me 323: mechanics of materials homework 7 fall 2020 problem … · 2020. 10. 27. · me 323:...
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ME 323: Mechanics of MaterialsFall 2020
Homework 7Due Wednesday, Oct 21
Problem 7.1 (10 Points)The beam ABC is loaded as shown below. Let the beam’s elastic modulus be E, the second moment ofarea be I, and the uniform distributed load be q = 4P/L.
Determine the following using the second-order integration method:
(a) The deflection at the free end A in terms of P ,L,E and I.(b) Non-dimensional plot of the deflection along the beam ABCD (i.e. x/L vs. v/ |vA|, where |vA| is the
magnitude of deflection at the free end A). You can use Matlab or other tools for the plot.(c) True or False: The maximum deflection (absolute value) occur at point A.
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fBD_P gof b TITIA X IB ei t TRftp.RB ic
4LEquilibrium
of Efg o PatRB 914L Re o CD
EMB o PL 91446L L ReBL ORc Pg 49
Substituting into
RB 431 8931for thedeflections we need section thebeaminto two partsSection 1 0 EXELmmmm
p
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If5559bpenA
Emo 0 M PA 19X Iz 0M PX 921
Integrating Mic W.r.tl
EIzicxk PzI 9zx3 GIntegrating EIU Lx W r t x
EIU PII 9141 9 4section 2 LS X yL
Grn
Ifsb b bb bbtgqm.ge'DA X B OfKI RB 1
II
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4 EM o O Mz PX 9X Iz RBI D O
Max PX 9214RB X D
Integrating MEX W r.tl
EI2zEX7 PzI 9zx3 RBXILX tC3Integrating EI VIX W r t xEI UCH PII 921ftRB 94314144 Unknowns Ge Ca Ge
boundary compatibility condition
Boundary conditionsa point B is a roller joint
241 4 0EIU X L EI 2k O
EI 2 CX L givesEIU ex L EI 924 G Ltc z OAD
EI Zz KL gives
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EI U2CX L3 941 RBz.LI GL Cy 0
point D is a pinned jointZz 11 44 0
EI Zz at 44 3231 323914 8kg3446344 0
compatibilityconditionslope at point B should the same as thebeam is a singlecontinuous member2,9 4 2541 4
Eg 9 79 2911 RBI EfSolving the Egs GD to 44
9 8279Et E PE Z RB E
Cz 27944 Pet Z RB B
as Eff 9 B PLZ RB EC 4 259Et log Pet RB B
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Using 3 and substituting for RB we get
a 49.8 Z PE2 12914 43 PLS
C 3 739.23 13PLZ
at 9Et HqPR
The final deflection equation
IIe IIe EIE t3E t EEE4PLSBEI
for OfXp LIEEE.EE
IiEEII 4IE 3EIIEsEIf1 BEET FEIL Hope for KK 4
a The deflection at the free end
ax 07 2 07 ff tHaEEP
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Given 9 41
VA 2CX G 2123BE I
b The deflection of the beam is
c false which is obvious fromthedisplacement plot
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ME 323: Mechanics of MaterialsFall 2020
Homework 7Due Wednesday, Oct 21
Problem 7.2 (10 Points)The cantilever beam ABC, shown below, is fixed to ground at A and pinned at B to the rod BD. A uniformload wo is distributed along the beam between A and B. The beam is made from a material with an elasticmodulus E1 and a cross-sectional second area moment I1, with a length L1 + L2. The rod has a uniformcross-sectional area A2, length L3, and modulus of elasticity E2. For this problem, use the following pa-rameters: L1 = 2L2 = 2 m, L3 = 1.5 m, E1 = 3E2 = 70 GPa, A2 = 25 cm2, I1 = 85 cm4, and wo = 1 kN/m.
Determine the following using the second-order integration method:
(a) The reactions at point A.(b) The tension in the rod BC.(c) The deflection at the free end C.
