hw4 - 1‘ l l l l’l‘oblem 6.5 (a) Let the dimension is...

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Unformatted text preview: 1‘ l l l l’l‘oblem 6.5 (a) Let the dimension is z 0.1 m. Deter- mine the axial forces in the members, and show that in this case this truss is equivalent to the one in Problem 6.4. (b) Let the dimension h = 0.5 m. Determine the axial forces in the members. Compare the results to (a), and observe the dramatic effect of this simple Change in de- sign on the maximum tensile and compressive forces to which the members are subjected. Solution: To get the force components we use equations'of the form TPQ = TPQGPQ : Tpri + Tpgyj where P and Q take on the designations A, B, C, and D as needed. Equilibrium yields At joint A: Eli‘s : TABX + TACX = 0, andZFy :TABY+TACY—1kN=0. — ‘TABX + TBCX + TBDX = 0. ’TABY + TBCY + TBDY = 0- —TBCX — TACX + TCDX : 0» —TBCY - TACY + TCDY + CY 3 0- ‘TCDX * TBDX + Dx = 0, = ’TCDY - TBDY + DY = 0- Solve simultaneously to get TAB = TED : 2.43m, = —2.78 kN, = 0, TOD = —2.88 kN. Note that with appropriate changes in the designation of points, the forces here are the same as those in Problem 6.4. This can be explained by noting from the unit vectors that AB and BC are parallel. Also note that in this configuration, B C carries no load. This geometry is the same as in Problem 6.4 except for the joint at B and member BC which carries no load. Remember member BC in this geometryiwe wiil encounter things like it again, will give it a special name, and will learn to recognize it on sight. (b) For this part of the problem, we set h. r 0.5 In. The unit vectors change because h is involved in the coordinates of point B. The new unit vectors are eAB = —O.986i+ 0.164j, BAG E —0.864i — 0.504j, 630 : 0i * lj. 9313 = -D.768i — 0.64%, and egg 3 —0.8321 + 0.555} We get the force components as above, and the equilibrium forces at the joints remain the same. Solving the equilibrium equations simultaneously for this situation yields ' TAB : 1.35 kN, TAG = —1.54 kN, THC = —1.33, TED = 1.74 kN, and TOD = mine kN. These numbers differ significantly from (a). Most significantly, member BB is new carrying a compressive load and this has reduced the loads in all members except member BD. “Sharing the load" among more members seems to have worked in this case. Problem 6.2.2 The Warren truss supporting the walk- way is designed to support vertical 50-kN loads at B, D, F, and H. If the truss is subjected to these loads, what are the resulting axial forces in members BC, CD, and CE? Solution: Assume vertical loads at A and I Find the external loads at A and I, then use the method of goints to work through the structure to the members needed. Sin-IT x [Y Ay+Iy“4{50)=0(kN} AB 2—180.3kN “350 ~95O —1550 —2}. 50 +241 :0 ()()()<) y 9:33.69, Solving A1, = 100 kN 1,, : IDOkN BCcosH+BD—ABCOSH =0 —50—ABsin9—BCsin0:0 Solving, BC : 90.1 kN (T) B1) = —225 lcN (0} a m 33.69“ A0 = 150 kN (T) AB 0036’ +AC = 0 BC 2 90.1 kN (T) ABSin0+Ay =0 ZFw: CE—AC-l—C'DcosfliBCcosBZO Solving, AB =—180.3kN C ( ) 21:3,: CDsin€+BCsinfl=0 A0 = 150 kN (T) 50“ng CE = 300 kN (T) y' CD = -—90.1kN(C) Hence BC 2 90.1 N (T) CD : —90.1kN(C) CE : 300 kN (T) Problem 6.31 For the truss in Problem 6.30, use the method of sections to determine the axial forces in mem- bers BO, CF, and FG; " I 501mm“: , a Solving BC = 300 kN (T) .. . CF : 7141.4 kN (C') :; FG z —200 kN (C) —BO—CFcos45—FG:0 g —C’F 511145“ — 100 : o ' —(1)FG’ — 2(100) = o E g n, Problem 6.32 Use the method of sections to determine the axial forces in members AB, BC, and CE. Solution: First, determine the forces at the supports 1 B D Ay+Gy—3F=u —1{F) _ 2(2F) + 39y : 0 Solving Aw : 0 Gy = 1‘67F Ay z 1.33F Method of Sections: CE+AB=0 o +ZMB: (—1)Ay + (1)03 = 0 Solving, we get AB : 71.331? (0) CE = 1.33F (T) BC = —0.33F (0) J .l .'l i .1 Problem 6.45 Use the method of sections to determine the axial force in member EF. Solution: The included angle at the apex BAG is 12 a =tan‘1 (—) c 36.37“. 