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Unformatted text preview: STATICS AND MECHANICS OF MATERIALS, 2’“1 Edition RILEY, STURGES AND MORRIS 6—44 Determine all forces acting on memberABE of the frame of Fig. PIS44.
SOLUTION First, from a freebody diagram ofthe complete frame, the equilibrium equations give the support forces
OEMA : 0: D(3oo)e150(300) : 0
D : 150 N
92F, =0: A,+150=0
A, : ,150 N
T 2F, = 0 : A, +1) = 0 15o” =—150N Next, from a freebody diagan] ofrnember BCD,
the equilibrium equations give the pin forces Omfo; 150(300)+Cy(100):0 F” c _,
g . It Cy=—450N W
DEA/10:0: 150(200)—B,(100) o efl c3 "ﬂ” By : 300 N
Finally, from a freebody diagram of member ARE, the equilibrium equations g've the pin forces
02M, : o: 300(100)+150(1oo)7150(200)731(100): o
B, = 150 N
—>EFI :0 : Exe150e150:0
E, : 300 N
Ey—300—150= o
E, : 450 N Therefore, the forces acting on member ABE are: A: 212 N F 45°
B = 335 N P 63.43”
E=541N d 56.310 647 Forces of 25 lb are applied to the handles of the pipe pliers shown in Fig. P647. Determine the force exerted
on the pipe at D and the force exerted on handle DAB by the pin at A. SOLUTION
From a freebody diagram of handle DAB, the equilibrium equations g've the forces 0221450: FD(1.25)—25(9)=0 ;
FD=180 lb , , ,. » tit35‘
FDsiu38°eA, : 0
AK :110.8191b
TZF}:O: AyezseFDoos380:0
A, =166.842 lb
A=2001b L 56.410 The force exerted on the pipe at D is equal and opposite to the force exerted on the handle
FD =1801b 3 52° 6—83 A 4000—lb crate is attached by light, inextensible cables to the truss of Fig. P6S3. Determine the force in
each member of the truss using the method of joints. SOLUTION
From a freebody diagram of the entire truss, the equilibrium equations give
omfo: 6E—4(2000)—8(2000):0 T"?
4—— —)EF7t = 0 : E — A1 : 0 Ax.
TEFy20: Ay—4000=0 a. E : 4000 lb A), = 4000 ll) 5100315 5
From a ﬂeebody diagram for joint A, the equilibrium equations give
a2}; = 0 : (4/5) TAB 7 4000 = 0 T 4000
TEFy20: 4000—TAE—(3/5)TAB=0 40w 0
I 0"
TAB = 5000 lb (T) ................................................................. Ans. _ ; 1' \ n+5
T : 1000 lb (T) ................................................................. Ans. '
AE . . . . . . . . Tie From a freebody diagram for Jomt E, the equlllbnum equations glve
—>EFI : 0: TM. +(4/5)T3E +4000 : 0 ._ .TgT/Whe
TZFy:0: TH +(3/5)TBE:0 ____* ___., 1;” = —1666.67 1b E 1667 lb (C) ......................................... Ans. ‘“’°° 73:9 205 STATICS AND MECHANICS OF MATERIALS, 2““ Edition RILEY, STURGES AND MORRIS TDE : 72666.67 lb E 2670 lb (C) ........................................ Ans. 1‘D
From a ﬂeebody diagram for joint D, the equilibrium equations give  T
—>EF;=O: T 7T =0 *"—' a
CD DE "iiDE" 121)
T2117, :0: TED—200020
.1000
TC), : —2666.67 1b 2 2670 lb (C) ........................................... Ans.
TB), : 2000 lb (T) ....................................................................... Ans Tue
Finally, from a freebody diagram for joint C, the equilibrium equations give ' \
‘— 0
T214}:0: (3/5)];C—2000:0 17.6 l
TBC = 3333.33 lb = 3330 lb (T) .............................................. Ans. ' ‘ 6—84 Use the method of joints and determine the force in each member of the truss shown in Fig. P684. All
members are 3 m long. SOLUTION
From a ﬂeebody diagram of the enﬁre truss, the equilibrium equations give exago: a4,:0 3b” UZMF:0: 3Ay—3(3):0 Ay:3kN T211320: F—Ay—5—320 lelkN From a ﬂeebody diagram for joint D, the equilibrium equations give T211320: —3—Tmsin60°:0 aEﬂ=Oz —TCD—TDECOS6OO=O
rm 2—3.46410 kN z 3.46 kN (C) ........................................ Ans.
rm = 1.73205 kN ; 1.732 kN (T) .......................................... Ans. From a ﬂeebody diagram for joint C, the equilibrium equations give
aEF; = 0: Tan +TCE coséOO—Ijﬂc 605600 = 0 T214; :0: —TCEsin60°—Tgcsin60°—5:0
TBC : 71.15470 kN a 1.155 kN (C) Ans
TC, : —4.61880 kN : 4.62 kN (C) ........................................ Ans. From a ﬂeebody diagram for joint A,
the equilibrium equations give T211320: TABsin6OO—320
aEF;=O: TAF+TAEc0360°=O
TA, 23.46410 kN z 3.46 kN (T) ........................................... Ans. rm. = 71.73205 kN E 1.732 kN (C) Ans From a freebody diagram for joint 3,
the equilibrium equations give T227, : 0: THC ammo—7:3,. sin600—TAB sin60° : 0 206 STATICS AND MECHANICS OF MATERIALS, 2mi Edition RILEY, STURGES AND MORRIS 0m : tan" (10/12) : 39.8060
60 : tan" (11/4) : 59.0360
h = (2/3)(10) = 6.66667 m Then= the equilibrium equations give OEMA : 0: —3(4)—4(8)—(Twsin6w)(12) : 0 TC, 2—41.28 kN : 4.28 kN (C) .................................................................................... Ans.
