i 3 i 2 Problem 8.31 Shear flow in circular section The centroidal axes are located at the center of the circle as shown above.

Transcription

1 Problem 8.31 Sher flow in circulr section i 3 R θ s i 2 t Remove@"Globl` "D H remove ll symbols L The centroidl xes re locted t the center of the circle s shown bove. (1) Find bending stiffness: From symmetry, = H 33 nd H 23 =. = 2 Ε πê2 πê2 HR Sin@θDL 2 tr θ π R 3 t Ε (2) Find sher flow distribution due to : The loding H ) is lso symmetric nd therefore the sher flow must vnish on the symmetry xis i 3. As result, f q Hq= pê2l = n the problem cn be nlyzed s two bck-to-bck semicirculr sections (see problem 8.18). Rther thn mke use of this importnt simplifiction, we will go hed nd construct "nive" solution, mking cut t q= insted. Sher flow f o in open (cut) system: θ Q 2 θ =Ε R Sin@αD tr α R 2 t Ε H 1 + Cos@θDL f oθ = Q 2 θ H 1 + Cos@θDL π R

2 2 chpter8esolutions.nb Closing sher flow f c : 2 π f oθ + f c wrp = R θ Gt 2 π Rf c 2 Gt 8f c < = 8f c <ê. Solve@wrp, f c D@@1DD : π R > Sher flow f o + f c : f θ = f oθ + f c êê Simplify Cos@θD π R NOTE: This is identicl to the expression for sher flow in semicirculr section, -p/2 q p/2. As result, the sher flow is zero on symmetry xis i 3 s pointed out t the beginning. (3) Find loction nd mgnitude of mx sher flow: It is cler from the expression for f q tht it is mximum t q=, p (where section crosses xis i 2 ), nd the vlue is: f θmx = f θ ê. θ π R

3 14 chpter8esolutions.nb Problem 8.34 Sher flow in circ. tube with flnges t i 3 i 2 R t t s3 s4 B θ s 2 A s 1 cut Remove@"Globl` "D H remove ll symbols L Sher is pplied to section. =αr; H simplifies the lgebr L (1) Find centroid nd bending stiffnesses: Since both xes re symmetry xes, the centroid is t the center of the circle s shown in the figure bove. 2 π =Ε HR Sin@θDL 2 tr θ π R 3 t Ε 2 π H 33 =Ε HR Cos@θDL 2 tr θ t + t R êê Simplify 1 3 R3 t I3 π+2 α I3 + 3 α+α 2 MM Ε H 23 = ; H due to symmetry L (2) FInd sher flow on section: The presence of the flnges will hve no effect becuse they both lie on xis i 2 nd therefore their stiffness sttic moment, Q, is zero. As result, this problem reduces to s simple circulr tube. Nonetheless, we will illustrte the full "nive" solution tht ssumes we didn't relize this. Sher flow f oi in open (cut) section: Cut tube s indicted in figure to crete n open section described by s i where s 2 = R q 2 nd s 4 = R q 4. Use nottion where Q ij refers to xis i i nd perimetric coordinte s j.

4 chpter8esolutions.nb 15 Q 21 =Εs 1 t; H right flnge L f o1 = Q 21 θ 2R Q 22 =Ε Sin@θD tr θ H upper semicircle L R 2 t Ε H 1 + Cos@θ 2 DL f o2 = Q 22 + f o1 ê.s 1 H 1 + Cos@θ 2 DL π R Q 23 =Εs 3 t; H left flnge L f o3 = Q 23 θ 4 R Q 24 =Ε Sin@θD tr θ H lower semicircle L R 2 t Ε H 1 + Cos@θ 4 DL f o4 = Q 24 + f o2 + f o3 ê. 8θ 2 π,s 3 < êêsimplify H1 + Cos@θ 4DL π R

5 16 chpter8esolutions.nb Closing sher flow, f c : π f o2 + f c wrp = Gt π f o4 + f c R θ 2 + R θ 4 êê Simplify Gt 2 π Rf c 2 Gt 8f c < = 8f c <ê. Solve@wrp, f c D@@1DD : π R > f 1 = f o1 f 2 = f o2 + f c êê Simplify Cos@θ 2 D π R f 3 = f o3 f 4 = f o4 + f c êê Simplify Cos@θ 4D π R (3) Verify equilibrium t A & B nd flnge ends: f 1 ê.s 1

