sanswer to selected problems in ch 5 and 8

8/3/2019 Sanswer to Selected Problems in Ch 5 and 8

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26Solutions - Chapter 5

(b) Show that on the boundary the maximum shearing stress is given by2Mt .1 2 2 2 2(Ts) max = Z 3 vb + x3(a - b ) so that the greatest shearing stress does occur at thena b t... ,. ~ end of the minor axis.

, -2 Z] VZ . '[2Mt x2 x3ADs. (a) We had in Section 5.14, Eq. (5.14.10), (Ts)max = nab a4 + b4 Now, for[XZ=kr3, we have

2Mt [kZ 1] V2(Ts)max = nab a4 + b4 X3 Thus,(Ts)max is the largest on boundary.Z Z ( X ~ ) 2Mt [ 4 Z Z 2 VZ(b) On the boundary,xz=a 1 -2 " ' thus, (Ts)max = 2 3 b +x3(a -b)] .b nab

2MtIf b > a, then (Ts)max is largest at x sub 3 =0 and (Ts)max = -z- .nab5.60. Consider the torsion of a cylindrical bar with an equilateral triangular cross-section as inFig.P5.4. (a) Show that a warping function SO = a ( 3 x ~ X 3 - )generates an equilibrium stressfield. (b) Determine the constant a in order to satisfy the traction-free lateral boundarycondition. Demonstrate that the entire lateral surface is traction-free. (c) Write out explicitlythe stress distribution generated by this warping function. Evaluate the maximum shearingstress at the triangular comers and along the line X3 = 0 in a cross-section. Along the linex3 = 0 where does the greatest shearing stress occur?~ ~ ? -ADs. (a) To satisfy the equations of e ~ H 1 : 1 f f i , SO must satisfy the Laplaace equation [see Eq.

aZSO _(5.14.3)]. Now, -" -6a:x3 anddri X3aZtp _ i!:. iP.._-z - - 6a:x3 Thus 2 + Z -0.ax3 axz aX3(b) Consider the lateral boundary xz=a. Theunit normal is n=ez. The traction free con (-2a,0> (a,O>dition requires that [Eq. (5.14.4](Vtp) n = e ' ( n ~ 3 - n 3 X z ) . Thus,a [ 6 a : r 3 ~ + 3 ( a z - x ~ ) e : 3 ] ~ = e ' x 3 ' Le.;--} e'6aa:r3=e'X3 a= 00'

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Solutions Chapter 527

(C) From Eq. (i) preceding Eq. (5.14.4) , we have, on the lateral surface, t=[.u8'( - n ~ 3 + n J X V + , . u ( V < p ) nJel' = ,.u[6aa( - n ~ 3 + n J X V + ( V < p ) 'l1Jel' (A), V < p = a 6 x ~ 3 e z + a ( 3 x ~ - 3 x ~ ) e : , "'. (0 On x2=a and n=e:2J Eq. (A) becomes t = ,.ua[6a(-X3)+6ax3]el =0. (li) On x3=iTc2a+xv, n=4-< -e2+v'3e:,).

a . r-; 2 2 a 6x2 . r- ; 2 1 2V


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Solutions - Ci..apter 8, I\

Ans. Since the spring and the dashpot are connected in parallel, therefore, the total force ris given by: r=rsp+rdashand the total displacement e is given bye=esp=edash

deNow, rsp=Ge and rdash=l1 tit' thereforeder=Ge+'1 tit '

To find the relaxation function, we let e =eof!(t), where H(t) is the Heaviside function. Thenr=Geo H(t) +11 eoo(t), where OCt) is the Dirac function. Thus, the relaxation function isre=G H(t) +11 o(t),o

8.4. Obtain the force-displacement relationship for a dashpot (damping coefficient 110) and aKelvin-Voigt solid ( damping coefficient '1 and spring constant G, see the previous problem)connected in series. Also, obtain the relaxation function.Ans. Let rlev be the force transmitted by the Kelvin-Voight element and rd be the forcetransmitted by the dasspot. Also let elev be the elongation of the Kelvin-Voight element anded be the elongation of the dashpot with damping coefficient 110' Then, we haver=rd = rlev 0) and e=ed+elev (li)

d ed where rd =110"dt = r (iii)delevand rlev=G elev+l1 -;;;- = r (iv)

.. de ded elev r 1 r r GFrom Eq, (ll), tit = tit +di = 110 +7j (r-G elev) = 110+7j-Tj (e-ed)2. d e= (-L+l) dr _ G (de _ ded) = 110+l1dr G de G r

I.e., tit2 110 '1 tit '1 tit tit 11 110 tit Tj tit + 11 110' 11 '10 ('1+110) d r deG tit2 = G tit 110 tit +r. That is

VIII -,3


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Solutions Chapte: ti

('I]+110)d , de 'I] '1]0 d2c,+ G dt = '1]0 dt +G dt2'This is the force ,vs elongation relationship.To obtain the relaxation function, let E=o H(/), then

('I] +'1]0) d , '1]'1]0 d,+ G dt = Eo '1]0 O(/)+co G dt(/).d , G Eo'l]oG Eo '1]110 dor "di+ 1]+'1]0' = '1]+'1]0 0(/)+ '1]+'1]0 dt(/).

The integration factor for this first order differential equation is exp ['I]Z'I]/]' i.e.,d G EGt]o G G '1]'1]0 ddi< ,exp['I]+'I]/D = '1]+'1]0 exp['I]+'I]otJco O(/)+Eo exp['I] +'I]ot] 'I] +'I]odt(t)., G Gt]0 It G It G '1]'1]0 deo exp['I]+'1]/] ='1]+'1]0 _:Xp['I]+'I]ol] o(t)dt+_coexp['I]+'I]/J'I]+'I]odt(t)dt

Gt]o 11'1]0 II G d='1]+'1]0 + '1]+'1]0 _!XP['I]+'I]ol]d;O(/)dtGt]o '1]'1]0 G I '1]'1]0 It G G

= '1]+'1]0 + '1]+'1]0 [o(t)exp['I]+'I]otJ- co - '1]+'1]0 -co'l]+'I]o exp['I]+'I]ot]O(/)]dtGt]0 '1]'1]0 G '1]'1]0 G= '1]+'1]0 + '1]+'1]0 [o(/)exp['I]+'I]/J '1]+'1]0'1]+'1]0' Thus,,G '1]'1]0 G '1]'1]0Eo = '1]+'1]) '1]0 - '1]+'1]0 ] exp[ 'I]+'I]olJ + '1]+'1]0 0(/)8.5. A linear Maxwell fluid, defined by Eq.(8.1.1), is between two parallel plates which are oneunit apart. Starting from rest, at time I = 0, the top plate is given a displacement u = Vo I whilethe bottom plate remains fixed. Neglect inertia effects, obtain the shear stress history.Ans. The velocity field for the fluid in this motion is given by (inertia negelcted)VI = volf(t)Xz, V2 = V3 = 0 (ii)

where H(/) is the Heaviside function. The only non-zero rate of deformation component isDu = V ~ ( t ) . Thus, from the constitutive equation for the linear Maxwell fluid, Eq. (8.1.1b),

d,uwe obtain 'u +A( f t =fl voH(t)

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'.. ; '-. ; . ~ Solutions - Chapteri

8.6. Obtain Eq.(8.3.1) by solving the linear, nonhomogeneous ordinary differential equationEq.(8.1. Ib).Ans. The integration factor for the differential equation I ' + A. d&a =2#D is i l )'. That ist -

t lthe equation can be written as fr

sanswer to selected problems in ch 5 and 8

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