This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Prob. 1.48. A weight W is supported by a cable attached to beam
BE at E, as shown in Fig. PIA8. The beam is supported by a
ﬁxed. frictionless pin at B and by an inclined cable from A to D.
(:1) Determine the tension in cable AD, and determine the reaction
force at pin 3. (b) Determine the internal resultants (axial force. shear force. and bending moment) on the cross section at point C. P1 .48
Solwion ‘ ta)+cmion 0¥ cable so ovt' (tad10H 'n' :
m: beam . PP EGAN—Bum LEFXSO: BﬁF'ﬁs'TAD so
4
sz'l'f'rho +4555“ 5W '2" TAP w=0 (mime/had resulmwls
pap: laws—ham beam semen EarU'liibViW I LZPX eozay +Fceo Fc= “ex—W
3
*4‘Zﬁ$=o= VL r, O VL5—W
i 12(ZMlc‘ 0‘ BM?) ‘Mc = O Prob. 1.41.1. Beam AD in Fig. 131.441 has a frictionlesspin
suppon al A and a roller support at D. The beam has a
linearly varying distributed load. of maximum intensity w. {farce per unit length). over twothirds of its length. (a) m H C
Determine the reactions at a and D, and (b) determine if
the internal resultants [axial force. shear force, and bending U3—+—u3 U3
moment) on the cross section at B. [14.11 and P1 .442
52%: 52M 3 kﬁl (4} etecf/Oaf ﬁ/ﬁﬁwa/ﬂ ‘57 i
I W Eel(‘A'Jrr'mm o jc’ﬁm 59.. 4)éfffn4/ Kerk/{1026'
frr'ﬂ‘gn'um of Kram fﬂimdu/ $2790 éfﬂtﬁo 6:0 so
.244
’4? r {z #Vé ‘0 vanﬂu Prob. 1.41. Each member in the pinjointed planar trusa in
Fig. P1.41 is 6 ft long. The truss is attached to a ﬁrm base
by a frictionless pin at A, and it rests on a roller support at
B. For the loading shown, (a) determine the reactions at A
and B, and (b) determine. the force in each of the three
members, labeled (1) through (3). 33/6425}??? .'
(a) ﬂfécfr’oru' 4/1? atria/6::
Jjélkbr r” ﬂ; r 761/ 444: r
5} = +4.70 éw r (/35) ﬁxr'ﬁ/ féfrI r'a mamﬁ'r—r (U '6’). :3; ‘42??? war 6=ﬁ+¢£4§v ff: ﬁféé’éWz’rl
6: 445. £02: [a] +T P1 .4—1 and P1 .42 aX/r'én'am .'
51»; F559 9/ m/m ﬁrm,
+( {2'qu r 0
— 344w ( 6ffJKrr5 66")— ééaér/é/J’F H3 MW
4; ' iffy £911.!
$ 275 c 0
a; n’éo'u =0
19:: = '5? +7“ ‘56 "0
,% ﬂ 6, 4+; 1 ﬁ; = o
I?) = A 70 / MW Ezmﬂ' 031»: 5
him Fiﬁ Jr" 27%;"? M?” 1/154 Kara; +f(Z/W/y :52 gKf/HAH m .t is; (4,?) =0
1?:% #454?! Zr? :0 é/ffln 40") — :54ng ~03  O
6" /. 764 1:}; Few. Fm) afﬂm/ﬁ) €018 60") 1t 5, =0 —: 4:; “ﬁg 455;, ﬂE/iw we: *ﬁz’ww‘) «2 5 2. m 4!» Prob. 1.43. For the pinhinted truss in Fig. PIA—3, (a) deter
mine the reactions at the pin supports at C and E, and (b)
detennine the axial force in each of the following members:
F3 (in member AD), F. (in member CD), and F3 (in mem
ber 3D}. ﬁ/Qéloq " W W Phi3 and P2.213 f ﬁI'AJr/‘erm a (3/3}? ﬂu.” .' F{(Z7’?)C =0
ZéN{i?~m}fjiﬁ/{EW} ff} (who
4? “ﬂabby
19273; =0
[MEo mama;
4‘27; to
warm/=0
fgmﬁ'in'um a)“, {laall :5 .'