6 chpter8esolutions.nb 17 f 1 + f 4 f 2 ê. 8s 1, θ 4 π, θ 2 < True f 3 ê.s 3 f 2 + f 3 f 4 ê. 8θ 2 π,s 3, θ 4 < True (4) Find loction nd mgnitude of mx sher flow: By inspection of sher flow expressions, mx sher flow is in the circulr tube t A nd B. Using A: f mx = f 2 ê. θ 2 π R

7 24 chpter8esolutions.nb Problem 8.36 Sher flow in box-z cross section t i 3 i 2 s 2 i 3 s 1 s 5 s 4 cut i 2 s 6 c b s 3 Remove@"Globl` "D H remove ll symbols L Sher is pplied to section. α=β=1 ê 2; H problem specifiction L b =β; c =α; H simplifies the lgebr L (1) Find centroid nd bending stiffnesses: For equl flnge sizes, the centroid clerly is t the geometric center of the box. =Ε t + 2t t Ε H 33 =Ε t + t t t Ε H 23 =Ε t t t Ε

8 chpter8esolutions.nb 25 H = H 33 HH 23 L t 2 Ε 2 Find sher flow distribution in section: It is necessry to mke cut to crete n open section. It is generlly esier to mke the cut t corner or junction s shown in the figure bove. Sher flow f oi in cut section : NOTE: The first index defines the xis nd the second defines the perimetric coordinte number. Q 21 =Εs 1 t 2 ; Q 31 =Εs 1 t b 2 s 1 2 ; Q 22 =Εs 2 t 2 s 2 2 ; Q 32 =Εs 2 t b 2 ; Q 23 =Εs 3 t 2 ; Q 33 =Εs 3 t b 2 + s 3 2 ; Q 24 =Εs 4 t 2 + s 4 2 ; Q 34 =Εs 4 t b 2 ; Q 25 =Εs 5 t 2 ;

9 26 chpter8esolutions.nb Q 35 =Εs 5 t b 2 c + s 5 2 ; Q 26 =Εs 6 t 2 ; Q 36 =Εs 6 t b 2 + c s 6 2 ; f o5 = HQ 35 H 23 Q 25 H 33 L H êê Simplify 3s 5 H + 6s 5 L 31 3 f o6 = HQ 36 H 23 Q 26 H 33 L H êê Simplify 3s 6 H + 6s 6 L 31 3 f o1 = HQ 31 H 23 Q 21 H 33 L H + f o6 ê.s 6 c êê Simplify 3 I s 1 + 6s 2 1 M 31 3 f o2 = HQ 32 H 23 Q 22 H 33 L H + f o1 ê.s 1 b êê Simplify 3 I s 2 1 s 2 2 M 31 3 f o3 = HQ 33 H 23 Q 23 H 33 L H + f o2 + f o5 ê. 8s 2, s 5 c< êêsimplify 3 I s 3 + 6s 3 2 M 31 3

10 chpter8esolutions.nb 27 f o4 = HQ 34 H 23 Q 24 H 33 L + f o3 ê.s 3 b êê Simplify H 3 I s s 2 4 M 31 3 Continuity t cut in box: b f o1 + fc wrp = Gt f o2 + fc b f o3 + fc s 1 + Gt s 2 + G t f o4 + fc s 3 + s 4 G t 3fc Gt G t fc =. 8fc< = 8fc< ê. Solve@wrp, fcd@@1dd : > % êê N : > Compute finl sher flows: f 1 = f o1 + fc êê Simplify 3 I3 2 7s 1 6s 1 2 M 31 3 f 2 = f o2 + fc êê Simplify 3 I s 2 1 s 2 2 M 31 3

11 28 chpter8esolutions.nb f 3 = f o3 + fc êê Simplify 3 I s 3 + 6s 3 2 M 31 3 f 4 = f o4 + fc êê Simplify 3 I s 4 1 s 4 2 M 31 3 f 5 = f o5 êê Simplify 3s 5 H + 6s 5 L 31 3 f 6 = f o6 êê Simplify 3s 6 H + 6s 6 L 31 3 (3) Verify equilibrium t corners nd ends: f 6 ê.s 6 f 6 + f 4 f 1 ê. 8s 6 c, s 4, s 1 < êêsimplify True f 1 f 2 ê. 8s 1 b, s 2 < f 2 + f 5 f 3 ê. 8s 2, s 5 c, s 3 < êêsimplify True

