Problem 5-1: Consider the clamped-clamped elastic beam loaded by a uniformly distributed line load q. l EI

Transcription

1 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 E 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(l) 0 w'(0) w'(l) 0 (b) Find the deflected shape use direct integration We use the 4 th order differentiated equation: Ew V q ntegrate 3 times 3 ql l Ew' l 0 C1 ClC 3 6 1

2 Use B.C. w'(0) w'(l) 0 We get C 3 0 Ew' 0 ql 3 l 1 l C C l (1) 6 ntegrate equation Ew' qx 3 x 1 C 6 C x We get qx 4 x 3 x Ew C 1 C 4 6 C 4 Use B.C. w(0) w(l) 0 We get C 4 0 Combine equations (1) and () to solve for C 4 3 ql l l Ew l 0 C1 C () 4 6 and C 1 ql C 1 ql C 1

3 Finally qx wx ( ) lxl 4E or x ql 4 x 4 x 3 x wl () 4E l l l (c) The maximum deflection magnitude occurs at the mid-span q l l l wmax l l x 4E l w max l x ql 4 384E (d) Determine location and magnitude of the maximum stress Stress distribution in the beam is Mz The maximum stress locates where moment and z magnitude are the maximum: max M max z max Recall M Ew'' Substitute wx into the above equation M q lx l 1 x 3

4 Let s consider moment at x L and x 0 M 1 l l l l ql l q x 1 4 M ql x0 1 M M ql max x0 1 ql max 1 z max 4

5 Problem 5-: 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- Solution: (a) Solid rectangular cross section of width b and height h. x yda A h h ybdy bh 3 x 1 (b) Thin-walled square box section of width and height b. 4 square s s t s st 1 1 4ts 4t s 4ts 4t 1 4 s s t 5

6 Eliminate higher order terms of t square 1 4 t 1 s 4t s 4ts 4t 1 ts 3 4 t s 3 t b 3 square 3 (c) Thin-walled square box section of width and height b. y x da A r da rd r 0 0 r sin rdrd sin drd r 4 1 sin 4 r J x x r 3 y r 3 y 4 6

7 Problem 5-3: n 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 d L/360 0 xx xx z 0 z z/180 dx L / The maximum strain will be at the outmost fiber at z h h xx 360 Typically the wooden beams used in house construction are in 6in. Use those values 6 1 xx

8 The strain seen by the extreme fiber of the wood is also seen by the plaster board. 8

9 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 ya A i i ya 1 1 y A A A i 1 1 t L t L t t L t tl Lt tlt 3L 3tLt 4L t 9

10 f we assume L t 3L 4 b) f we assume L t,and don t consider stress variation over the thickness in the flange, the second moment of inertia is i Ad i i tl 3 Lt 3 L Lt L Lt 1 1 tl L Lt L L Lt L tl 3 tl tL 3 4 Shear stress distribution First, let s define two parts of the cross section: Part the horizontal part and Part the vertical part Part Q z da where z is the moment are to the shaded region 10

11 t z L constant da tdy Q Q t L tdy L y t L tl y Part Q Q y0 z da where z is the moment are to the shaded region z z da tdz Q L t Q Q d y0 zt z z t L t tl L t z check Q 0, yes! z VQ = t Finally, the shear stress distribution is showed as below picture 11

12 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 n 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/ 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 b a R A P P 1 l l The corresponding bending moments and shear forces are a R B P (1) l Pbx R Ax l 0 l x R B l x l a x L Pb 0 xa l Pa a x L l M x Pa V x () 1

13 ntegrating governing equation Combing with equation (), we have E d w dx M x (3) 3 Pbx Ew C1x C 6l 0 xa Pa lx x 3 Ew CxC 4 l 6 a x L (4) Boundary conditions and continuity conditions w0 wl 0 w a w a dw dx x a dw dx x a (5) Substitute (4) into (5), we have C 0 Pal 3 Cl 3 C4 0 3 Pba 3 Pa la a 3 Ca 1 CaC 3 6l l 6 Pba Pa a C1 la C3 l l 4 (6) Solve for (6), we get the four unknown integration constants Pa C1 a lla 6l C 0 C 3 C 4 6 Pa Pl l 6l Pa 3 a l aa 13

14 (b) Check continuity condition (i) Check continuity of moment M at x=a, referring equation () M Pba M M 0 l l Pa l a (ii) Check continuity of shear force V at x=a, referring equation () Pb Pa V V V P l l (iii) Check continuity of angle at x=a dw Pba Pa alla dx xa l 6l dw Pa a Pa P lalaa la dx xa l 6l 0 (iv) Check continuity of deflection w Pba 3 Pa w all a 6l 6l 3 3 Pa la a Pa Plalaa Pa 3 w l 6 6l 6 w w w 0 All continuity conditions are satisfied. (c) Referring to equation (4) w 3 1 Pbx C1x C E 6l L L a, b w w Px 48E o w x L 3l 4x They are the same as was derived in lecture notes. 3 PL 48E 14

15 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 R xa, R and R xb ya are reaction forces 15

16 Moment equilibrium about point a R xb L PL0 P R xb Horizontal and vertical forces balance R ya P P R xa R xb Beam 1 Let s look at cross section O + Where P V P M L Also, let s look at cross section O - 16

17 Where P V P M L Distribution of shear force Distribution of moment 17

18 Beam At the left cross-section V P M PL Distribution of shear force Distribution of moment 18

19 (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 P M 1 L y 0 y L Governing equation dw 1 P E M 1 L y 0 y L dy ntegrate twice P Ly y 3 w1 C1y C 0 y L E 6 Apply boundary conditions w w 0 0 L 1 y 1 y C PL, C 0 6E 1 19

20 P Ly y 3 PL w 1 y 0 y L E 6 6E At y=0 PL 6E Vertical displacement of point O u o ntegrate governing equation of beam We have Use angle continuity condition o o P PL dy dy xl xlea EA dw E M dy PL x dw x E PLx C dx Lx x 3 Ew P CxC 1 6 dw PL E 1 x 6E dx 0 x0 PL C PL w u x 0 0 EA PL C A Finally w P Lx x 3 PL PL x E 6 6E EA 0

21 At x=l, the tip deflection is PL 3 PL w p E EA Now, let s compare the magnitude of the two terms in w p, for a square cross-section h 3 PL w p EA w PL p LA Lh 6 l E For hl1 0, w w p p w p, which is the deflection caused by compression, is three orders of magnitudes smaller than w p h, which is the deflection caused by bending. 1

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