*§-56 The 1-in.-diameter rod is subjected to the loads shown Determine the state of stress at point A, and show
the results on a differential element located at this point
IR =0 +1020; Y= -100b EE =0 N-75=0; N= 75 R= Y-80=0; ¥ = sb TM, = 0, MM, +808) = 0, MM = — 640b- in, IM =0; T +803)20 [a - 240b-in, aM, =Œ My + 1008) ~ 75(3)=0; A6 = — 575lb-in, 405) 1 (Q)s =¥A= —— =(9)(0.57) = 0.08333 in? az 2 =R=%f25F = i h=h=or = 705") = 0.015625» in* ^ Nomalsray: ơ = Ê „ Ä€Y „ À62 + af 4 + O= 6.61 ksi (1) Abs 1 (ta = 00( 0.08333) + 24005) 001561) 0.0312" = 1.39 ksi Ans Op Zé-É/ k5: a" z+ t6)2 139 bt ‘hy la = 0 Aus
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 28-57 The 1-in.-diameter rod is subjected to the loads shown Determine the state of stress at point B, and show
the results on a differential element located at this point tz, xứ V+10=Œ Kz ~ 10016
R20; Ma 1510 M7401 E20; 1 = 80 = 0; ¥ = 80!b IM, = 0; Mi + 808) 0; Mp = — 640 Ib in ib TM, s0, % +803) 20, T= — A0lb-in, IM, =0, Mẹ +100(8 ~ 723) = © M= ~5151b-in Ax ed = ec) = iat gaia z (059 “= 003125 # trẻ 2 = MOD ˆ(2)(19) = 0008333 02 (are = SS co in hehe m" = F (0s) # 0015625 bể Nonnal stress: P My, Me erat TG
3 4g ~ SOD - ste isi = 576K An 4 : 0.015625" %* Shear svess : v= Baar TẾ w > Te VQ _ 24003) 80.0833) (ty)a =~ ~~ se + + 6 si As J ok 003125 « 0.015625{1) (tale = 0 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 38-58 The post has a circular cross section of radius c Determine the maximum radius e at which the load can be applied so that no part of the post experiences a tensile stress Neglect the weight of the post
M:P< tt Require o, = 0 O04 = = PLM = _P (Pee ne c1 e= £ Ans 4
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 48-59 The masonry pier is subjected to the 800-kN load For the range y>0 x>0 determine the equation of the line y = f(x) along which the
load can be placed without causing a tensile stress in the pier Neglect the weight of the pier
4 iG * B00 Kn wa BK, x A = 3(4.5) = 13.5m? k= -L(3)(4.5) = 2.78125 m! 12 " 4 a) Mu= BOX LÊN 5 (4508) = 10.125 m* M.=24 — Lo x
Normal stress : Require o4 = 0
Pe My, me a= “"A kL A | 0= ~800(10) + 800(10”)y(2.25) + 800(10?)x(1.5) 13.5 22.78125 10.125 0 = 0.14&x + 0.0988y — 0.0741 y=075~ 15x Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 5*8-60 The masonry pier is subjected to the 800-kN load If
x = 0.25 m and y = 0.5 m, determine the normal stress at # each corner A, 8, C, D (not shown) and plot the stress dis-
tribution over the cross section Neglect the weight of the — —zz#”m Dier oo Kn) 800 KN dare ~ 5m fos” II + C = cc uw » ° s x = > A = 3(4.5) = 13.5m? ~ " (345°) = 22.78125 m‘ 1l E 249G”) = 10.125 m' - ~800( 10°) + 400( 107) (2.25) + 200(10”)(1.5) Ø, 4 13.5 22.78125 10.125 = 9.88 kPa (T) Ans on = 80010) „ 400(10°)(2.25) _ 200(10°)(1.5) 5 13.5 22.78125 10.125
= —494kPa = 49.4kPa(C) Ans
a = -800(10)) _ 400(10)(2.25) _ 200(10)(1.5)
© 13.5 22.78125 10.125
=~ 128kPa = 128 kPa (C) Ans
—800(10) 400(107)(2.25) — 200(10°)(1.5)
Op = - +
13.5 22.78125 10.125
=~ 69.1 kPa = 69.1 kPa (C) Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,