12 chpter8esolutions.nb 29 f 3 f 4 ê. 8s 3 b, s 4 < f 5 ê.s 5 (4) Find loction nd mgnitude of mx sher flow: The points of mx sher flow cn be identified from plots. f1n = f 1 ê.s 1 ηb êê Simplify 3 62 I η+3 η2 M p1 = PlotBf1n, 8η,,1<, AxesLbel :"s 1 ê", " f 1 ">, ImgeSize SmllF; f2n = f 2 ê.s 2 η êê Simplify 3 31 I 2 13 η+1 η2 M p2 = PlotBf2n, 8η,,1<, AxesLbel :"s 2 ", " f 2 ">, ImgeSize SmllF; f3n = f 3 ê.s 3 ηb êê Simplify 3 62 I η+3 η2 M p3 = PlotBf3n, 8η,,1<, AxesLbel :"s 3 ê", " f 3 ">, ImgeSize SmllF;

13 3 chpter8esolutions.nb f4n = f 4 ê.s 4 ηêê Simplify 3 31 I 2 13 η+1 η2 M p4 = PlotBf4n, 8η,,1<, AxesLbel :"s 4 ", " f 4 ">, ImgeSize SmllF; f5n = f 5 ê.s 5 ηc êê Simplify 3 η H1 + 3 ηl 62 p5 = PlotBf5n, 8η,,1<, AxesLbel :"s 5 ", " f 5 ">, ImgeSize SmllF; f6n = f 6 ê.s 6 ηc êê Simplify 3 η H1 + 3 ηl 62 p6 = PlotBf6n, 8η,,1<, AxesLbel :"s 6 ", " f 6 ">, ImgeSize SmllF;

14 chpter8esolutions.nb 31 p2<, 8p3, p4<, 8p5, p6<<d f 1 f s 1ê s 2 f 3 f s 3 ê s 4 f 5 f s s 6 Mx sher flow is t opposite points in the two sides H f 2 & f 4 ). Using f 4 for clcultions: 8s crit < = 8s 4 <ê. Solve@ s4 f 4, s 4 D@@1DD : 13 2 > f mx = f 4 ê.s 4 s crit f mx êê N.62419

15 Problem 8.43 Sher center of double-box section s 3 b s 6 t 1 i 3 t 1 i 2 t 2 t2 i 3 t w i2 i 3 s 2 f c1 s 7 s1 f c2 s 5 c b s4 Remove@"Globl` "D H remove ll symbols L In order to work this problem, vlues must be specified for the geometry. Following problem 8.38, we will use the following geometry: b = ; c = 3; t 1 = t 2 = t; t w = 2t; H problem specifiction L Axis i 2 is symmetry xis for this section nd therefore the sher center lies on this xis Hx 3 k = L. To determine x 2 k it is only necessry to pply, determine the resulting sher flow, nd use moment equipollence to find the sher center. The sher flow is determined in problem 8.38 nd is repeted here. ü Find sher flow distribution on section: The centroid is on the horizontl centerline of the section which is symmetry xis. As result, H 23 =, nd since only is pplied, it is only necessry to clculte (knowledge of b is not required). 1 =Ε 12 3 t t w t 2 + 2ct t Ε 2 + 2bt ü Sher flow f oi in open (cut) section: It is necessry to mke 2 cuts to crete n open section. Mking cuts t corner or junction simplify the selection of perimetric coordintes - see bove figure. In the nottion below, the first subscript refers to the xis nd the second to the perimetric coordinte. Q 21 =Εs 1 t 1 2 ; f o1 = Q 21 3s Q 22 =Εs 2 t s 2 2 ;

16 2 p8.43-f9.nb f o2 = Q 22 + f o1 ê.s 1 c êê Simplify 3 I3 2 + s 2 s 2 2 M 14 3 Q 23 =Εs 3 t 1 2 ; f o3 = Q 23 + f o2 ê.s 2 êê Simplify 3 H3 s 3 L 14 2 Q 24 =Εs 4 t 2 2 ; f o4 = Q 24 3s Q 25 =Εs 5 t s 5 2 ; f o5 = Q 25 + f o4 ê.s 4 b êê Simplify 3 I 2 + s 5 s 5 2 M 14 3 Q 26 =Εs 6 t 2 2 ; f o6 = Q 26 + f o5 ê.s 5 êê Simplify 3 H s 6 L 14 2 Q 27 =Εs 7 t w 2 + s 7 2 ; f o7 = Q 27 6 I 2 + s7 2 M s ü Find closing sher flows f c1 nd f c2 from comptibility t cuts: c f o1 f c1 f o2 f c1 c f o3 f c1 f o7 + f c1 f c2 wrp1 = s 1 + s 2 + s 3 s 7 êê Simplify Gt 1 Gt 1 Gt 1 Gt w 15 f c1 + 7f c G t

