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12-61 Determine the maximum slope and the maximum deflection of the beam E/ is constant

1 £1I mM &L Mo L MoL Ôc/A n3) 2EI Oc = Ocia + đạ 0= Mok + 0, 2EI —=MọoL MẹL Ans 6, ax = 6A = ——_ ~ ¬rự m 2EI 2EI TT ns

Amax = ltacl = EI 2 4 SEI

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 2

12-62 The rod is constructed from two shafts for which

the moment of inertia of AB is J and of BC is 2/ Determine the maximum slope and deflection of the rod due to the loading The modulus of elasticity is E

: 1 -PL_oL -PL_L =PL LL -5PL? _ SPL?

đục = (FED * Ge? * (Sepa) * Teer — 16EI

OO, = Bac + Oc

SPL’ o= SPL?

Omax = max 4 TOE! = + 16Eƒ Ans

Âmx = ÂA# lwcl -PL LL -PL LoL “PL LLL = —— — _ —=}+(——)(-)(—- _ 20006) + GPP * 5) Sep PG +P _ 3PL3 16EI

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 3

12-63 Determine the deflection and slope at C E/ is con- stant c[= A c CM, F= = == aT] 3m ) |, 1,—M, Mạt? ta = 1G ML) =~ "ser Ac= HA —2| taal _7MoL* Mo To

tora = NLL + 5 D+ yng )= SEI 2 1MạLˆ Mọi? _ 5MạL A = -(2X4——)=— ns Ae = ee OC = EFT 6, = \taval = MoL aL 6EI _3MoL _ 3MoL

6c = nang + Ay = SEI SEI

Oc = Ocr + Og

ge = Mol _ Mol 4MoL Ans

2EI 6EI 3EI

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 4

*12-64 The shalt supports the pulley at its end C Determine the deflection at C and the slopes at the bearings

A and B Elis constant

Yz “2 1,-PL ul -PL’ Ly -PL ~ i spt MS) = Ser L M Ac = ltcial — (z) Ital # 2 = PL? ~2 PL = PLP Ans

SEI - 48EI 12EI

: 2 = Pk

8, = {taal _ eel -Ö PL Ans Si

aE Lo 24EI

1 -PL L._ -PL? _ PL

{REI 2° 8El 8EI

8 =—— -~ — = Ans

b” SÉT - 24E1 12EI

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 5

12-65 The shaft supports the pulley at its end C Determine its maximum deflection within region AB ET is constant The bearings exert only vertical reactions on the shaft

tana ta Ob1a >= ——- (ÿ) Icky Py lyk Ty = 220) x = 0.288675 L 2 El (@) 1,P(0 Amax = 1 P(0.288675 L) 288675 ») 0.288675 1)(2) (0.288675 L) 2 El 3 3 Am, = 0.00882PL Ans EI

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 6

12-66 Determine the deflection at C and the slope of the beam at A, B, and C EJ is constant

1 (=8)¢6)(2) = 8 tua = 5G OO = Fr 18 = = 18 tora = 5 Gp Oe +2)+ G 3)01.5) Er 9, ,_156 948 84 ng b Ac =Itc/Al~ CHeAl = ý EN El 2 tanc 5 M — teal _ 8 Ans EL 6 ET ; T 1-8 24 _ 24 | = +( = = O54 np 6) El EI -8 Ez On = Opa t 94 24 8 _ l6 Ans 98-7) EI El 1-8 -8 48 48 =) (3y6+ (3) === ==

%c¿A sư” d+ EPO) EI El Oc = Oca + 94

48 8 _ 40

awe Zee Ans

c= Fr" Er EI

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 7

12-67 The flat spring is made of A-36 steel and has a rec- tangular cross section as shown Determine the maximum elastic load P that can be applied What is the deflection at 8 when P reaches its maximum value? Assume that the spring is fixed supported at A