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Trang 6861 The eye hook has the dimensions shown If it supports a cable loading of 80kN, determine the maximum normal stress at section a—a and sketch the stress distribution acting over the cross section
#oolb 2S hip Boob dA J & = 20 3.125- (125)? - 0.625) ) = 0.395707 r 2 R = A = BO65)" — 3.09343 in, J44 0.396707 ¿kỉ M = 800(3.125) = 2.5(10°) o= MR-D LP ArữŒ—R) A (0.Gdsé Ư - 3 (Ø)mx = 2.540 X3.09343-2.5) + 800 =16.0ksi Ans # (0.625)?(2.5)(3.125 - 3.09343) (0.625)? 3 2.5(10° )(3.09343 - 3.75) + 800 = -10.6 ksi A 7 (0.625)? (3.75)(3.125 - 3.09343) 7 (0.625) " (Ge max =
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 78-62 The C-clamp applies a compressive stress on the cylindrical block of 80 psi Determine the maximum normal stress developed in the clamp
35.3429 A, if.otzs cin, 35.3429 46 { 42025m> = 0.055786 ar 4 A _ 10.25) R=— = ————— = 4.48142 #4 0055786 P= 04A = 80n (0.375)" = 35.3429 Ib M = 35.3429(9) = 318.0863 kip- in = M(R ~ r) + P ArŒ-—R) A 318.0863(4.48142 - 4) 35.3429 (Op)nax = mm (1Á0.25X4)4.5-4.48142) (10.25) 838si BE AB 142-4), 35.3429 na i Ans 318.0863(4.48142 - 34
(Ge max = ¢ 2-5) 35.3429 = -6.96 ksi 1(0.25)(5)(4.5- 4.48142) (10.25)
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 88-63 The handle of the press is subjected to a force of 20 Ib Due to internal gearing, this causes the block to be subjected to a compressive force of 80 Ib Determine the normal stress acting in the frame at points along the outside flanges A and A, Use the curved-beam formula to compute the bending stress
Normal stress due to axial force :
A = 2[0.5(3)] + 5(0.5) = 5.5 in” oth l8 >I} On = = 10.9090 psi (1) th th
Nomal stress đue to bending :
r=l5imn rạ=l2m rạ = l8im
dA Tạ 12.5 17.5 18 _
“= R= <2 +4 0.5in— + 3in—— = 0.3752 in
x] r bin ry 3in 12 12.5 17.5 R= A _53_ = 14.6583 in [#4 0.3752 7-R = 0.3417 in _ M(R - ra) _ 59.0(12)(14.6583 - 12) = 83.4468 psi (D (ØA), = Ara ~ R) 5.5(12)(0.3417)
- MA - 1a) _ 52042046383 18) _ _ 69.9342 psi = 69.9342 psi (C)
(On), = AraŒ — R) 5.5(18)(0.3417)
0, = 83.4468 + 10.9090 = 94.4 psi (T) Ans Og = 69.9342 — 10.9090 = 59.0 psi (C) Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 9*8-64 The block is subjected to the three axial loads shown Determine the normal stress developed at points A and B Neglect the weight of the block
M, = ~ 250(1.5) — 100(1.5) + 50(6.5) = - 2001b- in M, = 250(4) + 50(2) - 100(4) = 700 Ib- in 1 3 1 3 „ 4 I = — (43) + 2(—)(2)3) 1p )+ (2X X3) = 741.33 in = 741 1 3 1 3 „ 4 = — (3)8 2— 4°) = 181 ỳ 1 0 )+ por ) 81.33 in A = 4(13) + 2(2Á3) = 64in? oa PME, Mey A fs L 400 7004) | -— 200 (- 1.5) ƠA Z — —— —~ —— + ———————— 64 — 181.33 741.33 = —~2l3psi Ans 400 7002) -200 6.5) Op = - — ———— + ———————— 64 18133 741.33 = -12.2psi Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 10865 The C-frame is used in a riveting machine If the force at the ram on the clamp at D is P = 8 KN, sketch the stress distribution acting over the section a—a
ZxA _ (0.005)(0.04)(0.01) + 0.04(0.06)(0.01) ~*“—= = 0.026m XA 0.04(0.01) + 0.06(0.01) x= A = 0.04(0.01) + 0.06(0.01) = 0.001 m? 1= = (0.04)(0.01°) + (0.04)(0.01)(0.026 - 0.005)? + = (0.01)(0.06") + 0.01(0.06) (0.040 — 0.026)? = 0.4773(107°) m* 1 M+8(22⁄) * =18opknm 8 EN ap, Mx 8(10°) 1.808(10”)(0.07-0.26) (mx =2 TT F "0001 0.4773(105)
