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Unformatted text preview: .se__ Eq (74) and (75) to solve Problem 72.24. TION:
_ raw the F30 for the beam and determine the reactions at
{rants A and B. ' EURN’N 4J<N/m EMA = 0 = (6,000N)(6m) + 20,000N . m — (4,000N)(6m)(3m) + By(6m) — B(4,000Nxm)(6m)](8m) B,=13.667N T 213 = 0 = —6,000N — (4,000Nf m)(6m) — [%(4,000me)(6m)] + A), +18,66’FN Ay = 23,333 N T fort} < x < 6 m: w= 0
X X
V=sjwdx= 060:0“:1 @x=0,v=6000N,so: C1=6000N
0 0
V{x] =6000 N
I 1
M = J'm = J1 600%: = 4000; + c2 @ x = 0, M = — 20,000 N—m, so: 02 = — 20.000 Nrn .' o 0
_'ANS M): =  [Ix—20 5For6 m < x <12 m:
' w=4000 Nim V=—J'wdx= —I4000dx= 4000): + C3 @ X: 6 v— — 17 333 N so: ca = 41333 N ;ANs Vx=11D—{ 4Qggx+41,3331N MI =I(—4000x + 41,333)dx 40003.:1 + 41,3333: + c4 @ x = 6m. M = 56.000 Nm. so: 0
C4 = 232,000 Nm g__ANs M =2 00x2+4 33x—232 Nm For12m<x< 18m:
w = (666.7 Nlm) (1B — x) m = (12.000  666.7 x) N I X
= — [mix = « {(12,000 — 666.7x)dx = 42,0003: + 3333;:2 +18,66? + cs @ x = 12 m. v =  666.7 N. 50: 1 c5 = 39,331 N.
. ANS V(x) = 1 12.000 x + 333.3 x2 + 89.381) N M = Ide: J'(333.3x2 “12,000“ 89,331)dx=111.1x3 — 0,0003:2 +108,000x + Co @ 3: =18 m, M = 0. so: (35 = 648,000
_ ANS M(x1=(111.1 xa— 6,000 x2 +103,000 x + §4B,UOO} Nm 401 PROB M 1.1
The beam consists of material with modulus of elasticity
E = 70 GPa and is subjected to couples M = 250 kNm at its ends. (a) What is the resulting radius of curvature ._..__.' l.6m—————1I ‘ l in
of the neutral axis? (in) Determine the maximum tensile c mam
stress due to bending. you man
§QLQTIQN: The moment of inertia for the crosssection is:
5&3 _ (0.16m)(o.32m)“ 1:..._._ 12 12
(a) Using Equation (3.10) to determine the magnitude of the radius of curvature: = 4.369.210“ m“ i=£=—92§9W———mﬁ28.174x10_3m4
p 51 (70x10 Nr'm )(4.369x10' m ) ANS g; 1.223 m (b)_The maximum normal stress due to the bending moment is: a H = Mym = (250,000N — m}(0.16m_)
M 1 4.369x10“m“ ANS omx = 91.6 MPQ WLE . 4.2 The material of the beam in Problem 81.1 will safely
support a tensile stress of 180 MPa and a compressive
stress of 200 MPa. Based on these criteria. what is the
largest couple M to which the beam can be subjected? §QLQTIQN:
The symmetry of the crosssection tells us that: (UWMNSILE = (0m)coMFResswE The moment of inertia for the crosssection is: 5.53 (0.16m)(o.32m)3 I=—= 12 12 = 4.369x10‘4m‘ The maximum normal stress due to the bending moment is: a, _ Mymx _) M “ 0M! _ (130x105Nrm=)(4.3s9xlo*m4)
W — I  ————~
yMAX 0.16m ANS M = 492 kNm W  The beam is subjected to a uniformly distributed load W0 = 300 Iblin. Determine the maximum tensile stress
due to bending at x = 20 in. if the beam has the cross
section (a); if it has the cross (b). (The two crosssections
have approximately the same area.) 85in Y By
we:
The moments of inertia for the crosssections in the two cases are: (4.47m)(4.47in)3 . 4 (6in)(6m)3 (490mm)3 , 4 I =—————=33.27 I =————=86.6? _.____ '
a 12 m b 12 12 m r— 300lb/ll’i
l
Summing moments about the cut through the beam at x = 20 in: 1‘ j”
M = (12,750 lb) (20 in) + (6,000 lb) (10 in) = 195,000 inIb EOin——lv
We see that the maximum tensile occurs at the bottom of each crosssection. p, y
. (1 95,0001):  rah47%)
In case (a), the maximum tensile stress is: (aﬁmx = ——;‘——
33.27:}:
ANS [QIJMMH .1 k iatth b ttom ofth r 5—5
in case (b). the maximum tensile stress is: (0?)JW = W
86.67:}: ANS [£2th = .75 k i the bott fthe r  ction 410 W19
The maximum anticipated load on the beam is shown.
Choose a material from Appendix B and design a cross section for the beam so that it has a factor of safety 8 = 2. Free 39d! Diagram: , / ED’DDON/n Q] It?” kam
I ’ i _
/ I
3m—L—3m A! Byr
SQLQTIQN:
From Appendix B for 7'075T6 aluminum, or, = 480 MPa. A crosssection is chosen such that its height is twice its width.
The moment of inertia for the crosssection is:
3
I =3}: = x(2.r)3 = 0.6I57x4m4
12 12
Ex
Summing moments about point B to determine the reaction at point A: i
1
EM = 0 = [420mm mlJm)](4m) — A (6m)
3 2 ’ —x——l A, = 20.000 N T The bending moment is maximum at the location where the shear force is zero. or the sum of vertical forces
is zero. Summing vertical forces on the FBD: 2F), = o = 20,000N — [1(6,66?Nim2 xm)](xm) ~+ x = 2.45 m
2 Summing moments on the F30: 2'” = 0 = (20.000N)(2.45m)  [%(6,667NFMX2.45mX2.45m) 1435”]+ M M=32,659 Nm The maximum allowable normal stress due to bending is found using the factor of safety. 0' 6 J 2
arm =§=ﬂxliﬂl=z40xw6mmz Now using Equation (8:12) to determine the dimension, x: 240x106Mm2 = w —» x3 = 0.000204 m3 0.667x‘m‘ ANS X = 9.0589 rn = 539 mm 42‘! ...
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 Spring '06
 Keil

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