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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. PIS-44.
SOLUTION First, from a free-body 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 free-body 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 free-body 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. P6-47. Determine the force exerted
on the pipe at D and the force exerted on handle DAB by the pin at A. SOLUTION
From a free-body diagram of handle DAB, the equilibrium equations g've the forces 0221450: FD(1.25)—25(9)=0 ;
FD=180 lb , , ,. » tit-35‘
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. P6-S3. Determine the force in
each member of the truss using the method of joints. SOLUTION
From a free-body 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 flee-body 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 free-body 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 flee-body diagram for joint D, the equilibrium equations give - T
—>EF;=O: T 7T =0 *"—' a
CD DE "ii-DE" 121)
T2117, :0: TED—200020
.1000
TC), : —2666.67 1b 2 2670 lb (C) ........................................... Ans.
TB), : 2000 lb (T) ....................................................................... Ans Tue
Finally, from a free-body 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. P6-84. All
members are 3 m long. SOLUTION
From a flee-body 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 flee-body 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 flee-body 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 flee-body 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 free-body 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. P6-88 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 flee-body 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'ee-body 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 flee-body 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 flee-body diagram of the lefl-hand 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 6-351- fiéfl—E'
rm : 309.30 N 2 0.809 kN (T) ....................................... Ans. Tue
Finally, fi'oma free-body 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. P6-89. 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 flee-body 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 flee-body diagram
of the right-hand side of the truss. The equilibrium equations give 6—91 The truss shown in Fig. P6-91 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 one-third of the load. All
members of the truss are made of structural steel, and each has a cross-sectional 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 free-body 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. P6-97 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 flee-body 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 free-body 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 free-body
diagram of the lefi-hand 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 free-body 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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