17 p8.43-f9.nb 3 b f o4 + f c2 f o5 + f c2 b f o6 + f c2 f o7 + f c1 f c2 wrp2 = s 4 + s 5 + s 6 s 7 êê Simplify Gt 2 Gt 2 G t 2 G t w 7 f c f c G t 8f c1,f c2 < = 8f c1,f c2 <ê. Solve@8wrp1, wrp2 <, 8f c1,f c2 <D@@1DD : , > ü Find totl sher flow: f 1 = f o1 f c1 êê Simplify 3 H s 1 L f 2 = f o2 f c1 êê Simplify 3 I s 2 26 s 2 2 M f 3 = f o3 f c1 êê Simplify 3 H37 26 s 3 L f 4 = f o4 + f c2 êê Simplify 3 H s 4 L f 5 = f o5 + f c2 êê Simplify 3 I s 5 26 s 2 5 M f 6 = f o6 + f c2 êê Simplify 3 H17 26 s 6 L f 7 = f o7 + f c1 f c2 êê Simplify 3 I s 7 26 s 7 2 M ü Find sher center: The lternte method given in eq. 8.4 or is much esier if equipollence is enforced t the top of the center web, thus eliminting the contributions of f 3, f 6 nd f 7 to the moment. We will use this s point "" in the following clcultions.

18 4 p8.43-f9.nb b = 5 6 c t 1 c + 2ct t 2 H bl + 2bt 2 b 2 ì HH2 c+ L t 1 + H2 b+ L t 2 + t w LêêSimplify x 2 = b ; H loction of "" t top of center web L c R 1 = f 1 s R 2 = f 2 s b R 4 = f 4 s R 5 = f 5 s M = R 1 cr 2 + R 4 + br 5 êê Simplify x 2k = x 2 + M ê. 1 H sher center horiz. loction L x 2k êê N H numeric result L x 2 x 2k êê N H sher center distnce to left of center web L

19 Untitled-3 5 Find sher center: The lternte method given in eq. 8.4 or is much esier if equipollence is enforced t the top of the center web, thus eliminting the contributions of f 3, f 6 nd f 7 to the moment. We will use this s point "" in the following clcultions. c In[28]:= b = t 1 c + 2ct t 2 H bl + 2bt 2 b 2 ì HH2 c+ L t 1 + H2 b+ L t 2 + t w LêêSimplify Out[28]= 5 6 In[29]:= x 2 = b ; H loction of "" t top of center web L c In[3]:= R 1 = f 1 s 1 Out[3]= In[31]:= R 2 = f 2 s 2 Out[31]= b In[32]:= R 4 = f 4 s 4 Out[32]= In[33]:= R 5 = f 5 s 5 Out[33]= In[34]:= M = R 1 cr 2 + R 4 + br 5 êê Simplify Out[34]=

20 6 Untitled-3 In[35]:= x 2k = x 2 + M ê. 1 H sher center horiz. loction L Out[35]= In[36]:= x 2k êê N H numeric result L Out[36]= In[37]:= x 2 x 2k êê N H sher center distnce to left of center web L Out[37]=.76989

21 Problem 8.45 Strength of cnt. bem with C-chnnel section i 3 L P s 1 b i e 1 i 3 M K N i 2 h s 3 d P s2 In[38]:= Remove@"Globl` "D H remove ll symbols L In[39]:= L = 2.; h =.4; b =.2; t =.4; H bem & section L In[4]:= Ε= ; σ y = ; ν=.25; H mteril unitsd L In[41]:= P = ; L In[42]:= G = Ε ; H sher modulus L 2 H1 +νl (1) Find loc. & mg. of mx bending moment, sher, & torque: From sttics, the mximum bending moment for tip-loded cntilever is t the root. Both the sher nd torque re constnt long the length of the bem. Thus the criticl section is t the root. For the given geometry nd tip lod: In[43]:= 8M 2 mx = PL, mx = P, M 1 mx = P Hd el< H use list to group for ssignment L Out[43]= 81., 5, 5 Hd el< All other sectionl lods re zero. (2) Find xil nd sher distribution on the criticl section: Initil clcultions: In[44]:= e = 3b H sher center loction from web using eq.h8.41l L 6 + h ê b Out[44]=.75