/4P | tr ` | ——+ on a i # S él I= (1500) = 0.125(107?) in* 6, = Mc 36(10°) = 146009) P=643lb Ans I 0.1250) 1 ~-90 Ap = tara B= tava = ~(——)(14)(9.333+3 5! Ei (14) ) = = = 2.14 in Ans El 29(10*)(0.125)(1073)

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 8

*12-68 The acrobat has a weight of 150 Ih, and suspends himself uniformly from the center of the high bar Determine the maximum bending stress in the pipe (bar) and its maxi- mum deflection The pipe is made of L2 steel and has an outer diameter of 1 in and a wall thickness of 0.125 in

131 _†+ +— 3d TT ¢ E—:a S 4 4 Zf {is 3t Toth = 75th 7sIb 7/6 428 eft (FF Mmax = 75(3) = 225 Ib- ft I= Zo" ~ 0.375%) = 0.033556 in*

Omax = = —————— = 40.2 ksi Ans

I 0.033556 40.2 ksi < oy =102ksi OK

225 1,225 1244.53 lb - fr Amax = tac = (—~)(0.75)(3.375) + —(—~)(3)(2) = —— AIC (7 )( ) 5 ep! )(2) ET

3

¬— ` Ans

29(10°)(0.033556)

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 9

12-69 Determine the siope at C and the deflection at B E7 is constant Gey tana ™— ey, M "Er tang Ỉ Ị ==za 4 apa Ft ET FI 2Pa 1/ Pa ĐcA =[—“—” z ( EI }2+{ m9 -|—— _ _5Pa! _ 5Pa? ) 2EI 2EI Ốc = cra Œa = Pa? = “` 2E 1 Pa 2a lf Pa 2a 2Pa a ẢAn= =—Ì|—— —_— —-|—— — —— - 3 = {inal ( mJ'2(Ÿ)*3( a) (at s+ EI }2(3) 3 - 23Pa + Ans 6EI 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 10

12-70 The bar is supported by a rolier constraint at 8, which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown, determine the slope at A and the deflection at C El is

constant LPL LPL EL _3PL?

Oye = (= KEO)“ "sEï Ủ

I Oa = On 2£ 2 + , = PL Ans | 8EI

1 PL LOL PL LOL LỘ 1iPL3 = ~(——\(-\(-) + — Ee +=

tare = 3 (DG) * er 99 * = ABET

fore = Fer DD? = Tent

Ac et _1iPL? PL? _ PL?

c = tae lca Teer 16EI GEI 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 11

12-71 Determine the maximum deflection of the shaft ET is constant The bearings exert only vertical reactions

on the shaft [-i-t —* ——+—i—]| E te P t * ơ Fz “2% Ì 3¿ $ ⁄ P P P P (a) -tan A | “22c fh Er (b) La 282 (cy Ana = tac = Pa : Co + 2) + 1gG)) a a 1 Pa a a = 11Pa° 48E7 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 12

*12 72 The beam is subjected to the load P as shown Determine the magnitude of force F that must be applied at the end of the overhang C so that the deflection at C is zero El is constant 1, Pa Pa? 2Fa? =^(— *(-—^X2a\(Ša)=——— ‘ava 2E 422)4)+ 1 mí Š a= 2EI 3EI

tem = AES y(20)(20) + 2 1 yaaya + 22) +

2a Pa” 2Fa` 2C Xa a )= EH EF 3

Ac = tera — tA =0

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 13

12-73 Determine the slope at B and deflection at C EJ is constant fame 82S J fname 94 mamxei

Onc = 2C 2@)+ LP Fe y0) + Paya) = EE

2 CE EI 4EI 7Pa?* = = Al Os = Onc = ms 1,P

Ac =limcl =: a ys oars a+22)+

9Pa?

