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Unformatted text preview: Problem 2.43 Three prismatic bars. two of material A and one of material B.
transmit a tensile load P (see ﬁgure). The two outer bars (material A) are identical.
The crosssectional area of the middle bar (material 3) is 50% larger than the crosssectional area of one of the outer bars. Also. the modulus of elasticity of material A is twice that of material B. (a) What fraction of the load P is transmitted by the middle bar? (b) What is the ratio of the stress in the middle bar to the stress in the outer bars?
(c) What is the ratio of the strain in the middle bar to the strain in the outer bars? Solution 2.43 Prismatic bars in tension EQUATION OF EQUILIBRIUM 2Fh0ﬁ2=0 PA+PB—P=O EQUATION OF COMPATIBILITY
6A 2 6}; FORCEDISPLACEMENT RELATIONS A A = total area of both outer bars PAL PBL
5A = 53 = EAAk ERAS
Substitute into Eq. (2]:
PAL __ PBL
EA AA EBAB SOLUTION OF THE EQUATIONS Solve simultaneously Eqs. (1) and (4]: EAAAP EBABP
PA — P3 —
EAAA + EBAB EAAA + EBAB
Substitute into Eq. (3]:
PL 53 Zam+%@ (l (7 (3) (4) (5) (6) ) ) STRESSEs:
PA EAP
(TA : — = —
AA EAAA + EBAB
PB EBP
=—=— 7
W AB EMA+QMB ()
(a) LOAD lN MIDDLE BAR
ﬂ__iﬁL___J__
P EAAA 'l‘ EBAB EAAA +1
EBAB
EA AA 1 + l 4
G' :— = 2 — = = —
Ive“ Ea AB 1.5 3
P3 _ l _ l _ 3 (_ 'P E 8 ll
(aw—0+1 a M 3 (b) RATIO OF STRESSES 32%; (_ 034 EA 2 (c) RATIO OF STRAINS All bars have the same strain Ratio = l (— Problem 2.46 A plastic rod AB of length L : 0.5 111 has a
diameter all 2 30 111111 (see ﬁgure). A plastic sleeve CD of length r : 0.3 111 and outer diameter d; : 45 111111 is securely bonded to the
rod so that no slippage can occur between the rod and the sleeve.
The rod is made of an ac1ylic with modulus of elasticity E1 = 3.1
GPa and the sleeve is made of a polyamide with E; = 2.5 GPa. (a) Calculate the elongation 8 of the rod when it is pulled by
axial forces P : l2 kN. (b) If the sleeve is extended for the full length of the rod, what is
the elongation? (c) If the sleeve is removed, what is the elongation? Solution 2.46 Plastic rod with sleeve Adi P : '12 kN
L = 500111111
Rod: E1 = 3.1 GPa
Sleeve: E3 : 2.5 GPa ad? r,
Rod: A1 : : 706.86111111‘ b : 100111111 (71 : 30111111 (1’: = 45 111111 c = 300 111111 Sleeve: A2 : £11411 + E3143 = 4.400 IVIN — (1’12) : 883.5711n112 (a) ELONGATION 0F ROD Part AC: SAC : — : 05476111111 1‘4 1 PC Part CD: 5CD = m
: 081815111111
(From Eq. 218 of Example 26)
6 = 25AC + 6CD = 1.91111111 ‘— (b) SLEEVE AT FULL LENGTH 5 _6 (15)_(08181S )(500111111
— CD 0 — ' ‘ mm 300111111
: 1.36111111 <—
(c) SLEEvE REMOVED
5 : : 2.74111111 *— £1.41 ) Problem 2.410 A nonprismatic bar ABC is composed of two segments: A8 of length
L. and cross—sectional area A.: and BC of length L2 and cross—sectional area A2. The modulus of elasticity E. mass density p. and acceleration of gravity g are constants.