= 106.48 MPa = 106 MPa ¡S4 Mpa
MPa y= B_ Me _ 810) 1.808(10°)(0.070- 0.026) qi’ (maxbc TC TT F 001 0.4773(10-5) = ~158.66MPa = ~159 MPa = ~ = W=*, x= 419mm 158.66 106.4
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,
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Trang 118-66 Determine the maximum ram force P that can be applied to the ‘clamp at D if the allowable normal stress for the material is oyjow = 180 MPa
ZxA (0.005)(0.04)(0.01) + 0.04(0.06)(0.01) = 0.026 m A 0.04(0.01) + 0.06(0.01) x= A = 0.04(0.01) + 0.06(0.01) = 0.001 m? I= = (0.04(0.01°) + (0.04)(0.01)(0.026~ 0.005)? 1 + 1a(0.019(0.06)) + 0.01(0.06)(0.040~0.026)? = 0.4773(105) m° đ= + >i"t Mx I
Assume tension failure,
180(10°) = —P— 4 2226 P(0.026) 0001 0.4773(10-)
P = 13524N = 13.5kN
Assume compression failure,
~ 180(10°) = —P_ ~ 9.226 P(0.070~ 0.026) 0.001 04773(105)
P = 9076N = 9.08kN (controls) Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 128-67 Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm If the cylinder has a wall thickness of 2 mm, deter- mine the state of stress in the wall of the cylinder
Pp > 4} mm P _ 210) ™~ P =— A “ E00) — 31438013Pa =—_— —_ = P pr _ 314 380.13(0.045)
Ởi ree z# — = ————_——_Ặ St 3003 = 7.07 MPa Ans
đ = 0 Ans
The pressure p is supported by the surface of the pistons in the longitudinal direction
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 13*8-68 Determine the maximum force P that can be exert- | ed on each of the two pistons so that the circumferential stress ; component in the cylinder does not exceed 3 MPa Each pis- 1 ton has a radius of 45 mm and the cylinder has a wall thick- | ness of 2 mm = PỊ, fy = 200.045) a= 300) =o 002 p= 133.3kPa Ans P = pA = 133.3( 10°) (#) (0.045)? = 848 N Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 148-69 The wall hanger has a thickness of 0.25 in and is used to support the vertical reactions of the beam that is loaded as shown If the load is transferred uniformly to each strap of the hanger, determine the state of stress at pomts C and D on the strap at A Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown
10 kip t0Kịp 26): /2 Aip mewenw beni 2Z:Ô0 kịp lín S80 kipin 6 Kip GEM, =0; — 12(3)+10(8)—F,(10) =0 Fy = 11.60kip 1 1 = 2[55(0.25)(2)"] = 0.333 in‘ A = 2(0.25)(2) = 1 in? At point C, P _ 2(5.80) Ớc c # — = A i ~ = 11.6 ksi + Ans tT =0 A ns At point D,
ap = 2 MeL 245.80) _ [2(5.80)1(1) A I 1 "0333 = -23.2 ksi Ans
tr = 0 A
ns
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 158-70 The wail hanger has a thickness of 0.25 in and is usea to support the vertical reactions of the beam that is loaded as shown If the load is transferred uniformly to each strap of the hanger, determine the state of stress at points C and D of the strap at B Assume the vertical reaction F at this end acts in the center and on the edge of the bracket as shown 10 kip lin SD Kipin
(+E My = 0; Fa(10) - 10(2) - 12(7) = 0; Fy = 10.40 kip
= 2 (0.25)(2)*) = 0.333 in*; A = 2(025)(2) = lin?