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Equilibriumy
WI offsx
TFA
pet4
Fy o FA Wo4 53 0 I
EMA 0 MA two421 5134 0 2
3Unknowns 2 earns Statically indeterminate
Elongationez fBL3_
EL Az
4URNS MA FA c FB e ee3 egns d
Deflection
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Sections 03 XEL
F 8 4 mAFAtaxMeo M IwoXENA FAX 0
MEX lzwox4 FAX NA
QLXKEI.SN DXI l
SEIWOXIFAX MA DXI I
f f WoXIE FAKEMAX 9but 0407 0 69 0
440 5 DX
E t woXI f FAIREMARTEbut 2,02 0 062 0
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So
ycxkfe.tt woxIlzfaxEIMAx3
compatibilityV CLI ez
5 UKNS MAeFA FB ez VCH5 e9Us d E 4 CED
solvesKD
ITI Wo4 f FA f IMA43 fBL3E2AZG
Dc c 6
HITEight two4 FB ECEWolf FIBfBL3_Ez Az
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thatE f wait C It FB LieEi IFBB
4 Ez Az
WEI histGILLI FB
FB 344L 749.56N843 24511132Eu Az
a
FA Woh FB 1250.44N
MA IzWolf FB 4 500.88 NM
b FB 749.56NC section Z L E X S Lz
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Wolz FB
MAfgyat FB ofMEX
VIX
x
EMo 0
MEX FBA 4 two MA FAX
MIX fBLXLD Woh X Hz MATFAX0240 1 fMdxdXGIT
IBCE 4 7 WOLKE
MAX FLAIR c
but 0 443 0 14
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LIEB MAX FAITH Cf f WolffA MAX
CF FB f WolfQdD
qfBC 4 2 WOLKE
MAX FALI LIEB fwolf22 FWdx
e EB 413 Wo4 473
MAIL.IE t EIfB fWo43XtCzJbutcUzC47
2iC4213wqLI MALIifafe ezzywoftfFAl.FIMAY
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CE FBgI41
E
k EBEI LEIHEI Ew of 4 t 441 4 tLifMAI FA X
3
2241 4 Lz On 0028 M
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ME 323: Mechanics of MaterialsFall 2020
Homework 7Due Wednesday, Oct 21
Problem 7.3 (10 Points)A cantilever beam ABC is loaded as shown below. The beam is made from a material with a modu-lus of elasticity E and has a circular cross-sectional area with a diameter d. For this problem, use theparameters: L1 = 1 m, L2 = 2 m, d = 40 mm, E = 200 GPa, P = 150 N,Mo = 100 N.m, and wo = 80 N/m.
Determine the following using the superposition method:
(a) An analytical expression for the deflection curve, vx, and the deflection at point C.(b) The reaction at point B.
Note: Please refer to the attached table for the necessary beam deflection equations.
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By inspection the beam is staticallyindeterminate
Roller supportatB redundant
principle of superpositionWo Wo
m WuA B C A Bf at A B C
mo TF Mo TP 193gWo
IA Btfb
Mo MpIl Wo
ftp.t 0IxIz8fEIL54xIzxIx42wot2Ex7tEtEEx4t4oi 180 3
ATIA B e zExg fEf
6 YXtx7OEXFI
1 8eI4X D LEX
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VI Ext Foxy05 71
Y E it
A B C 031 2 1502
1 Ap6
94
If2SX7225X
IA BAnoof 501905 51
100 1FIR X DHEX
2 74
3osXE
EXKEtgv.azt100Xt5o
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Jack ELIE x715 9 cos Xfl3
If His Rt45KYEA B C
2 BEFITSX O El
f 3X DLEX
ZEN tf BEEFBIB ow
BIX BZ I
To find By
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2131 293 050
FEI 157750 b
By 5
EExk ER2sx osxs25 1 231 E3
U UCxH2Ex a
a
Tofindthedeflectionatpointe
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E If co0234 E 2004109
UCXmFO0l03MT
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ME 323: Mechanics of MaterialsFall 2020
Homework 7Due Wednesday, Oct 21
Problem 7.4 (5 points)
(A) Indicate which of the schematics presented below depicts the deflection curve of the following beam:(2 points)
Briefly justify your answer!!
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O
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ME 323: Mechanics of MaterialsFall 2020
Homework 7Due Wednesday, Oct 21
(B) Beam (i) and (ii) shown below are identical, except that beam (i) is made of steel, and beam (ii) ismade of aluminum. Note that Esteel > Ealuminum.
(a) Beam (i) - steel. (b) Beam (ii) - aluminum.
1. Let v⇤1 be the maximum deflection in beam (i) and v⇤2 the maximum deflection in beam (ii). Which
of the following is a true statement about the maximum deflection in each beam? (1.5 points)(a) v⇤1 = v
⇤2
(b) v⇤1 > v⇤2
(c) v⇤1 < v⇤2
2. Let vmax be the maximum deflection in beam (i). If the length of beam (i) increases from its originalvalue L to a new value 2L, and the same load is applied at the free end. The new value of themaximum deflection becomes v⇤max. Circle the correct answer: (1.5 points)(a) v⇤max = vmax(b) v⇤max = 2vmax(c) v⇤max = 4vmax(d) v⇤max = 8vmax(e) v⇤max = 16vmax
Briefly justify your answers!!
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O0
IEM 00 6
Vmax PIO SEIvmax PULI
SEIMax