=)= A

= aa a+ 2)” GET ns

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

This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher

Trang 14

12-74 The A-36 steel shaft is subjected to the loadings developed in the belts passing over the two pulleys If the bearings at A and & exert only vertical reactions on the shaft, determine the slope at A The shaft has a diameter of

0.75 in, | 300 th 300 Ih 342-861 J0tb — 67/16 beolb 1,6171.48 1,6171.48 ¡ =2 0) + = 3(11.08)(20.31) bia = 3! zl )(18)(30) 5! aT 1 -7200 2160231.8 *(Cˆ“—)(1292)(431= ———— + 2C a lai _ 5143441 _ 51434.1 Os = TT ET — 29059(70(0.375) = 0.114rad = 6.54° 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 15

12-7§ The A-36 steel shaft is subjected to the loadings developed in the belts passing over the two pulleys If the bearings at A and B exert only vertical reactions on the shaft, determine the deflection at C The shaft has a diameter of

0.75 in

3#2861Á 4;erb HEFL lá booth

1 6171.48 1 6171.48 tya BA = TT )(18)0) 3! EI )(11.08) (20.31) = —( 18)(30) + - 1 _~7200 + z(C——)(1292)(441) = 71602318 aH EI , 54 2 tạ = (tgA)(—) = 2777441 42 EI 1 6171.48 1 6171.48 tia 1A = - 56 EI )(18)(42) + 5! EI )(11.08) (32.31) 18)(42 - 1 -7200 1 -7200 2333287.6 + -( 7L g7 )(12.92)(16.31)+ -(——— )( ) 5! EI )(12)(8) =o E Ac =ien — g —— -4153 29(10)(7)(0375)1 Ac = 0.987in 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 16

*12-76 The 25-mm-diameter A-36 steel shaft is supported at A and B by bearings If the tension in the belt on the pul- ley at C is 0.75 kN, determine the largest belt tension 7 on the pulley at D so that the slope of the shaft at A or B does not exceed 0.02 rad The bearings exert only vertical reac-

tions on the shaft

10.247 10.247 = -(— 5° El )(0.2)(0.1333) + 3! El )(0.3)¢ ~(——)(0.3)(02 + 01 ) 7 ao 1 -150 00147 12.5 ~(——)(0.5)(0.3333) = ~——— * 2 EI 0-5) ) EI EI 6, = teal _„ 00287 _ 25 qd) 0.5 EI EI 10.24 T 1 0.247 = —(———)(0.2) + —(——)(0.43 65/4 5! El )(0.2) 53! El )(0.3) 1-150 0.06T 37.5 mm 5) =— ——— 15 kN 2T * 3! EI (0.5) EI EI Og = Opa + 94 6 = Er er er TH 12.5 _ 00327 _ 125 @ EI EI

For O4 = 0.02 rad mon

EI From Eq (1) : | 0.02(ED = 0.028 T - 25 2247 0.02(200)(10°)(19.175)(10°) = 0.028 T — 25 T = 3632N For 63 = 0.02 rad From Eq (2) : 0.02(E D = 0.032 T - 12.5 0.02(200)(107)(19.175)(10”) = 0.032 T — 12.5; T = 2787 N = 2.79 kN controls 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 17

12-77 The 25-mm-diameter A-36 steel shaft is supported al A and B by bearings If the tension in the belt on the pul- ley at C is 0.75 kN, determine the largest belt tension T on the pulley at D so that the slope of the shaft at A is zero The bearings exert only vertical reactions on the shaft

100 mm 300 mm ——T†— 200 mm | Ụ “1 ao 200mm | 2BT+Lôo | 19 KA T LAT~0-5 MỸ EL tana

oi, nĂ Tana

Require tg, = 0 1 —1500 2 1 0.247 ft BIA = —(——)(0.5)(-)(0.5) + -(———)(0.3)(0.3 5! Ei )( Gre ) 5! Bĩ )(0.3)(0.3) 1,0247 2 + ~(——)(02)(-)(0.2)= 0 5! Er )( IGM ) T = 8928.6N = 8.93 kN 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 18

12-78 The beam is subjected to the loading shown Determine the slope al B and deflection at C E/ is constant

The slope : 1 —Moa

wa = leash < DI a)(2a)