Initially. bar ABC is horizontal and then is restrained at A and C and rotated to a vertical
position. The bar then hangs vertically under its own weight (see figure}. Let A. = 2/1; =
AandL. =§L,L2=§L. A
(a) Obtain formulas for the reactions RA and RC at supports A and C, respectively. due to l
gravity. L1
(b) Derive a formula for the downward displacement 53 of point B.
(c) Find expressions for the axial stresses a small distance above points 8 and C.
respectively. '
A
Stress
elements
L2 Solution 2.410 (a) find reactions in lidegree statically indeterminate structure
use superposition; select RA as the redundant
compatibility: 5A. + 5A2 = 0 WAR 2 gAlLl WBC = 21°1sz segment weights: find axial forces in each segment: use variable 5 measured from C toward A NAB: —pgA1(L +L2—§) L25§5L1+L2
NBC: —lWAB+PgA2(L2_ 5]] 03 Q'ng displacement at A in released structure due to self weight L} L + L
_ NBC ' 3 NAB
8A1 2/ + / o EAz L2 EAI
8 _ Lz—ngA1L1+ P2A2( L2 _ t)‘ d I L'+L2— EAK Ll + L2 _ g) d
Al 0 EAZ g ' L2 EA] ‘2 — 1 2A1L1+A2L2 —1 L12+2L1L2 + L2" 1 2L1+ L2
8A1: 2 9st BAG l 2 Pg E + 29va —Pg
6A] 2 (2L2A1Ll + A2L22+ AQLIZ) Next. displacement at A in released structure due to redundant RA ‘ L1 L2
5A2 = RAlfAB + Use) 8A2 2 RA — + EA] EA2
enforce compatibility: 5A, + 5A2 = 0 solve for RA
Pg 2 2
ZLAL ——AL +AL
R 2(EA2)( 211 22 21)
A_
1:11’113nfB
1 A3L13+ MILIL2 + 121sz
RA—ngAl “
g L1A2+ L2A1
statics: RC = WAB + WBC — RA
1 A1
R = AL + AL —— (ZLAL +AL2+AL2—
C P21110222 2P2 211 :22 21JL1A2+LZZAI
R 71 A A1L12+2A2L1L2+A1L22
(1—29” 2 L1A3+L2A1
F A —A A —3 L —£ L —£
or I— 2— 2 1— 5 2— 5
A(1L)3 stL 14(1)2
lZA I
R1A25 552% RBTAL 3705,,9
: 2— r3 (— —: A
A 2Pg 3LA 2L A 709“ 70 ‘
——+—A
52 5
3L 2 A3L 2L 2L 2
1( )12 11( )
R—1 A 5 255 5 R—2 LA <— 2—0271
C zpsz 3LA 2L C 70‘)“ 70 ‘
——+—A
52 5 03) use superposition to ﬁnd displacement at point B SB = 53. + 532 where 531 is due to gravity and 532 is
due to RA L3 N
5 B] = f % dg < due to shortening of BC
0 2 — “L3
5 = D “ 2AL +AL
B] 2(EA2)( 1 1 2 2) ] A2L12+ 2A1L1L2 + L2 L1A2+ LgA, EA2 _ —l AlL] + Ang
2 (LiAz + MADE (e) expressions for the c‘n‘emge axial stresses :1 small distance above points B and C NB = axial force near B = RA — WAB NB= gAL — pgAi—L N3 = j—JpgAL
19
NC:—RC O'C: —(ﬁng ) 0C: —19 6L <— A _ 35 .0:
2— pipes having a crosssectional area A : 3500 null2 and a modulus of
elasticity E = 210 GPa. Member BC is of length L = 2.5 111. and the
angle between members AC and AB is known to be 60°. Member AC
length is Z) = 0.71L. Loads P = 185 kN and 2P = 370 kN act vertically
and horizontally at joint C. as shown. Joints .4 and B are pinned supports.
