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Lecture 5 Solution Method for Beam Deflection Problem 5-1: Consider the clamped-clamped elastic beam loaded by a uniformly distributed line load q. q x l EI a) Formulate the boundary conditions. b) Find the deflected shape of the beam using the direct integration method. c) Find the maximum deflection magnitude and location. d) Determine the location and magnitude of the maximum stress in the beam. Problem 5-1 Solution: (a) Boundary conditions w(0) w '(0) w(l) 0 w '(l) 0 (b) Find the deflected shape use direct integration We use the 4th order differentiated equation: EIwIV q ql 3 6 l2 2 Integrate 3 times EIw ' l 0 C1 C2l C3 1 Use B.C. w '(0) w '(l) 0 We get C3 EIw' l ql 3 6 0 l2 C1 2 0 C2l (1) Integrate equation qx3 6 EIw ' C1 x2 2 C2 x We get EIw qx 4 24 x3 C1 6 x2 C2 2 C4 Use B.C. w(0) w(l) 0 We get C4 EIw l 0 ql 4 24 0 l3 l2 C1 C2 6 2 (2) Combine equations (1) and (2) to solve for C1 and C2 C1 ql 2 C2 ql 2 12 2 Finally w( x ) qx 2 x2 24EI or ql 4 24EI w(l ) 4 x l 2lx l 2 x 2 l 3 x l (c) The maximum deflection magnitude occurs at the mid-span wmax x q l 24EI 2 l 2 wmax l x 2 2 l 2 2 2l l 2 l2 ql 4 384EI (d) Determine location and magnitude of the maximum stress Stress distribution in the beam is Mz I The maximum stress locates where moment and z magnitude are the maximum: max M max zmax I Recall M EIw '' Substitute w x into the above equation M q x2 2 lx l 2 2 12 3 Let’s consider moment at x M L 2 l x 2 and x 0 q M M 2 1 l 2 2 max max ql 2 24 ql 2 12 x 0 M l2 12 l l 2 2 x 0 ql 2 12 ql 2 zmax 12 I 4 Problem 5-2: Calculate the second moment of inertia of the beam cross section for: a) Solid rectangular cross section of width b and height h. b) Thin-walled square box section of width and height b. c) Solid circular cross section of radius r. Problem 5-2 Solution: (a) Solid rectangular cross section of width b and height h. Ix A y 2 dA h 2 h 2 Ix y 2bdy bh3 12 (b) Thin-walled square box section of width and height b. I square s s4 12 1 2 s 12 1 4ts 12 2t 4 12 s 2t 2 s2 s 2t 2 4t 2 2s 2 4ts 4t 2 5 Eliminate higher order terms of t I square 1 4ts 12 1 2ts 3 3 4t 2 2s 2 4ts 4t 2 4t 2 s 2 2 3 tb 3 I square (c) Thin-walled square box section of width and height b. dA rd Ix A y 2 dA 2 r 0 0 2 rdrd r4 sin 2 drd 4 2 0 r4 4 2 r 4 r sin 1 sin 2 4 2 0 3 J Ix Ix Iy r3 2 Iy r3 4 6 Problem 5-3: In wood construction building codes the beam deflections cannot exceed L/360 where L is the length of the beam. Where do you think this requirement comes from? Choose a typical beam example and state clearly the formulation and your assumptions on the boundary conditions and loading. Using the deflection criteria estimate the fracture strain of the plaster board which is nailed directly to the ceiling beams (joist) in single home construction. Problem 5-3 Solution: The L/360 constraint is required to prevent the plaster board from cracking We assume a simply supported beam with a distributed load The deflected shape and subsequently curvature can be simplified as 0 xx xx z d dx 0 z L / 360 L/2 z /180 The maximum strain will be at the outmost fiber h at z h xx 2 360 Typically the wooden beams used in house construction are 2in×6in. Use those values 6 xx 2 360 1 60 7 The strain seen by the extreme fiber of the wood is also seen by the plaster board. 8 Problem 5-4: Given a beam with a “T” section subjected to pure bending shown in Figure 1, calculate: a) the location of the neutral axis b) the second moment of inertia c) Find the shear stress distribution in the “T” section. Figure 1 Problem 5-4 Solution: a) The location of neutral axis is yi Ai Ai y1 A1 y2 A2 A1 A2 1 L t L t t L t 2 L t t Lt t tL 2 3L2 3tL t 2 4 L 2t 9 If we assume L t 3L 4 b) t ,and don’t consider stress variation over the thickness in the flange, the If we assume L second moment of inertia is I Ii Ai di tL3 12 Lt L tL3 12 Lt 3 L L 4 tL3 12 tL3 8 Lt 3 12 2 I Lt 2 Lt 2 L 2 3 L 4 L 2 2 5tL3 24 Shear stress distribution First, let’s define two parts of the cross section: Part I the horizontal part and Part II the vertical part Part I QI z I dA where z I is the moment are to the shaded region I 10 zI t constant 2 dA t dy L L 2 y L t L t 2 QI QI t 2 tdy L 2 y Part II QII 2QI z II dA y 0 where zII is the moment are to the shaded region II zI QI QII t L 2QI t 2 z d A tdz L t y 0 z L 2 t 2 check QII = z ztdz