16 The interior angles BCA. DEG, FGE, HIG are!)’ = 90° — o: = 5313". The length of the member ED is LED : 8 tan a x 6 ft. The interior angle DEF is x3 =tan*1(~4—)= 33.69“. LED The complete structure as a free body: The moment about H is M H = —1o(12) — 10(16) + 1(12) 2 o, from which I z % 2 23.33 kip. The sum of forces: XE?! : Hy + I : 0) from which Hy z —I : w23.33 kip. EFT, :H..+20:o, from which H}, = —2() Rip. Make the cut through EC, EF, and DE. Consider the upper section only. Denote the axial force in a memberjoiningI I, K by IK. The section as afree body: The sum of the moments about E is ME = —10(4) — 10(8) + DF(LED} : D, from which DF : % : 20 kip. The sum of forces: EFL. : —EFsin,6-—EGsin'y-—DF=0, ZR“ = —EFcosfi+EGcosry+2U: D, from which the two simultaneous equations: 0.5547EF + 0.8EG : “20, and 0.8320EF — 0.6EG = 20. Solve: EF = 4.0 kip (T) 10kip_ $0.10? -l i ‘5 ! Problem 6.69 The mass m : 12 kg. Determine the forces on member ODE. Solution: The equations of equiiibrium for the entire truss are: ZFX :AX+CX =0, ZFY sz —mg=0, and 2 MA = 0.40): — 0.7mg : 0. From these equations we get Ax : 7206.0 N, AY = 117.7 N, and 0:1: 2 206.0 N. Member ABD: The equations are ZFX :AX+BX+DX+T=0, ZFY =Ay+By+Dy=0, and EMA = 0.2131, + 0.2BX + 0.2Dy + DADX + 0.1T = 0. Member CDE: The equations are ZFX :meDX +EX =0. ZFY = "DY +EY = 0, and 2 MD = 0.4133, — 0.2EX : 0. Member BE: The equations are “BX+PX”‘EX =0, “BY+PY;'EY:0: and 2 ME = 0.43,, = n. The Pulley: The equations are V 7T i PX z 0. The Weight: The equation is 2 FY = T — mg : 0. Soiving the equations simultaneously. we get BX = 117.7 N, BY = G,DX = -—29.4 N, DY : —117.7 N, Ex=w117 N,Ey=0 ,T=117.7N, PX = —117.7N, Py : 7117.7N Problem 6.80 The weight W : 60 kip. What is the magnitude of the force the members exert on each other at D? Solution: Assume that a tong half will carry half the weight, and denote the vertical reaction to the weight at A by R. The complete structure as a free body: The sum of the forces: 217g =R—W=0, fromwhichflzw Tong-Hay’ACD: Elemem AC: The sum of the moments about A: (1) (2) (3) (4) (5) ZMA=30y+3Cszo The sum of the forces: R 25;, = §+Cy+Ay :0,and ZFw=cm+Amzo Element CD: The sum of the forces: 2F :DmiP703:O,and W ZEJzoy—Oy—Em20‘ The sum of the moments: 3 2MB :20: ~30y —3P+§W20 Element AB: The sum of the forces: ZFy=—Ay+%—By:0,and ZF$=—A$—Bm :0. Element BD: The sum of the forces: W ZFy=By—Dy—3:O,and ZFm=Bw—DE+P:U. These are ten equations in ten unknowns. These have the soiutiou R : 60 kip. Cheek, A1: = —-30 kip, Ag 2 0, Bx = 30 kip, By = 30 kip, CI = 30 kip, Cy : 73D kip, m r 110 kip. Dy r 0, and P = 80 kip. The magnitude of the force the members exert on each other at D is D = 110 kip. L E 4 i i ‘ Problem 6.81 Figure 3 is a diagram of the bones and biceps muscle of a person’s arm supporting a mass. Ten- sion in the biceps muscle holds the forearm in the hor- izontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N, and the mass m = 2 kg. (a) Determine the tension in the biceps muscle AB. (1:) Determine the magnitude of the force exerted on the upper arm by the forearm at the elbow joint 0. Solution: Make a cut through AB and BC just above the elbow joint C. The angle formed by the biceps muscie with respect to the forearm is on = tan—1 (%) : 802°. The weight of the mass is W : 