(Use the law of sines and law of co sines to ﬁnd missing dimensions and
angles in the ﬁgure.) Problem 2.414 Threebar truss ABC (see ﬁgure) is constructed of steel J: P1 (a) Find the support reactions at joints A and B. Use horizontal reac
tion BX as the redundant. (b) What is the maximum permissible value of load variable P if the
allowable normal stress in each tluss member is 150 MPa‘.) Solution 2.4—14 NUMERICAL DATA L : 2.5 111 b : 0.71 L : 1.775111 E : 210 GPa .4 : 3500 1111112 P : 185 kN 6A : 60°
arr, : 150 MPa FIND MIssING DIMENsIONs AND ANGLEs IN PLANE TRuss FIGURE arc : boos (6A) : 0.8875111 ye : bsin(6A) : 1.5372111 Z) L _ bsinwA) 5111(63) = Smwd) so 63 = nsrn (L) = 3794306” 9C : 180° — (6A + 93) : 82.05694o r? = Sinai) sumac) = 285906111 or c = V 253 + 1:3 A 252 cosmc) = 285906111 (a) S—L CT B)r As TH R DU'NL)AN I: P.:R.ORM sUFERPOSITION ANALYsIs TO FIND BA. THEN USE STATICS TO FIND REMAINING
REACTIONS. FINALLY USE METHOD OF JOINTS To FIND MEL{BER FORCES (STE FXAixrFL': ll) 83x : displacement in xdirection in released structure acted upon by loads P and 2P at joint C: 535‘. = 1.2789911 111m < this displacement equals force in AB divided by ﬂexibility of AB . . . . . C
53x3 = displacement In xdrrection In released structure acted upon by redundant Bx: 53X; = Bx 5 —EA
CorIRATIBILIn’ EQUATION: 53)“ + 8m : 0 so BX : C STATICSI EFX = 0 AX = —BX — 2P 2 —41.21<N
1
EMA = 0 By 2 — [213(2) sin(HA)) + 13(2) cos(6A))] = 256.361 kN
C 2F}. 2 0 A REACTIONS: Ax = —41.2kN 14y: —71.4kN Bx = —3291;N a}, = 256kN (b) FIND MAXIM1 PERMISSIBLE VALUE OF LOAD VARIABLE P IF ALLOWABLE NORFLAL STRESS Is 150 MPA
(1) Use reactions and Method of Joints to ﬁnd member forces in each member for above loading. Results: FAB = 0 FBC = —416.929 kN FAQ = 82.40 kN (2) Compute member stresses: —416.93 kN _ _ a 82.4 kN ﬂ a
QB = 0 03C 2 +4 2 —119.123 MPa QC 2 T = 23.543 MPa (3) Maximum stress occurs in member BC. For linear analysis, the stress is proportional to the load so 0'0 P = 233 W So when downward load P = 233 W is applied at C and
horizontal load 2P 2 466 kN is applied to the right at (I
the stress in BC is 150 MPa lIinax : 0'30 Problem 2510 A rigid bar ABCD is pinned at end A and supported by
two cables at points 8 and C (see figure). The cable at B has nominal
diameter d5 = 12 mm and the cable at C has nominal diameter dc = 20 mm. A load P acts at end D of the bar. What is the allowable load P if the temperature rises by 60°C and each
cable is required to have a factor Of safety of at least 5 against its ultimate
load? (Note: The cables have effective modulus of elasticity E = 140 GPa and
coefficient Of thermal expansion 0: = 12 X 10—6/°C. Other properties of
the cables can be found in Table 2—1. Section 2.2.] Solution 2.510 Rigid bar supported by two cables FREE—BODY DIAGRAM OF BAR ABCD From Table 2— 1:
AB = 76.7 mm2 E = 140 GPa AT : 60°C AC : 173 1111112
0: =12 >< 10—5700 EQUATION OF EQUILIBRIUM
EMA = 0 m (2‘ T3121?) + Tct4b) — P(5b) = U or 2TB + 4TC = 5P {Eq. 1]
TB : force in cable B TC : force in cable C
(is = 12 mm dc = 20111111
DISPLACEMENT DIAGRAM SUBSTITUTE NUMERICAL VALUES INTO EQ. (S):
A 2b B 2b C b D T3846) — 12:06.7) = —1.338.000 (Eq. 6)
in which T3 and TC have units of newtons.