At point C :
Ớc = : = =m = 10.4 ksi Ans
tT =0 Ans
At point D :
Op == - “te = - eee = ~20.8ksi Ans
fp=0 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 16871 A bar having a square cross section of 30 mm by 30mm is 2 m long and is held upward If it has a mass of 5 kg/m, determine the largest angle @, measured from the vertical, at which it can be supported before it is subjected to a ten- sile stress along its axis near the grip
A = 0.03(0.03) = 0.9(107°) m? 2m 8 I= = (0.03)(0.03") = 67.5(10°°) m4 > Require 04 = 0 = P Mc 0%, =O0=—+— A 1 - -#8.1cos Ø „ 98.1 sin 6(0/015) t2 § 6 0.902) 67.5(10) 59-81) (2) = 98.1 AJ V
0 = —1111.11 cos Ø + 222222.22 sin 6 =98./ sing
N = 98 lQse@ tan@ = 0.005; 6 = 0.286° Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 17*8-72 Solve Prob 8-71 if the bar has a circular cross sec- tion of 30-mm diameter A= 7 (0.03") = 0.225%(10”) mỸ 8 " = Z(00159 = 12.65625z (10) m' Require 04 = P Me ø =0= 2 — -981cosØ 98.1sin6(0.015) ~ 0.225m(10-3) 12.65625 m(10) 0 = — 4444.44cos @ + 1185185.185 sin 6 N>28.1c„ so tan 6 = 0.00375 @ = 0.215° Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 188-73 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm If the largest normal stress is not to exceed 150 MPa, determine the maximum pressure the tank can sustain Also, compute the number of bolts required to attach the cap to the tank if each bolt has a diameter of 20 mm The allowable stress for the
bolts is (Gatlow)b = 180 MPa
r 750 Hoop Stress for Cylindrical Tank : Since :“T
= đi.6 > 10, then ghờt w all analysis can be used Applying Eq.3~1 À 2 3 6oMfa
pr TT Ơi = Ø,ov = T 0) 150( 10) = 2750) 18 | 1 p=3.60 MPa Ans Force Equilibrium for the Cap:
+TEE =0; 360(105)[x(0.75°)]~ =0
F, = 6.3617( 10°) N
Allowable Normal Stress for Bolts : (Pew), = 1
6.3617(105) t80( tố) = n[ ‡(0.02) ]
n=t112.5
Use n= 113 bolts Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 198-74 The cap on the cylindrical tank is bolted to the tank along the flanges The tank has an inner diameter of 1.5 m and a wall thickness of 18 mm If the pressure in the tank is p = 1.20 MPa, determine the force in the 16 bolts that are used to attach the cap to the tank Also, specify the state of stress in the wall of the tank
Hoop Stress for Cylindrical Tank : Since = 750 18 = 41.6> 10, then thin wall analysis can be used Applying Eq.8~ 1
120 MPa
o, = — = ———————- = 50.0 MPa Ans
Longitudinal Stress for Cyiindrical Tank : |
lof % P 1.20 105) (750)
> 2018 = 25.0 MPa Ans
[[+28.0/^P^-
Force Equilibrium for the Cap: T=50.0MPa,
+TER 20; 1.20(10°)[*(0.75*)]-165 =0 F, = 132536 N= 133 KN Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 208-75 The crowbar is used to pull out the nail at A Ifa force of 8 Ib is required, determine the stress components in the bar at points D and E Show the results on a differential volume element located at each of these points The bar has a circular cross section with a diameter of 0.5 in No slipping occurs at B
Support Reactions:
(+, =0; 8(3—P(1697)=0 P=l4l4b
Internal Forces and Moment: grLE, =0; N=0 WHEE =0; V-1414=0 V=l4l4lb + =0; M~1.414(5)=0 M=7.071D-in Section Properties: A= (0.25?) =0.0625đ in” In : (0.25*) = 0.9765625"(10") in’ œ=0 Qœ=ÿA'= 4(0.25) 3 E (x) (0.25 |" 0.0104167 m 1 2)}_ 3
Normat Stress : Since N = 0, the normal stress is caused by bending stress only
1-414 tb Mc 7.071 (0.25) Sb*T “08765625z(107) T6 Pi(T) Am n My 7.071(0) af 70710) — _ Ter Fbo psc “:^~T * 99765615x(10-) Ans cv `
Shear Stress : Applying the shear formul, OD
vi T= 576 Pst Tp = + =0 Ans VQ; — 1414(00104167) 2 —— = = 29, 8 = “07656237(102)(05 OP AN
