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Unformatted text preview: Problem 10.16 The crosssectional area of the pris Free Body Diagram:
matic bar is A : 2 in2 and the axial force P : 20 kip. '
Determine the normal and shear stresses on the plane P.
Draw a diagram isolating the part of the bar to the right
of plane P and show the stresses. 20 ,OOOI b {—— Solution:
Thc area of plane P is: AP : (2 in2)/(cos 70°) AP : 5.8481n2
Summing vertical forces on the FBD: EFL, = D = —¢7Ap(sin 70°) + TAp(cos 70°) [1] ‘T : 2.75:
Summing horizontal forces on the FBD:
[2] SF3 z 0 : 20, 0001b ~— 0Ap(cos 70°) — TAP(SiI1700)
Substituting Equation [1] into Equation [2]:
20, 000 lb = aAp{cos 70°)+(2.75U)AP (sin 70°) 2 cr(5.848 in2){cos 70° )+(2.750) (5.848 11:2) (sin 70°) ANS: or = 1169 psi
Using the determined vaiue of 0' in Equation [1]: if) f}
a “P
.. 4 {
ANS: "r = 2.750 = ~3215 psi Note: The negative sign con—
forms to the sign convention given in Figure 3.15 T 2 2.75:: z —3215 psi r ProhleniVilfl'm Free Body Diagram:
‘1henormahstress’oh the plane P in Problem 1016 is
6000 psi, what is the axial force P? 6000A Eb re
20:. 70°
J_ TA Solution:
The area of plane P is: AP 2 (2 in2)/(cos 70°) AP = 5.848 in2 Summing vertical forces on the FBD: SF]; 2 0 = —0Ap(sin 70°) + TAP(CDS 70°) [1] 'r = 2.750'
Summing horizontal forces on the FED:
[2] SF“; : {J = P u 044120305700)“ TAP(sin 70°)
Suhtsituting Equation '[1] into Equation [2]:
P : 0AP(cos 70°)+{2.750)(AP)(Sin 70°} : (60001b)(5.848in2)(003 70°)+(2.75)(60001b)(5.848 in2)(sin 70°)
ANS: P = 102, 600 lb : 102.6 kip Problem 10.40 The bar has a circular cross section
and moduius of elasticity E = 70 GPa. Parts A and G
are 40 mm in diameter and part B is 80 mm in diameter.
If F1 = 60 kN and F2 : 30 kN, what is the normal
stress in part B? Free Body Diagram: «Mmmm—4hmmm4 é ZOOnnn ZOOWWH Solution: We must ﬁrst determine the reactions at the left and right walls, We
allow the right—hand side of the bar to “ﬂoat.“The displacement of the
“free” righthand side of the bar is: _ FlLa __ FQLE i F3LA
R _ AAEA ABEB AAEA 5
6R _ (60.000 N}(0.2 m) (30,000 N)(0.4 m} m 30,000 N)(0.2 m)
. («WENroman? N/m“) (WW) (Toxioﬂ N/mﬂ) «W13 j(70x106 mm?) 63 = 0.0341 m = 34.1 mm The {caction at the right waii must be sufﬁcient to prevent ANY dis
placement. lts magnitude is: 7 RE I . 0.4 0.2 m
0‘0341 m “ 70x105 lb/in (R (0.04 r1112) "(0.0531135 ) + (1' (0.01m? 4
RR = 6000 N The reaction at the left wali is: 253; :0:60,000N_6,000N~w30,000N7RL RL = 24,000N F The stress in section A is: —24,000 N+60.000 N I 50.08 m 08: = Note: The negative sign indicates a compressive stress.
ANS: 0,4 = "7.16 MPa Problem 10.41 In Problem 10.40, if F1 = 60 kN, what force F2 will cause the normal stress in part 0 to
be zero? Free Body Diagram:
F1 ——+~ﬁ400mmml 200mm Solution: For the normal stress in section C to be zero, we see that the displace
ment of the intersection of sections B and C must be zero. The equation for the displacement of the intersection of sections B and
C‘ is: 530 =0: M_wm_— 71' ((0.04 uni/4) (70 x 109 N/mz) 71 ((0.04 1102/4} (70 x 109 N/mz) 11' ((0.08 1102/4} (70 x 109 mm?) ANS: F2 = 40, 000 N 2 40 kN Problem 14.9 Model the ladder rung as a simply sup
ported (pinsupported) beam and assume that the 200
1b load exerted by the person‘s shoe is uniformly dis tributed. Determine the internal forces an moment at
A. Solution:
The distributed load due to the weight exerted by the foot is: we = (200 1b)/{4 in) = 50 lb/in ' Draw the FED for the iadder rung and determine the reactions at the
Eeft and right ends. 2001b
<——— 10in 5in—— E 2M1; = U = —[(50lb/in){4in)i(101n)+Rzy{15in) —} 3/9 = 133.311) T 25;; = o : —[(501b/in}(4in)]+Ry+Ly = —2001b+133.3lb+Ly —> Ly = 66.6711) T
Since no forces are exerted in the as~directiom L3 : Rm : 0
Cut the beam through point A and draw the FBD: EUUIb lin * MA loin—4iﬁ5 
Ly
EMA z 0 : m(65.67’1b}(10in)+£{501b/in)(2in)}{1in)+MA I ANS: MA : 567m in: SE}, = o = Ly—(EOIb/in)(2in)+VA = 66.67ib—1001b+VA
ANS: VA : 33.31bT 251m : 0: PA
ANS: PA =0 Problem 14.28 (a) Determine the internal forces and
moment as functions of m. (b) Draw the shear force and
bending moment diagrams. Solution:
Draw the FBD and determine the reactions at point A.