+ te yaya + 2 -) 2 tas b)

_ Mo(b? + 3abt - 2a")

~ 6EKa + b)

0g =

a+b 6EKa+ by? E7

_ : —® —)(b\( 2 Mob tcp = Aen b) 6El(a + b) b Ac = >? tạp —ÍC(B , Mạb (bÈ + 3ab - 24) — — Mạb - 6El{a + b} 6El(a + b)

- Moa b(b-a) Ans 3EKa + b)

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 19

12-79 If the bearings at A and B exert only vertical reac- lions on the ’shaft, determine the slope at A and the maxi- mum deflection +r 17Pa° 3EI 1,Pa a Pa = -(— 3 ~})+(—)¿ +a)= toa 5p a+) ( 7 a)(a+a

Ans Assume Amax is at point D located at 0 <x < 2a

1 Pa Pa Pa’ Pax

= ~(—— +(— =——+——

Sova 2 ET a) 7 1) 2EI EI

Op = 0 = Ops + G4

Pa? Pax ~17Pa’

0 = — + —— +(—_—— 2EI EI 12EI lt x= a@ 12 Pa lL (2a-t4a) 481Pa? = =(— 2 ~—— —— ooo Ans

Amax Itapl ps a 1221 2 a] 288EI

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 20

*12-80 The two bars are pin connected at D Determine the stope at A and the deflection at D E/ is constant

M tr £r fa SEL + L 1 | T T ¥ Ỷ

Am Pa “4` 2'2EI 4EI

Os = On, + 9 2 0= =Èa + 8ạ 4EI 2 = Pa" Ans 4EI Pa?

"Oy = 04 = Bees SEI

Pa An = Đ›a=—— Ans

p= 7885 xpy

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 21

1281 A beam having a constant £/ is supported as shown Attached to the beam at A is a pointer, free of load Both the beam and pointer are originally horizontal when no load is applied to the beam Determine the distance between the end of the beam and the pointer after each has been dis- placed by the loading shown

{5S kN 20kN | CT : — | = — ali? ame bid iS BN 20kN 1 >»k im I am {SEN T ce 4» ting Determine t CLA „ 1/30 2/2 toa= 3 gp HD2 40 †C/A = — Ans EI

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 22

12-82 The two A-36 steel bars have a thickness of 1 in and a width of 4 in They are designed to act as a spring for the machine which exerts a force of 4 kip on them at 4 and B If the supports exert only vertical forces on the bars, determine the maximum deflection of the bottom bar

for Setram plate

4rr 4F? A | (F | 4 ship 4*ï orn Smax c1 M | #@ % _ 48 1,48

Anas = tye B/C (EPUB + 12)+ 2œ?2/) = (— sy

_ 20448 20448

El” 200 %ENaK) Ảng

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 23

12-83 Beams made of fiber-reinforced plastic may one day replace many of those made of A-36 steel since they are onc- fourth the weight of steel and are corrosion resistant Using the table in Appendix B, with cajow = 22 ksi and tajiow = 12 ksi,

select the lightcst-weight steel wide-flange beam that will 5 kip safely support the S-kip load, then compute its maximum

deflection What would be the maximum deflection of 4 € 2 this beam if it were made of a fiber-reinforced plastic with

Ep = 18(102) ksi and had the same moment of inertia as the

# ⁄2z/( steel beam? Zs 22 - 7/4 ) a, 5 kíp tana taf) AS he 2 0 - tane (b) be-—— 100 4 lon- -— vá) z.e Mmax = 25 kip: ft M(-Ft) Z§

Steq'd = Mes = 2 = 13.63 in? a N

Santow 22 ¬

Cy

Select W 12x 14

(S,=149in? 7, =886in' d=119lin 4, = 0.200in.)