2(9‘81}: 19.62 N. The section as a free body: The sum of the moments about 0 is 2 Mo = ~50Tsin a + 150(9) + 350W : o, from which T 2 166.76 N is the tension exerted by the biceps muscle AB. The sum of the forces on the section is ZFX =C$+TCGSD£:O, from which CE = —28.33 N. 23F}; :Cy+Tsincv—9—W=G, from which Cy 2 —135.72. The magnitude of the force exerted by the forearm on the upper arm at joint 0 is F:1/C§+CE=138.65N Problem 6.90 For the bolt cutters in Problem 6.89, determine the magnitude of the force the members exert on each other at the pin connection B and the axial force in the two—force member OD. Solution: From the solution to 6.107, we know BX : —695 N, and By : —435 N. We also know that 0;; : 695 N, and Cy : 535 N, from which the axial load in member CD can be caiculated. The load in CD is given by Top : 1/0} + 03, : 877N Problem 6.91 The device is designed to exert a large force on the horizontal bar at A for a stamping operation. If the hydraulic cylinder DE exerts an axial force of 800 N and or : 80°, what horizontal force is exerted on the horizontal bar at A? Solution: Define the ac-y coordinate system with origin at C. The projection of the point D on the coordinate system is Ry = 2505ina : 246.2 mm, and R3 = 250 (:0st: = 43.4 mm. The angie formed by member DE with the positive :1: axis is 9 = 180 — tan—1 = 145.38°. The components of the force produced by DE are Fm = Fcosfi = —658.3 N, and F‘ = F sinG : 454.5 N. The angie of the element AB with the positive 3: axis is ,6 = 180 —— 90 — a- : 10°, and the components of the force for this member are Pm = P cos ,6 and P.” = P sin ,9, where P is to be determined. The angle of the arm BC with the {:05“ itive :1: axis is 'y = 90 + a 2 170°. The projection of point B is Lm = 250 6057 : 7246.2 mm, and Ly = 2503in‘y = 43.4 mm. Sum the moments about 0: 2 MC : RmFy 7 Rye. + mpg — Lyra... = 0. Substitute and solve: P : 2126.36 N, and Pa: = P c0553 3 2094N is the horizontal force exerted at A. Problem 6.104 Determine the force exerted on the bolt by the bolt cutters and the magnitude of the force the members exert on each other at the pin connection A. Solution: Element AB: The moment about A is EMA : 71013 w 54F m o, where F 2 90 N. From which B r 7486 N. The sum of the forces: Zr}, =A+B—F=U, from which A = 576 N Element BC: The moment about U: 2M0 = —lfiB—8Fc :0, from which the cutting force is For : 972 N Problem 6.105 The 600-113 weight of the scoop acts at a point 1 ft 6 in. to the right of the vertical line CE. The line ADE is horizontal. The hydrauiic actuator AB can be treated as a two-force member. Determine the axial force in the hydraulic actuator AB and the forces exerted on the scoop at C and E. SOIHfiOl‘l: The free body diagrams are shown at the right. Place the coordinate origin at A with the m axis horizontai. The coordinates (in ft) of the points necessary to write the needed unit vectors are A (0, 0). B (6, 2), C {8.5, L5), and D (5, G). The unit vectors needed for this problem are my, = w0.9491 — 0.3163, uBc = 0.981i — 0.196j, and LIED : 70.447i H 0.894j. The scoop: The equilibrium equations for the seoop are ZFX = —TCB“BCX + Ex 3 0, 2 FY = -TCBUBGY + Ev — 500 : 0, and 2M0 :1.5EX w 1.5(6001b) = 0. Solving, we get EX = 6001b, Ey z 4801b, and T03 : 611.9 lb. Joint B: The equilibrium equations for the scoop are 2 F): = TBAuBAX + TBDfl-BDX + TGBUBCX : 0, and 2 FY S TBABBAY +TBD'uBDY + TCBUBCY 3 0. Solving, we get TBA = 835 lb, and TED : W429 lb. ._ lft6" 600 lb ...
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hw4 - 1‘ l l l l’l‘oblem 6.5 (a) Let the dimension is...

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