OEMC = 0: 3(4)79.33333(8)+TJK (6.66667) = 0 TJK : 9.40 kN (T) ........................................................................................................... Ans. 6—88 The Gambrel 111155 shown in Fig. P688 supports one side of a bridge; an identical truss supports the other
side. Floor beams carry vehicle loads to the truss joints. Calculate the forces in members BC, 3G, and DE
when a truck having a mass of 3500 kg is stopped in the middle of the bridge as shown. The center of gravity
of the truck is l m in front of the rear wheels. SOLUTION
W : 3500(9.81) : 34,335 N From a ﬂeebody diagram of the entire bridge (which is consists of two trusses), the equilibrium equations give —>EFI : 0 : 24.r : 0
UZME20: 34,335(4)—(2Ay)(10):0
0210150: (2E)(10)7(34,335)(6)=0
,4y : 6867.0 N E210,300.5 kN STATICS AND MECHANICS OF MATERIALS, 2’"1 Edition RILEY, STURGES AND MORRIS Half of the truck’s weight is supported by each of the two trusses. From a
ﬁ'eebody diagram of the truck. the equilibrium equations g've UEMR=0: 34,33S(1)—(2NF)(4)=0
OZMF :0: (2NR)(4)—34,335(3):0
NF :429188 N NR =12,875.6 N From a ﬂeebody diagram of the ﬂoor panel between joints H and G,
the equilibrium equations give UEMG:0: NF(2)—3H:O H:2861.25 N Next. cut a section through EC, BG, and GH, and draw
a ﬂeebody diagram of the leﬂhand side of the truss. 9: tan’1(2/3) 233.6900 ”Tag
q..v)=tan—i (3/3): 450
Then, the equilibrium equations give “A _ I .. T“
UEMG:0: 2861.25(3)—6867(5)—(TBcc039)(5):0 . .2 .
1.
TM :—6190 N:6.19 kN(C) ...................................... Ans. T l ‘H
n d
Tray :0: 6867g2361.25+rgc 51116)ng sin¢=0 6351 ﬁéﬂ—E'
rm : 309.30 N 2 0.809 kN (T) ....................................... Ans. Tue
Finally, ﬁ'oma freebody diagram for joimE, a
the equilibrium equations give ‘
Tmzo: 10,300.5+TDEcost9:0 Té:"“@ ' "
TDE :—12,380 N: 12.38 kN (C) Ans  T
to, 300.53 6—89 A truss is loaded and supported as shown in Fig. P689. Determine
(a) The normal stress in member CD if it has a diameter of V2 in. (b) The change in length of member CF if it. has a diameter of 1/2 in. and a modulus of elasticity of 29(106) psi.
(c) The change in length of member EF if it has the same diameter and modulus of elasticity as member CF.
SOLUTION From a ﬂeebody diagram of the entire truss. the equilibrium
equations give OEMA =0: F(18)71000(6)72000(12)
—2000(18)—1000(24):0
F:50001b Cut a section through CD. CF. and FG. and draw a ﬂeebody diagram
of the righthand side of the truss. The equilibrium equations give 6—91 The truss shown in Fig. P691 supports a sign that weighs 3000 lb. The sign is connected to the truss at joints
E, G, and H, and the connecting links are adjusted so that each joint carries onethird of the load. All
members of the truss are made of structural steel, and each has a crosssectional area of 0.564 inz. Determine
the axial stresses and strains in members CD, CF, CG, and F G of the truss. SOLUTION From a freebody diagram of the entire truss, the equilibrium
equations give 021M, : 0: r, (48)—1000(8)—1000(24)
—1000(40): 0 210 6—97 Snow on a roof supported by the Howe truss of Fig. P697 can be approximated as a distributed load of 20
lbfft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of
the member. Determine the forces in members BC, BG, CG, and GH. SOLUTION From a ﬂeebody diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end
of the roof panel as F, = F, = W/2 = (20)(Jﬁ)/2 = 89.4427 1b Then, from a freebody diagam of the entire truss,
the equilibrium equations give —>EF,:0: 34,:0
T2134: Ay+E—8F1:0
om,=0: E(32)7(2Fl)(8)7(2F)(16)
—(2m(24)—(m(32)=o
A, :E:3F1 : 357.7711b Next, cut a section through members BC, BG, and GH, and draw a freebody
diagram of the leﬁhand portion of the truss. a =tan'18/16 = 26.5650 Then equilibrium equations give ,_. 02M, = 0: 7(2Fl)(8)i(TBG sin¢)(16) = 0 77m = 7200 lb = 200 1b (C) .......................................................................................... Ans.
02MB = 0: TGH (4) +5; (8)7 (357.771)(8) = 0 TGH = 536.656 1b a 537 lb (T) .................................................................................... Ans. 216 STATICS AND MECHANICS OF MATERIALS, 2“‘1 Edition RILEY, STURGES AND MORRIS OEMG = o: (2E)(8)+Fl (16)7(357.771)(16)7(Tmsin¢)(16) = 0 TBC : —400 lb : 400 1b (C) .......................................................................................... Ans.
Finally, from a freebody diagram for joint C, the equilibrium equations give 1 JP.
$21220: Tacosgé—Ilgccosqa520 é
Tram: —2F{—TCG—TCDsm¢—Tgcsm¢20 ; i 72
T3,; 5
TCD = —400 lb
Tea Tea 2178.9 lb (T) ................................................................ Ans. ...
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 Fall '08
 Brody

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