Check shear :

Tmax = Mass = —— = 1.05 ksi < Tayow = 12 ksi OK

A, 11.91(0.200)

Use W 12x 14 Ans

1 25 2 833.33 kip - ft

Amax = Ít, [tact = -(—)(10)(—-)(10) = —————— 3 ep! Gt ) EI

For the A-~36 steel beam :

833.33(122)

= —————- = 0.560 in Ans 29( 103)( 88.6)

max

For fiber - reinforced plastic beam :

833.33(12?) _ = 3 in Ans

18(103)(88.6)

Amax =

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 24

*12-84 Determine the slope at C and deflection at B EI is constant

Support Reactions and Elastic Curve : As shown

Wa,

M/E! Diagram : As shown

A ef 4 Moment-Area Theorems : The slope at support A is zero The ae le

slope at C is a

a

Z'¢

1( wat wa?

6.=18, c~I®eal =2|-E }a+(-se}a =~-l—— —

wa?

eT Ans

The displacement at B is

_ _ tf wa 2 wat

As = [fal = air )o+34}+($8 ales 5) at _ wa? 3 3L zz/2 (52) 4lwa*$ = “Ee L 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 25

12-85 The A-36 steel shaft is used to support a rotor that exerts a uniform load of 5 KN/m within the region CD of the shaft Determine the slope of the shaft at the bearings A and B.The bearings exert only vertical reactions on the shaft 22m“m sm 8 ITTIIIIItLLL aH form Ph £t Kc wet |? | | Soe mm | ama 759 750" 156 4.805 Ben = 2G)04)+ cee — (0.15) + + NÓ 19-52 4805 4.805 aS = —— 2 = 0.00306 rad = 0.175° El ¥)£(0.01)* 64 = 8g = 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 26

1286 The beam is subjected to the loading shown Determine the slope at 8 and deflection at C £/is constant

| +—+— — 2 2 3 wa’ a 2.wa Twa

Sw (-) + -(-— la) = me = TD * 3 DEP * 12p

7wa”

6p = Đg¿c a = Onc = 12ET A ns 2 2 Ke = tyre = 4 (5Xa+ 3+ 24 € = đực 2011 3 3'agr)G2) Yayv>

25wa*

= Ans 48EI

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 27

12-87 The, two bars are pin connected at D, Determine the slope at A and the deflection at D E† is constant

| AVR EME oe le ail am D a | ;———i———t— A t ề £ 2% £ z (4) whe 2EL te, -PL?

t BIA = = —(—— sư XG by = DET

2

A, = oval „ Pr Ans L 12EI

- :

ae -PL LoL tia DIA = A SL 5 =(—— —+— 3)† 24 sea)? -——

_ Pv ~ 4EI ‡L r3 Ap = Ital - " )fmal = ——- - -(——) = —— Ans 4EI 2 12EI SEI

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Trang 28

"12-88 Determine the maximum deflection of the beam E/ is constant ~- wor - wae 2£? E1 2

~wa ava ~wa

tye = (—Y\(-\(-) = —~ ve = Oe OG * Ten 4 wa Ag =| fee] = —— 16EI 2 2 4

“Wwa“ a a, 1.-wa 3a Twa

toe = (S=— oat -) += 78 EP 009 6:00 —)=——— r NT:

Áp +1 tovel —ltwel = Twa" _ wa" = wae 16EI l16EI 8EI 3wa†

§EI

Amax = Ap =

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 29

12-89 The W8 x 48 cantilevered beam is made of A-36

steel and is subjected to the loading shown Determine the

deflection at its end A

10 kip tui ===: | 6A —| - Sf: + lo FÌ? †„/@#7a'° A Cr L 8 £ , io Ki? a “ee | 1S xip-ft ~<—— ú, SPL’ _ 5(0)(12') _ 1800 ;

(Aah = 48EI — 48EI “REI BÍ EI

ML? = 15(127) _ 1080 1080 | (Ase = 957281 ET 100 1080 2880 +4 Ag =(Aah + Aah = TT =F El 2880(1728) _ 9.933 in Ans ='39(10°)(184)