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 21*§-76 The steel bracket is used to connect the ends ot two cables If the applied force P = 500 Ib, determine the maximum normal stress in the bracket The bracket has a thickness of 0.5 in and a width of 0.75 in
Internal Force and Moment: As shown on FED
Section Properties ;
A =0.5(0.75) = 0.375 in? 1
T= 55 (0.5) (0.75*) = 0.01758 in’
Maximum Normal Stress ¢ The maxim: ura normal s Occurs at the bottom of the steel bracket ms
M2/187-5 fb-in N Me ơ, =e y+ oe wana = 500 + 1187,5(0.375) 500 tb | — [237s i 0.375 0.01758 = 26.7 ksi Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 228-77 The clamp is made from members AB and AC, which are pin connected at A If the compressive force at C and B is 180 N, determine the state of stress at point F, and indicate the results on a differential volume element The screw DE is subjected only to a tensile force along its axis
Support Reactions :
Grom, =0; — 180(0.07) ~ T_(0.03) =0 Toe = 420 N
Internal Forces and Moment :
SEF =0, 420-180-V=0 V=240N +T EF =0; N=0 +EM@,=0; 180(0.055)~420(0.015)—~/=0 M=3.60N-m * Section Properties : A =0.015(0.015) = 0.225( 107) m? +(e" o-6dm 1 - £= 55 (0.015)( 0.0157) = 4.21875( 10°) m* œ=0 ooiŠm + 420N Vit Normal Stress : Since N = 0, the normal stress is caused +
by bending stress only A
= 6-40 MPa
4, = ME = 3:60(0.0075) =640MPa(C) A
FT ~ 42187505) m
Shear Stress : Applying shear formula, we have
vớ;
tra =0 Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 238-78 The clamp is made from members AB and AC, which are pin-connected at A Hf the compressive force at ˆ C and B is 180 N, determine the state of stress at point G, and indicate the results on a differential volume element The screw DE is subjected only to a tensile force along its axis
Support Reactions :
É EM, =0; 180(007)—=1pg(0/03) =0 Tpg = 420N
Internal Forces and Mfoment :
“»E =0; 420-180-V=0 V=240N
—— on —
+TEE =0, N=0 \ Tt —r-sø
+EMạ =0; 180(0.055)—420(0.0159)-M=0 — ——_
Ni M=3.60N-m Tyr 60 MPa fe |e" oo
Section Properties: Ay
A =0,.015(0.015) = 0.225( 10°) m Is 75 (0-015)( 0.015") = 4,21875(10°°) mí
Qs = F°A’ = 0.00375 (0.0075) (0.015) = 0.421875( 10) m*
Normal Stress : Since N = 0, the normal stress is caused Shear Stress : Applying shear formula, we have by bending stress only
te va, 2⁄20 0.421875(105)] zat = =1, 1.60 MP A
s=?2 3.60(0) Ans oh ~ 82187510) 0.015) a ans
TT 7221082009) —
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 248-79 The wide-flange beam is subjected to the loading shown Determine the state of stress at points A and B, and show the results on a differential volume element located at each of these points
Support Reactions ; As shown on FBD Internal Forces and Moment : As shown on FBD Section Properties: A =4(7) —3.5(6) = 7.00 in? t I ; 1=1z(9(?) =5) (6) = 51333 in @ =0 Œœ = 7A” = 3.25(0.5) (4) +2.00(0.5) (2) = 8.50 in? Normal Stress : Since N = 0, the normal stress is contributed by bending stress only
Me — 7.20(12) (3.5) : 2S ee ee eS, A oO, T 31333 5.89 ksi (C) ns My — 7.20(12)(1) : =—=—————=l.6ä8ks A a'r $1.333 1) ma
Shear Stress : Applying the shear formula
o—— A Tạ i 0 ns VQ, _ 1.20(8.50) „1Ó 1200820) _ 0.597 ks A eT = 51333005) , ™ 0306)*4-80 kip mm +” 6£ 8£ 2-40 kip 2.40 KiP 0-304) =)20 Kip
CTO mMa7Zo kip fe Nz0 ailapll V= 120 Kip 240 kip A bos in #246 ||! | Bin ÿ-2zz —T ain =F in Fzz3 ] yap Gs! i Tatlbbkx SF T0397 KS
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 25
9-1 Prove that the sum cf the sormal stresses
o, +o, = oy + oy is constant sr, Tey
Stress Transformation Equations : Applying Eqs 9-1 and 9-3
of the text
CG +9, o,- 3,
9, +O, = T+ ¬ “cos 20 +1,,sin 20 0.78, 0, -6, ya cos 2 ~t,,sin 20
+O, =O,+0, (Q.E.D.›
