI = constant 10,000 ksi = 10X106 psi 8 ff -gbin А 5 EL с नी B x=L Y=0 21 - 2x Miln) (osas) M₂ (W) = wna WL, (<<L) Partla) We have dur) da? EI for section AC d²u(a) M, (2) - 2 wr EI du? Integrating with respect to a

3 d uli) du =G- W x2 G = constant EI > once again integrate the below equation dulu) da wn? EI ta U(x) = -wwe tante BEI Note : 0 is slope equation and ③ is deflection equation Boundary conditions: VW) & deflection =o) put in ② o tot 27 (6=0 at A lat XO also at n=k Ow=0 at si U (W)=0 3 ) در بیا cil t O to 34 EI WLZ - > 19= la- 27 EI Wrve Hence We have slope and deflection for A to C section 052343

putting i and a in 6 and ② Slope vin) - wn² EI WLZ 2761 + W = 205 lb Lago in 22.135 inu I- E = 107 lb lin? 2osx 96² => U'(x)=-205*? 22-135 X 107 27x107x22.135 U'60) = -9.261x107x?+ 3.168x104 osn<3 (xin inches) Deflection we? V(X) = WL²x 27EL +0 3 EL 27x107x 22.135 U (W) = -205 x² 205*962xne + 3x107x22.135 V(x) = 3.06x107x3+ .30161810n OSX43 (B) How for section c to B, 1 x K5 L M₂ (1) Wh-LW EL M₂() = Wro w We have d²v(x) du? - Integrating once EI

El dv(x). ww² WLr ta du) Integration again EBV) = wr . WLx? taxts - Boundary conditions at r= 1 v(x) = 0 puting in Deflection at n= 1 is o قريبا iL ابيا ta - 18 3 27x6 5 at + 6 = 4 WEB 81 that the beam is a single unit hence slope x = 1/5 is same for both section AC and CB Note at W' W) Ac) (CB) x=43 x=4 using slope equations. - wie WL? u) หา ta + WLR EL - EL 26L ET 2761 WL2 في حيا + -WI? JEI Wh? 27E1 18 EI ta BEL EL => WL² 54 put in S 2 = 4 WE 1 X WE -25 کر یا 81 su 162

Put ci and ca in 9 and 5 Slope El du(K) ww² นง LM + d'n) I w22 54 .2 U'(X) 205x² +(205x 96 x +11/205 x 962 2x107x 22.135 5uX 10722-13 107x 2 2.135 => '0'(x) = 4.63x107x²- 8.9x10x + 10738x10² S Reflection Wh u? EI Ulu) = ww² 6 "I wll ntch su Wha? =) EI V(K) = wh² itthu 11 w2²na -25 we 162 6 3 n VO) = 205 205x96x42 2x107x22.135 107x22.135X6 X205 X 962 x 162*2058.963 su 107x22-135 107x 22.13's -5 -3 R yuy'sx102x2 +1.738X10 lum) = 10543x107 x² V -001264 nin inches.

part (6) first Draw Bending moment diagram with given equations. of the beam RAP w 니 3 B FBD 21 3 PC Rc A B C 0 BMD 다 MCO N=L Moment at a 2 WL Mc = -2x W x L 3 method. According to moment area Deflection at B SB = moment of area about B + EISB = + 글쌓사서 능) + 불 () ) gw13 -ㅋw3 키 SB 1502 81 ET 8 EL 81 EI

Griven SB=0.250 in 1 → 0.250 = 15 X IN X 963 81x107x220135 mar allowable W = 337.753 lb π