L t z2 0, yes! VQ It Finally, the shear stress distribution is showed as below picture 11 Problem 5-5: Continuity Condition Solve the problem of a simply-simply supported beam loaded by a point force acting at eh symmetry plane, but at a distance a from the left support In the notes of lecture 5 the solution of this problem was outlined, but not completed, (a) Complete the derivation by calculating all four integration constants (b) Proof that all continuity conditions are satisfied at x=a (c) Show that in the limiting case of a=L/2 the solution is identical to one that was derived in class and you were asked to memorize Problem 5-5 Solution: (a) The reaction force are calculated from moment equilibrium RA P b l P RB P 1 a l a l (1) The corresponding bending moments and shear forces are RA x Pbx l RB l x M x V x Pb l Pa l Pa l x l 0 x a a x L 0 x a a x L (2) 12 Integrating governing equation EI d 2w dx 2 M x (3) Combing with equation (2), we have EIwI EIw II Pbx3 C1 x C2 6l Pa lx 2 x 3 C2 x C 4 l 2 6 0 x a (4) a x L Boundary conditions and continuity conditions w 0 w l 0 wI a w II a dw I dx dw II dx x a (5) x a Substitute (4) into (5) , we have C2 0 Pal 3 C3l C4 0 3 Pa la 2 Pba 3 C1a 6l l 2 Pba 2 2l C1 Pa la l a3 6 a2 2 C3 a C4 (6) C3 Solve for (6), we get the four unknown integration constants C1 C2 C3 C4 Pa a l 2l a 6l 0 Pa 2 2 3 Pa 6 Pl l a 2l a a 6l 13 (b) Check continuity condition (i) Check continuity of moment M at x=a, referring equation (2) M (ii) M Pa l a l Pba l 0 Check continuity of shear force V at x=a, referring equation (2) V (iii) M Check continuity of angle dw I dx x a II dw dx x a V V Pb l Pa l P at x=a Pba 2 2l Pa a l 2l a 6l a2 2 Pa la l Pa 2 2 P l a 2l a a 6l 0 (iv) Check continuity of deflection w w w w Pba 3 Pa 2 a l 2l a 6l 6l 2 Pa 3 P l a 2l a a Pa la a 3 l 2 6 2 6l w w Pa 3 6 0 All continuity conditions are satisfied. (c) Referring to equation (4) wI a L ,b 2 1 Pbx 3 EI 6l C1 x C2 L 2 wI wo Px 3l 2 4 x 2 48 EI PL3 I w L x 48 EI 2 They are the same as was derived in lecture notes. 14 Problem 5-6: Another problem to test your knowledge on continuity conditions. A system of two identical beams shown in the figure below is a statically determined problem. The beams are rigidly welded, so that the angle remains 90 degrees. Determine (a) The distribution of bending moment and shear forces in both segments. (b) Find the deflected shape of both beams and make sketch. (c) Find the relation between the tip-- load and the vertical displacement of the tip.(Hint: assume the rotations to be very small.) (Note: Do not solve this problem using the Castigliano’s theorem, which is much simpler, but has not been covered yet.) Problem 5-6 Solution: (a) Free body diagram as below, in which Rxa , Rxb and R ya are reaction forces 15 Moment equilibrium about point a Rxb 2 L PL 0 P 2 Rxb Horizontal and vertical forces balance Rya P Rxa Rxb P 2 Beam 1 Let’s look at cross section O+ Where V M P 2 P L 2 Also, let’s look at cross section O- 16 Where V M P 2 P L 2 Distribution of shear force Distribution of moment 17 Beam 2 At the left cross-section V M P PL Distribution of shear force Distribution of moment 18 (b) Deflection is anti-symmetric for beam 1, at point O, w 0 , deflection shapes are (c) From moment distribution of beam 1, we know M1 P L 2 d 2 w1 dy 2 M1 y 0 y L Governing equation EI P L y 2 0 y L 0 y Integrate twice w1 P Ly 2 2 EI 2 y3 6 C1 y C2 L Apply boundary conditions w1 y 0 C1 w1 y L PL2 , C2 6EI 0 0 19 P Ly 2 2 EI 2 w1 y3 6 PL2 y 6EI 0 y L At y=0 PL2 6EI Vertical displacement of point O o uo x L P dy L EA o dy x PL EA Integrate governing equation of beam 2 d 2 w2 EI dy 2 P L x M2 We have EI dw2 dx EIw2 P x2 2 P Lx Lx 2 2 x3 6 C1 C1 x C2 Use angle continuity condition EI dw2 dx PL2 6 C1 w2 1 x 0 x 0 PL2 6EI x 0 PL EA PIL A u0 C2 Finally w2 P Lx 2 EI 2 x3 6 PL2 x 6EI PL EA 20 At x=L, the tip deflection is PL3 2EI wp PL EA Now, let’s compare the magnitude of the two terms in w p , for a square cross-section wp II wp I II I For h L 1 20 , wp wp PL EA PL3 2EI 2I LA h3 12 Lh 2 2 1 h 6 l 2 0.25 10 3 . wp II , which is the deflection caused by compression, I is three orders of magnitudes smaller than wp , which is the deflection caused by bending. 21 MIT OpenCourseWare http://ocw.mit.edu 2.080J / 1.573J Structural Mechanics Fall 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.