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 269-2, The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 9.1
x’ ` ng AA Si H4 me oa Adasen’ 202/122 sia ude
N++ EE,, = Ú AF, + (8AAsin 40°)cos 40° ~ (SAAsin 40°)cos 50° — (3AAcos 40°)cos 40° +
(8AAcos 40°)cos 50° = 0 AF, =~4.052AA
4š =0 AF,-~(8AAsin 40°)sin 40° — (SAAsin 40°) sin 50° + (3AAcos 40°)sin 40° + (8AAcos 40°)sin 50° = 0
AF, = -0.4044AA
căn AF
Ø„= tim aso —— =~-4.05 ksi Ans
AF: :
Try = Em 2,0 22 =-0.404ksi Ans
The negative signs indicate that the sense of o,: and Ty: are opposite to that shown on FBD
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 279-3 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 9.1
68 MPa k©SÉP £05 2.77
4E F, =0; 6, AA+ 20AA sin 30° cos 30° + 20AA cos 30° cos 60° ~ 65AA cos 30° cos 30° =0
Oy = 31.4MPa Ans
\ LF, =0; + ¥ 2 Ty: AA xy + 20AA sin 1 30° sin 30° —20AA cos 30° sin 60° ~ 65 AA cos 30° sir n 30° =0
Try: = 38.1 MPa Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 28*9-4 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AB Solve the problem using the method of equilibrium described in Sec 9.1
440 Atos bo°
SEF, =0 AF,.—400(AAcos 60°)cos 60° + 650(AAsin 60°)cos 30° = 0 AF, = -387.5AA
\+Z 5 =0 AF,.—650(AAsin 60°)sin 30° — 400(A.Acos 60°)sin 60° = 0
AF, = 455 AA
AE
Oy Ans OAL =—388 psi Ans
AE
Tey = MT rry =455psi Ans
The negative sign indicates that the sense of G, is opposite to that shown on FBD,
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 299-5 The state of stress at a point in a member is shown on the clement Determine the stress components acting on the inclined plane AB Solve the problem using the method of
equilibrium described in Sec 9.1 ÑA Sinàø °
ÀA “ly
50 MPa 2~AACeS 30 °
60 MPa
+ iFy = 0; AF, + 60 AA cos 30° cos 30° — 28AA cos 30° cos 60° + 50AA sin 30° cos 60° — 28 AA sin 30° cos 30° = 0
AFy = — 33.251 AA
+} EFy =0, — AFy: —28AA cos 30° sin 60° — 60AA cos 30°sin 30°
+ 50AA sin 30°sin 60° + 28AA sin 30° sin 30° = 0 AF, = 18.33 AA
: AF
Oy = lima, 0 = — 33.3 MPa Ans
Try = limaasoef2 = 18.3 MPa Ans
The negative sign indicates that the sense of oy is opposite to that shown on FBD
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 309-6 The state of stress at a point in a member is shown on the element Determine the stress components acting on the inclined plane AJB Solve the problem using the method of equilibrium described in Sec 9.1
90 MPa 50 MPa Fo SA Gaze" Asingo* oot AÁz2 bhasy ÁAˆ ssuacioo Jn +“ Âm ’ Xy -2SAj/® * \X+E:=0 AH:~50AAsin 30° cos 30°~ 35AAsin 30° oos 60° +
90AAcos 30° sin 30° + 35AAcos 30° sin 60° = 0
AF, = ~34.82AA
+ =0 AF, —50AAsin 30° sin 30° + 35AAsin 30° sin 60° -
— 90AAcos 30° cos 30° + 35AA cos 30° cos 60° = AF, = 49.69 AA
: AF,
Oy = im, soa = =497MPa Ans
AF,:
Try =lim,, og TA —-=~-34.8MPa Ans
The negative signs indicate that the sense of o,- and T,y: are opposite to that shown on FBD
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 319-7 Solve Prob 9-2 using the stress-transformalion cqua- tions developed in Sec 9.2
5 ksi
Ø, = 5 ksi o, = 3 ksi Try = 8ksi @= 130°
O,+0, Ơ,—Ơ,
=o
2
= 5+3 + 2 cos 260° + 8 sin 260° = —4.05 ksi Ans 2
cos 26+ f,, sin 26
x
The negative sign indicates o,: is a compressive stress
Ớ,T—Ø,
Try = sin 26 + T,, cos 26
= -CC )sin 260° + 8cos 260° = -0.404 ksi Ans
The negative sign indicates T,,,- is in the — y' direction
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 32*9-8 Solve Prob 9-4 using the stress-transformation equa- tions developed in Sec 9.2
A 400 psi 650 psi R 6, = ~650 psi 6, =400 psi 1y =0 @ = 30° = Ox * Oy + TS ng 28+ 1+ sin 28 a 2 =
—— + eons 60° +0=—-388 psi Ans
The negative sign indicates o,- is a compressive stress