Summing vertical forces on the FED: EFy = o = — (100 Ib/ft) (6 m] + Ag Ag = 3001b Summing moments about point A: EMA : 0 = — (100 lb/ft) (6 ft)] (10 n) + MA MA =3000ft—1b
ForU <9: <6ft:
10:0 2: V=—/Udm=0+01 atm=0,V=3001b
0 0'1 x 3001b
V = 300 lb M=derszf300dw:300$+Cg
o o a: : 0, M = m3, 000 ftelb 02 = —3, 000 n—sn ANS: M : (3000: — 3,000) ft—lb {1]
Forﬁ < m < 12ft: w 2 (new) (we a) x 100) W + 100) d3; = $352+ 100w + 03 cc=12ft,V=0 03 = 0
ANS: V 2 ~33in + 100:1; 1: MAQ r—IUH / zj
,/’ I / 6Ft——i—— IGUlb/F‘t :n
2
M = dea; = f (—56ng + 100m) dm = (—359? + 50m2 +04) ft—lb
0 0 fromequationIl],a: = 6, M m —1.200ft—1b0‘4 : m2,400 ftmlb
ANS: M : (~%m3 + 503;2 ~ 2, 400) itwlb i
E
E
g i
E
i
E
L Problem 14.55 The loads F = 200 N and C =
800 Nem. (a) Determine the internal forces and mo—
ment as functions of w. (b) Draw the shear force and
bending moment diagrams. Solution:
Draw the FED for the beam and determine the reactions at points A
and B. EMA = o = —(200N)(16m)+800N—m+By(8m) ——> 3,, = 3mm T EFy = D = ~200N+By+Ay m —200N+300N+Ay —> All = IDDN J, For0<rs<4mzw=0 V=——/Odm=0+01 at$=D,V=—IUUN,30:01=—100N ANS: V = ~100 N ~4m+4m.~—v~8m Mmedm=f(—1ODN)dm= (—100:z:+02) N—m atm=D,M=0, SO:
02 z 0 ANS: M= ~100a: N—m 11]
For4m<nt<8rru 111:0 V:—fﬂdm=0+0’3atm=4,V=—EDDN, C3 = —100 N
ANS: V: 7100N
M: dex= f(—100N)d:c = (~100wN+C4) Nem from equation ii], at a: z 8, M = —160{) N—m, so: 04 = —800
ANS: M = (—1003 — 800) N—m
For8m<m < 16m: 10:0 V:idecc:D+C'5ata::16,V:200N, so: 05 : 200 N
ANS: V = 200 N M :demeOD N) do: : (200:1:+Cs) N—m ata: = 16, M : 0, so: 05 = —3,200 N—m
ANS: M : (20033 e 3, 200) Nem i
I
t
i Problem 15.9 The beam consists of material that
will safely support a tensile or compressive stress of
350 MPa. Based on this criterion, determine the largest
force F the beam will safely support if it has the cross
section (a); if it has the cross section (b). (The two cross
sections have approximately the same area.) Solution:
The moment of inertia of the crosssection in case (a) is: W M3 7 (0.0233 m)(0.060 m)3
“T 12 m 12 The moment of inertia for the crosssection in case (b) is: : 4.194x10—7 m4 It: (0.050 m)(0.060 m)3 2 [(0.020 rn) (0.040 m)4
"' 12 12 Summing moments about point B to ﬁnd the reaction at point A:
2MB = O : F(O.6 m) —— Ay(1.6 m) —t A1, = 0.375F Maximum bending moment occurs at the point where the concentrated
load is applied, so we calcuiate maximum bending stresses 1.0 m to
the right of point A. Maximum bending moment is: MMAX 2 (0.375F)(1.0 m} 2 0.375F N—m
For crosssection (a): 0.375F)(0.03 m)
195 N 2 ___ (—
350x /m 4.194x104 m4
ANS: F = 13.05 kN For crosssection (b): gamma N/mg 3 (0.375F)(0.03 m)
6.87x10—7 m4 ANS: F : 21.4 kN Problem 15.10 If the beam in Problem 15.9 is sub
jected to a force F = 6 kN, what is the maximum ten
sile stress due to bending at the cross section midway
between the beam‘s supports in cases (a) and (b)? Solution:
The moment of inertia of the cross~section in case (a) is: 3 I 2 I 3
ﬂ : 91% z (00 33mg? 060m) :4'194X104 m4 The moment of inertia for the cross—section in case (b) is: It 2 (0.050 m)(0.000 m)3 _2 [(0.020 m) (0.040 1104] 12 12 Summing moments about point B to ﬁnd the reaction at point A: 2MB = 0 = (0,000N)(0.6 m) — Ay{1.6m) —> 4., = 2250N T The bending moment at the midpoint of the beam is: M : (2256 N) (0.8 m) : 1800 Nam