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 30

12-9 The W12 x 45 simply supported beam is made of A-36 steel and is subjected to the loading shown Determine the deflection at its center C

PL? 12024) 3456 A, = FELL 1204) _ 3456 (Gch = tear al” ET Ỷ Mx f 1 A 2) = (2 -3Lx ri +2L) 2 Lh Kip-ft x = > Km At point C, rok oe „ 2 MC) L 2 A, — + (Ach = moe -tLễ )+2L’) ML? “soe _ 1800 | "I6 1681 EI _ 3256(1728)

=——————- 2200650 =0.895 in 895in.l — 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 31

1291 The W14 x 43 simply supported beam is made of A-36 steel and is subjected to the loading shown Determine the deflection at its center C

2 kip/It 40 kip-ft = = oa C B L 101 J 10 ft { LRP Lt 46 kự-k — ‹ & “2t 70 fe it 222/7 Se + (A), — 4erpkt 5w1ˆ 5(2)(20° 2083.33

(Ac), = ee = IC) 768EI — 768EI EI 1

Mx ; 2 2 40(10)„.„¿ ?

Ac)2 = x —3Lx4+2L’) = ————[10* ~ 3(20)(10) + 2(20

(Ac)2 cai | ) 520) ET! (20)(10) + 2(20)”]

_ 1000 EI 2083.33 + —- 1000 Ac = (Ac)i + (Ac)2 = EI EI - 3083.33 kip: £¢ EI

Numerical substitution for W 14 x 43, , = 428 in‘

083.33( 12?

c= BOP = 0.429 in, 29(102)(428) Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 32

*1291 The WI14 x 43 simply supported beam is made of A -36 steel and is subjected ta the loading shawn Determine the slope at A and B

Ar 1 ĩ 2 kinH | | | 40 kip-t 10 ñ † 10 -| 6u Ox It 6a, + Oa, 71wLŸ „ ML 384EI 6EI 7) (240°) „ 4012)(240) _ — 61,200 384 EI 6EI — 29(10)(428) 0.00493 rad = 0.283° Ans Ổp, + Op, 3w17 „ ML 128EI 3E 32240) „ 4012240) _ _ 92,400 128 ET 3EI ———«-29( 10) (428) 0.007444 rad = 0.427° 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 33

12-93 The W& x 24 simply supported beam is made of A-36 steel and is subjected to the loading shown Determine the deflection at its center C

5 kipft

1=82.8 mÝ )

+ TS (Acỳ ha

(ac); = SWE’ 5(6)(16*) _ 2560 768El 768EI EI

M Áa¿@=-——@?~31x+ 2L2) 6LEI At point C, - 2 MG) ob (Ac) =— 2 6LEI° 4 .(*⁄_ ~ 3G)+2L) + 2 - ML? _ 56) _ 80 l6 16EI ET Ac = (Ac); +(Ac)2 ~ 2560 | 80 _ 2640 El El EI = 76401728) _ 90; 29(10°(82.8) 77 Ảng

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 34

12-94 The beam supports the loading shown Code restric- tions, due to a plaster ceiling, require the maximum deflec- tion not to exceed 1/360 of the span length Select the light- est-weight A-36 steel wide-flange beam from Appendix B

that will satisfy this requirement and safely support the load

The allowable bending stress is @ajjow = 24 ksi and the allow- able shear stress is Tayjoy = 14 ksi Assume A is a roller and

Bisapin

đua =36 kip ? Sokip

Musas = 162 kip - ft Strength criterion : M lé Satiow = " Sreq'd a 4+ 162(12) % sp qd ầ sị” ° £ x 24= Sreqra = 81 in”

Choose W 16x50, S=810in, 4, =0.380in, d= 16.26in., I, = 659 in‘

Check shear : ŸạHow = —— web 36 142 ——— _ = 5.83 ksi (16.26)(0.380) ksi OK Deflection Criterion; wL* (41242) 1