Try = ~ oz 2 9 5in 164 „y cos 2Ø
~650~400 _
= —( 0) 2 cin 60° = 455 psi Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 339-9 Solve Prob 9-6 using the stress-transfarmation cqua- tions developed in Sec 9.2
90 MPa
Zig
Ø, = 90 MPa o, = 50 MPa Ty = -35 MPa 6 = —150°
—” + eos 28+ +„ sin 26 90+50 90—50 = : + —S—-œ0s(~300°)+ (~35)sin(~300°) = 49.7 MPa Ans vị Ø,—Ø
Tyy: =— + sin 26+ Tt, cos 20
= 0) sin(-300°) + (~35)cos(—300°) = -34.8 MPa The negative sign indicates t,,, acts in — y' direction
=+ Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
© 2005 R C Hibbeler Published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, NJ All rights reserved
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Trang 349-10 Determine the equivalent state of stress on an element if the element is oriented 30° clockwise from the element shown Use the stress-transformation equations
300 psi
—————
950 psi
|
o,=0 o, = —300 psi Try = 950 psi @=-30°
Oy = ane ta Seo 20+ 1,, sin 20 0-300 0-(-300
= = + cos (—60°) + 950sin (~60) = -898 psi Ans
0, — By
Try: = —( )sin 20 + 7„„ cos 2Ø 0—(~300)
=~( 2 )sin (—60°) + 950 cos (—60°)= 605 psi Ans
Ở,+Ø, —
oy = ST _ + 28~ 7„„ sin 28
- 0~300_,0-(-300) 2 2 )cos (-60°) — 950 sin (—60°) = 598 psi Ans
5% psi
505 PSC
838 ƒs(
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 359-11 Determine the equivalent state of stress on an element if it is oriented 50° counterclockwise from the element shown Use the stress-transformation equations
10 ksi
Ø; = —10 ksi ơy, =0 try =~-16 ksi 8=+50
Ơy +Ơ, -
Oy = =~ + — 28+ try sin 28 -10+0 —l0-0
== > + —— 100° + (—16)sin 100° =-19.9 ksi Ans
0; — `
Try: = (FE sin 28+ cos 20 -10-0
2
=-( sin 100° + (—16)cos 100° = 7.70 ksi Ans
= Ft Oy 0T Ơ,
2
-10+0 =10-0
= —T~ (——es 100° ~(—16)sin 100° =9.89ksi Ans
Oy: cos 20— %, sin 20
399K:
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 36*9-12 Solve Prob 9-6 using the stress-transformation equa- tions,
90 MPa
@= 120° o, = 50MPa =90MPa 1, = 35 MPa
Oy = a + Tran 28 + ty sin 20 50 + 90 + 30 - 90 2 = 49.7 MPa Ans cos 240° + (35)sin 240°
The negative sign indicates 0, is a compressive stress CO, - Oy
Ty = ~ —————>sin y 3 2Ø + 7,„cos 2Ø ¥ 50 ~
29 sin 240° + (35)cos 240° = ~34.83MPa Ans
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 379-13 The state of stress at a point is shown on the element Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point Specify the orientation of the element in each case
50 MPa ee 30 MPa 45 MPa
o, = 45 MPa Øy =~60 MPa Try = 30 MPa
+0, Øy ~ Ø, a) a= 7 ty (F485? 45-60 45 -(-60) 2 seo ty )? + (30)? 2
0, =53.0MPa Ans 6, =~68.0MPa Ans 7zzM/'!
Orientation of principal stress : 68.0M pa 15 MPa
¿ø5 MÑ
tan 20, = 2 = “05714 53-0MPA
PỮt(g.—-ơ)2 (45—(-60))/2
6,=1481, -75.13 sếp
Use Eq, 9 - | to determine the principal plane of 0; and Ø; :
oe = EA * 2 2 Mh c0§ 204 tysin2O, where @= 14.87° = AS +0) , 45~ (60) 5 29 749 +30 sin 29.74° = 53.0 MPa 2 2
Therefore 6), =14.9° = Ans and 6) =-75.1° Ans
b)
k mak mee = (a> — 2# +ry2 = 2 430 =60.5MPa Ans
og, 2 et 2 BECO) 7 50MPa Ans A
etre 2 75MPa
Orientation of maximum in- plane shear stress :
Bot’, 49°
GO-SMPa
68.0mpa
6, = -30.1° Ans and 6, = 59.9° Ans
53-0 MPa
tant26, = =(Ø, ~ 9,)/2 = —(45~- _ =~175 Tay 3
By observation, in order to preserve equilibrium along AB, Timax has to act in the direction shown
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 389-14 The state of stress at a point is shown on the clement Determine (a) the principal stresses and (h) the maximum in-plane shear stress and average normal stress at the point Specify the orientation of the element in cach casc
180 MPa
————— 150 MPa
Ø, = 180 MPa o, =0 Try = -150 MPa 0, +6, o,-G,
ay 2 a ty tt?