SOLVE SIMULTANEOUSLY EQS. {1] AND (6):
53 T3 = 0.2494 P i 3.480 (Eq. 7)
5 TC = 1.1253 F + 1.740 (Eq. 8)
c in which P has units of newtons. COMPATIBILITY: SOLVE EQS. (7") AND (8] FOR THE LOAD P: 6C 2 263 (Eq. 2) PB : 4.0096 TB + 13.953 (Eq. 9)
FORCEDISPLACEMENT AND TEMPERATURE—DISPLACEMENT ’05 Z 0888? TC _ L546 (Eq‘ l0)
RELATIONS ALLOWABLE LOADS T L . a A
63 = B + MATE (Eq 3) From Table E 1. EAB ITEM” = 102000 N (ram = 231000 N
6C 2 + “(ATE (Eq_ 4) Factor of safety = 5 C [T3)a”Dw = 20,400 N {TC}..an = 46.200 N
SUBSTITUTE EQS. (3) AND {4) INTO HQ. (2): From Eq. {9): P3 = (400961120400 N] + 13.953 N
TCL + A L _ 2TBL + q A L : 95.700 N
EAC 0“ n _ EAR ‘0‘" n ' From Eq. {10): PC = (0.8887)(46,200 N) — 1546 N
= 39500 N or Cable C governs. 2TBAC _ TC'AB : AC Fallow = (— Problem 2.51? Wires 3 and C are attached to a support at the left—hand
end and to a pin—supported rigid bar at the right—hand end (see ﬁgure]. Each wire has cross—sectional area A = 0.03 in.2 and modulus of elasticity
E : 30 X 106 psi. When the bar is in a vertical position, the length of each
wire is L = 80 in. However, before being attached to the bar. the length of
wire B was 79.98 in. and of wire Cwas 79.95 in. Find the tensile forces T3 and TC in the wires under the action of a force
F = 700 lb acting at the upper end of the bar. i—SO in. Solution 2.51? Wires B and 1: attached to a bar ,5 = 700 lb EQUILIBRIUM EQUATION Ell/{pin
Tc(b) + new = Pt3b) P=700lb :0 (— b TB 2TB+TC=3P {Eq.1)
[9
Tc
Pin
P = 700 lb
A = 0.03 in.2
E = 30 X 106 psi
LB : 79.98 in.
LC : 79.95 in.
DISPLACEMENT DIAGRAM Combine Eqs. (3) and {5):
S=80'.—L=0.02‘. TL
3 m B m —C = 5C + 6 (Eq. 7)
SC = 80in. — LC = 0.05 in. EA
Eliminate 5 between Eqs. (6) and {7):
T GT i EASE ZEASC (E 8
B A c — L L Q J
I 5 Solve simultaneously Eqs. {1) and (8):
. 6P EAS ZEAS TB 2 _ + B _ C
._ 5 5L 5L
L280”  T SF 254.33 + 4EASC
C : _ _
5 5L 5L
Elongation of wires:
63 2 SB + 26 {Eq' 2] SUBSTITUTE NUMERICAL VALUES:
EA
6.; : Sc + 6 lEq. 3) E = 2250 lbﬁn.
FORCE—DISPLACEMENT RELATIONS
TB:8401b+451b—225]b=660lb (—
a—Ea—E (gas) a (—
B EA C EA qs. . TC—4_Olb—901b+4501b—780lb
SOLUTION 0F EQUATIONS {Both forces are positive, which means tension. as
Combine Eqs‘ (2) and {4): requtred for Wires.)
TBL
— : 33 + 26 (Eq. 6) EA ...
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 Fall '07
 Rodin

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