Vinax = 0.006563 3 EI =0.006563(—————-“~.) =0.7875in « —— 2q0X69)) 7875 in < 34022) 0800 = OK Use W16 x50 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 35

12-95 The simply supported beam carries a uniform load of 2 kip/f Code restrictions, due to a plaster ceiling, require the maximum deflection not to exceed 1/360 of the span length Select the lightest-weight A-36 steel wide-flange beam from Appendix B that will satisfy this requirement and safely support the load The allowable bending stress iS Catow = 24 ksi and the allowable shear stress is Tallow = 14 ksi Assume A is a pin and B a roller support

the

Mraz = 96 kip - ft Shy

PUTT PGT A e/fs

Strength criterion : tH sy leet

24 by, ự 24 fp, ` Ổaliow = ———— M req'd 24 = 26412) Sreq'd Sega = 48 in’

Choose W 14x34, S=48.6in’, 1, = 0.285 in, dđ=1398in, 7=340in

Tàjiow # ———

'web

>———= 6.02 ksi (13.980.285) *S! OK

Deflection criterion; Maximum is at center

nn = ae 384E7 ` SETA) 4 (9 PCB) SEI lle) ~ (4)? ~ (8) 9112)? ~r3@6)1 „11 330 384E7 4.571108) * 39(10%)(340) ~ 000464 in < 35(16)(12) = 0.533 in OK JseW 14x34 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 36

*12-% The W10 x 30 steel cantilevered beam is made of A-36 stecf and is subjected to unsymmetrical bending causcd by the applied moment Determine the deflection of the centroid at its end A due to the loading, Hint: Resolve the moment into components and use superposition

kL=170in, 1 =167inf - (Msin 8 )Lˆ _ 4.5(sin 30°)(15°)(12)° 2E1, 229409167) 09032 in max _ (M cos 8)12 ~ 4:5(cos 30°)(152)(12) 2EI, 2(29)(101(170) A4=y0.9032?-0.1537?=0.9l6óin Ans max = 0.1537 in,

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 37

12-97 Determine the vertical deflection at the end A of the bracket Assume that the bracket is fixed supported at its base B and neglect axial deflection E/ is constant

Pa?b b El (Aa) = 64) = As = PL’ _ Pa?

(ah = Sey SEI

Path | Pa? _ Pa’(3b+a)

Aaa (Aah + (Bade = Er * agp” SET Ans

From Mechanics of Materials, Sixth Edition by R C Hibbeler, ISBN 0-13-191345-X

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Trang 38

12-98 The rod is pinned at its end A and attached to a tor-

sional spring having a stiffness k, which measures the torque

per radian of rotation of the spring If a force P is always applied perpendicular to the end of the rod, determine the

displacement of the force EJ is constant

ia

k

In order to maintain equilibrium, the rod has to fotate through an angle @ GEM =0 k-PL=0, 9 = PL k Hence, A’ = Le = Pe) _ PL k k Elastic deformation - 3 A” = PL 3EI Therfore, 2

A=A'+A”= TE PLY pal L k 3EI k

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 39

12-99 The pipe assembly consists of three equal-sized pipes with flexibility stiffness EJ and torsional stiffness GJ

Determine the vertical deflection at point A

_ PG? PL Ap = 3EI 24EI le Pty? 3 (Ay), = GX PLY P 3EI 24ET e e ( = gn Th _ PLIDG) _ PL Ce) JG JG AIG LJ ~~ ~ Ae, Pus ¬ 4, = SEI 2n, (84); = 6(È) = PL 23 8G Ø

Ag = Ag+ (Ag) +(Ag)2 † Ps

PD PL` PIẺ =

= MEI 24EI BIG + ~~ PT (a 4 (As),

1 1 B

= PL (—— + -—_ oe nh Ans = đa A

FE],

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 40

*12-100 Determine the vertical deflection and slope at the

end A of the bracket Assume that the bracket is fixed sup- ported at its base, and neglect the axial deformation of seg- ment AB ET is constant

44=——=——=—= Ans

= ea 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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