— (OD Caso
Mf
o, =265MPa = Ans 0, =-84.9MPa Ans ASM fa ne
Orientation of principal stress :
6, = 60.482° and ~29.518°
Use Eq 9- 1 w determine the pricipal plane of ơi and Ø; :
Ov = _— + See Shas 20+1,,sin20, where @= 60.482" = = + = Ô 9s 2(60.482") + (—150) sin 2(60-482°) = 84.9 MPa
Therefore 6,, =60.5° Ans and 6,.=-29.5° Ans
b) ~-~
= if( 22%) 2 =f -]502 = “^ wuA
Taek, wie 7 ( 3 )*+Ty” =U( 2 )?+(-150 =175 MPa = Ans ! } `
ayy = ES = TC” =900MPs Ans 155°
Orientation of maximum in - plane shear stress : tan 26, = ~(Ø, ~ Øy)/2 = (180 — 0)/2 =06
= 155° Ans and @=-74.5° Ans
By observation, in order to preserve equilibrium along AB, t,,, has to act in the direction shown
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X © 2005 R C Hibbeler Published by Pearson Prentice Hall,
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Trang 399-15 The state of stress at a point is shown on the element Determine (a) the principal stresses and (4) the maximum in-plane shear stress and average normal stress at the point Specify the orientation of the element in each case
30 ksi
G6, =-30 ksi og, =0 Try = —12 ksi
a)
_ oO, +9, Ở;—Ø, ¿ ;_ -30+0 -30-0.; A2= 1.2 2 t ự 7 + y= 2 tực 2 )? +(~12) 12)?
Øi =421ksi Ans Ø; =~34.2 ksi
Orientation of principal stress :
(đ,=øœ2 ` 30-0/2
6, =19.33° and ~70.67°
Use Eq 9- 1 to determine the principal plane of Ø, and ớ;
Ans Oy = Sst Oy ˆ Phos 20+ t,, sin 20 6= 19.33° -30+0 ~30- lạ = ” + “« 2(19,33°) + (—12)sin 2(19.33°) = ~34.2 ksi Therefore La =19.3° Ans and vã =-~?0.?° b)
Fant ag = SEI + ty? = (P+ ene =19.2ksi Ans
Ouyg = I = 1S ksi Ans Orientation of max in- plane shear stress :
tan 20, = OM Sy 2 _ (230-02
, to -12
6,=~257° and 6439 Ans -1.25
By observation, in order to preserve equilibrium along AB, Tmax has to act in the direction shown in the figure
42) KS S42 ES (4° /4+2 k4 257°
G2 KSt PORE + 2e E4; /S ksi ‘
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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Trang 40*9-16 The state of stress at a point is shown on the ele- ment, Determine (a) the principal stresses and (>) the max- imum in-plane shear stress and average normal stress at the paint Specify the orientation of the element in each case
250 MPa ———~T—* 115 MPa 200 MPa
8) Ø =—200MPA Ø =250MPa ty = 175 MPa
_ Ø +Ơ đc“ Ởy 1, 3 Øa = + f(t ˆ 2 7 3/0 MPa ~ 200 + 250 - 200 - =————— + (728-292 is: 11-1? 2 ~ 2
a = 310 MPa a, = - 260 MPa Ans
260 MPa Orienation of principal stress :
tr, 175
an 26, = gr = appa = - (0.7777
2 ?
8; =~ 18.94° and 71.06°
25.0MPa Use Eq 9- } to determine the principal plane of o; and oz
ay = % - Sy 7% os204 7, sin26 ZBSHP* 36.9 MPa
264° 6 = 6 = ~ 18.94°
- 200+ 250 -200 ~ 250
Or = —z—— + ————wi(- 37.88°) + 175 sin (~ 37.88°) = ~ 260 MPa = Ơy Therefore Øp, = 71.1” 6p, = —18.9° Ans
3/0 MPa
đ, ~ ở - -
b) Fang = (Sb + oy = (ee +175? =285MPa Ans io plane 2
Tht
285 MPa „+ -
Ou = oo = aes = 25.0 MPa Ans
<6-/°
Orientation of maximum in - plane shear stress : 184"
(a - = 200 ~ 250 25.0
mn 26, = ~ —2— =~ 20857 S-0Mfa 260 hPa
Ty 175
8, = 26.1? Ans and ~63.9° Ans By observation, in order to preserve equilibrium, tm,, = 285 MPa has to act in the direction shown in the figure
From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X
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