L25 :9
ﬁrst/6 =0 5 75:5
Hfgeo J r
513%»; =0 5; rf/gjﬁzmm
Lj jfwéﬂ
t;— Amo M! g 5. F" fgu.‘//Jn’un cg {EH31 [/95
3 {—
+756 ‘0 ‘ZW'§6 =0 5 = *aéwpzmw 6 5 ' EEﬂXIAJﬁ'fMm 9F {ha/5! x—g Frag} w awﬂgw €=—£m{ﬂ If. C} C Egg/A‘o/H'wa gF {leil! ’1 K
_ r x; c aim «a? 74 ,5 = may (a ﬁg: 2J0 Mm
25; = 3.00 44/ (t) Prob. 1.44. The pinjointed truss in Fig. PIA4 supports a
ioad of P = 2 kips at joint C. Determine the axial forces F3 through F5 in the truss members. fa/a 519/2} ﬂﬂl‘P/M/ﬂc ﬁrK/a/ fé/Lgf 5 IL. I D
a» 0] 5: A #7 M; m 6. Z,00 41/“! {7! g: ﬂé/é 49% {5} g, /,/H ér/«x [75
5, .— /./41;> £03; [7! {91.44 and MEIER éZMr/r'gﬂUM cfz 514/724? Zémrrf
T $27.. so a :0 24% ﬂat/I 'DJM/z‘] =0
Jr, 97 ’ ﬂ /¢’Z? 4?”
0; +FCZ—M/g 2 o a] my; 1%,; {who
5'; T 05577 Apr fk/ﬂXr/‘am a hiya/ﬂ;
5
+727;ch a} # 3:6 =0
2? —M‘Zﬁé AW
925: to 5 $90 6 ‘75 F. = #42,; MA. 55’%/@) Lip/Aim}... 0%j/‘01‘I1 / (’.' 277; :0 5:5; , / /4Zf/n//;ar E650 5’  54/“ ﬁguﬂén‘m 0; {WM 0 : wig.0 IFKﬂ/Z)*0y:0
53 ~ 170; = /.é/62 Mu Prob. 1.440. A unﬂame distributed vertical load 19.] = 3.2
erafm acts over the entire 25 m length of tho Horizontal
mambo: BC of frame ABC, as shown in Fig. PIA—20. The
ﬂame is suppmed by a frictionless pin at C and rests on a
rollcr support at A. (3} Datermine the mactﬁons at A and
C, and (1:) determine the internal resultants (axial force,
shear force, and bending moment) at arbitrary cross sections
of members AB and BC, using the rsspective notations
shown in the two cutaways. ﬁ‘éﬁ’é‘éﬂ.‘
5?} ﬂrécﬁfw; Q/ﬁmc/C’. r KEN“ rum/m [a A'Jnd/fe’fﬂ/f‘éw [If fin we = 3.2. thn 2.3 m PTA20, PEA40, and Pad24 Ezuxﬂ'én’um df [JP/amt *((gm)c=u
(43%“ ﬁéﬂénf/Mrn} =0
ﬂydﬂu Hit“; =0 {1}Jr 6’£N+ (.3 =0
Q,  mfg/u” 0 {f} :0 ﬂ; vim I"r are, 5,: HINT ffur'A'é/‘I'MM car/f man/r9 .' Militia 6'; 3i/gﬂcﬂ}:0 gbzmgu
“73'va %(£1féM)V,«z0 Mama: (K r
Wm
Me ME = 321’ r in; Mm; 3M: 1,4 JEAN/2,91%?) 1‘ mm :5, 2195.2,“ (km
)5 :0 ML “:2 rz.:xl)’{£)—rrJ’/z.f—X=J =0
“ ZN" fzﬁn *Kéxal {’W'M ...
View
Full Document
 Winter '07
 staff

Click to edit the document details