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# hw07 - STATICS AND MECHANICS OF MATERIALS 2’“1 Edition...

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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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