In case (a), the maximum bending stress is: (1800 N—m)(0.03 m)
4.194x10—7 m4 ANS: (a..)MAx : 128.8 MPa In case (b), the maximum bending stress is: (1800 N—m)(0.03 m)
6.87x 10"“r m4 ANS: (05mm = 78.6 MPa (Ja)MAX : (00)MAX = Free Body Diagram: _ 6.87x10"7m4 Free Body Diagram: l
l
l Problem 15.35 The beam whose cross section is shown consists of three
planks of wood glued together; At a given axial position
it is subjected to a shear force V = 2400 lb. What is the
average shear stress at the neutral axis y’ : 0? Solution:
Finding the centroid of the entire cross section (measuring from the
TOP): ‘4in.+IlintlI4in. I 7 _ m (2 in) (8 in) (4 in) + (2) [(2 in)(4 in)(7 111)} w 5 5 in
(2 in)(8 in) + (2) [(2 in)(4 in)] ’ 8 :n. Calculating E: Calculating A':
A' : {2 in)(5.5Inchc3) 2 11 in2
Calculating Q:
Q = (—2.75 in)(11in2) = —30.25 in3
Calculating the moment of inertia: I = [W + (2 in)(8 in)_(4 in — 5.5 inf] +2 [ 12 12 Now calculating the average shear stress: _ V_Q _ (2,4001b)(—3a.25 i113)
“VG _ bI ' (21n)(162.7in4) ANS: mm : 7223.11b/in2 Problem 15.36 In Problem 15.35, what are the mag
nitudes of the average shear stress acting on each glued
joint? Solution: Finding the centroid of the entire cross section {measuring from the TOP): ' (2 in)(8 in) (4 in) + (2) [(2 in) (4 in) [7 in)} (2 in)(8 in) it (2) [(2 in)(4 in)] Calcuiating the moment of inertia: I _ [(2in)(8in)3
_ ' 12 r _ a: =5.5in + (2 in)(8 in)(4 in — 5.5 inf] +2 [ 12 Calculating E:
if: 7in— 5.5m: 1.5m
Calculating A’:
A’ = (2 in)(4 in) = s in2
Calculating Q:
Q 2 EA’ : (1.5 in)(B inz] : 12 1113
Now calculating the average shear stress:  VQ (2, 400 lb}(12 1:33)
T = _ : __._
AVG 131 (2 in)(162.7 in4) ANS: TAVG m 88.51b/in2 (4 in)(2 in)3 (4 in)(2 in)3 J. R‘ﬁ 2m + (4 in}(2 in}(7 in — 5.511132] : 162.7in4 + (4 in)(2 in)(7 in v 5.5 i102] : 162.7in4 I Problem 15.46 The beam is subjected to a distrubuted load. For the
cross section at m z 0.6 m, determine the average shear
stress (a) at the neutral axis; (b) at y’ z 0.02 In. Free Body Diagram: 0.04
m 130 kN/rn M
.. erlLﬁ—z Cross section. ’,a13U,UOUN/m
1
 Solution: Summing the moments about point B to determine the reaction at point
A. 2 MB = 0 = — [gaseous N/m)(1.4 111)] (D.133m)+Ay(0.Bm) —> A5, =15,130N i Cut the beam at a: = 0.6 m and draw the FBD, (130,000N/1.4m)(x> / Summing the vertical fences to determine the shear force at :12 = 0.6 m: 1
:Fy = 0 z —15.130N [—§(55.714N/m)(0.6 m) +V a V = 31,844N A: (a) Using Equation (1518) to determine the average shear stress at the
neutral axis: 3V 3(—31,844 N T : EAT " 2(o.04 m){0.06 m) ANS: mm m —19.9 MPa (13) Using Equation (1517) to determine the average stress at y’ :
0.02 m: _ 51/ a 2 , 2 _ 6(——31,844N) 0.06m 2
TAVG * m _ (3") i " (0.94 m)(0.06 m)3 2 ) m (0'02 my] ANS: mm = "11.06 MPa ...
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This note was uploaded on 10/12/2009 for the course MAE 210306210 taught by Professor chatterjee during the Fall '08 term at UCLA.
 Fall '08
 CHATTERJEE

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