22 2 Untitled-4 In[45]:= b = 2bt b ì H2 bt+ htl H centroid loction from web L 2 Out[45]=.5 In[46]:= =Ε 1 12 h3 t + 2bt h 2 2 H bending stiffness L Out[46]= In[47]:= H 11 = 1 3 G H2 b+ hl t3 H torsionl stiffness L Out[47]= Bending stress clcultions: For purposes of clculting the xil nd sher stresses on the section, we will use the 3 perimetric coordintes s i in the figure. Recll tht s 1 = E M 2 x 3. In[48]:= σ 1 top =Ε M 2 mx h H bending stress in top flnge L 2 Out[48]= In[49]:= σ 1 web =Ε M 2 mx h 2 s 2 H bending stress in web L Out[49]= H.2 s 2 L In[5]:= σ 1 bot =Ε M 2 mx h 2 H bending stress in bottom flnge L Out[5]= Sher stress due to The sher stress is constnt through the wll thickness, t si = f i êt where the sher flow for C-chnnel is given by eqs. (8.26, 8.27 & 8.28) s: f 1 =- 1 2 Ehts 1 ê, f 2 =- 1 2 Es 2 t Hh - s 2 L ê nd f 3 = 1 2 Ehts 3 ê.

23 Untitled-4 3 In[51]:= f 1 = 1 2 Ε hts mx 1 ; H sher flow in top flnge L In[52]:= τ s1 = f 1 ê t H sher stress in top flnge L Out[52]= s 1 In[53]:= f 2 = 1 2 Ε s 2 t Hh s 2 L mx ; H sher flow in web L In[54]:= τ s2 = f 2 ê t H sher stress in web L Out[54]= H.4 s 2 L s 2 In[55]:= f 3 = 1 2 Ε hts 3 mx ; H sher flow in bottom flnge L In[56]:= τ s3 = f 3 ê t H sher stress in bottom flnge L Out[56]= s 3 Sher stress due to torque: The sher flow due to torsion is given by eq.(7.65) s t t = GtM 1 ê H 11 with right-hnd positive sense. In[57]:= τ t = GtM 1 mx ê H 11 H sher stress due to torque L Out[57]= H.75 + dl (3) Find P mx (d) using von Mises yield criterion: The xil nd sher stresses define plne stress problem with s 2 =. The von Mises equivlent stress must be computed s function of the perimetric coordintes long the flnges nd web. Since t t is of opposite sign on the opposite wlls of the section, the extreme sher stress is the sum of the mgnitudes of t s nd t t. In[58]:= σ e1 = Iσ 1 top M HAbs@τ s1 D + Abs@τ t DL 2 H von Mises stress in top flnge L Out[58]= I Abs@.75 + dd Abs@s 1 DM 2

24 4 Untitled-4 In[59]:= σ e2 = Hσ 1 web L HAbs@τ s2 D + Abs@τ t DL 2 H von Mises stress in web L Out[59]= - J3 I Abs@.75 + dd Abs@H.4 s 2 L s 2 DM H.2 s 2 L 2 N The bottom flnge is identicl to the top nd will not be clculted. Clerly, s e1 is mx for the lrgest s 1 which is t the corner, nd this must be equl to s e2 t the top of the web. As result, s e2 is the governing stress. For slender bem such s the present cse, the sher stress due to is smll compred to the xil stress. However, becuse the torsionl stiffness, H 11, is extremely smll for open sections, even smll torque will produce reltively lrge sher stresses which my dominte the equivlent stress. This is constnt stress over the entire section,so it preserves the symmetry of s e over the section. Thus, the extreme vlue of s e2 will occur t either t the end of the web, s 2 =, or t the midpoint, s 2 = hê2. Both re plotted below: In[6]:= se1 =σ e2 ê.s 2 ; se2 =σ e2 ê.s 2 h ê 2; H σ e2 t web top nd midpoints L In[61]:= Needs@"PlotLegends`"D In[62]:= Plot@8se1, se2<, 8d,, b<, AxesLbel 8"d", "σ e2 "<, PlotStyle 8Blck, Dshed<, PlotLegend 8"Web end", "Web mid"<, LegendPosition 8.7,.4<D 2.5 μ 1 8 s e2 2. μ 1 8 Out[62]= 1.5 μ μ μ 1 7 Web end Web mid d In[63]:= 8se1, se2< ê. d H web middle governs L Out[63]= , = In[64]:= 8se1, se2< ê. d e H web end governs L Out[64]= , =

25 Untitled-4 5 In[65]:= 8se1, se2< ê. d b H web middle governs L Out[65]= , =