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Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO Problem 1 and 2 will be marked out of 3 (each) and problem 3 will be marked out of 4. 1) The this rectangular plate is subjected to the 4 forces shown. Determine the equivalent force-couple system at O. Fcos30.(2b) Is the Resultant R perpendicular to Mo? Fcos30.b 2 1 3 Fsin30 Fcos30 Notes on Solution (what you should spot immediately): This is a 3D problem with simple forces (all forces are easily converted to be parallel to X, Y and Z). You should deal with this using scalar mathematics. Note also that X, Y and Z are oriented differently. This should not change any of our rules and equations. It only requires extra care on your behalf. Typical Solution Steps: 1. Express all forces in X, Y and Z directions. This means resolving the force at the tip (point 3) into F.sin 30 (in Y direction) and F.cos30 (in Z direction). 2. Find Fx, Fy and Fz. These are the components of the resultant force. The value and direction of force do not change when it is moved from one point to another. Force has the same vector representation no matter how they are moved around. 3. To find the equivalent moment, just find the moment of all forces around O. To expedite the calculations, just use scalar mathematics keeping in mind the right hand rule. When you are done with calculating moment values, just put them in i, j, k directions. Mo Mx My Mz F0 0 0 0 F1 0 F (2b) 0 F F2 Fb 0 0 F3 Fcos30 Fcos30 (b) -Fcos30 (2b) 0 Fsin30 0 0 Fsin30 (2b) Equivalent Force: Fx = 0 Fy= Fsin30 = 0.5F Fz= -F + Fcos30 = -0.13397F http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 1 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO Equivalent Couple: Mx = F.b + Fcos30.b = 1.866Fb My= F.2b – Fcos30.2b = 0.26795b Mz= Fsin30.2b = Fb F = 0.5F j – 0.13397F k |F| = 0.51763F M = 1.866Fb i + 0.26795 j + Fb k |M| = 2.134Fb The resultant F and M are perpendicular this should be as clear as the Sun. http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 2 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO 2) The turnbuckle is tightened until the tension in the cable AB is 1.2KN. Calculate the magnitude of the moment about point O of the force acting at point A. r FAB Notes on Solution (what you should spot immediately): The tension in AB is not parallel to any of the planes. This means that this is a true 3D problem. We have to find the force in terms of i, j, k (of course you can find Θx, Θy, Θz and then find the components of the tension and then use scalar mathematics. However, this may take longer time and, in some cases, could be confusing). Steps of Solution 1. Express relevant points in 3D. In this case, A, B and of course O. 2. Express relevant directions in 3D. In this case we will need OA and AB. 3. Use the direction to calculate the force. You need to find the length of the force vector (in this case, AB). Express the force using the λ formula. Note: you can use the components of force here and calculate moments in scalar mathematics. However, I recommend using determinants here. 4. Find M=rxF using determinants. 5. To find the final magnitude of the moment use the equation M o (Mx)2 (My)2 (Mz)2 A = (1.6, 0, 2)m B = (2.4, 1.5, 0) m OA= r = (1.6, 0, 2) m AB = (0.8, 1.5, -2) m |AB| = 2.624m AB λAB = | AB | T AB = TAB . λAB = (0.36573kN) i + (0.68574kN) j – (0.91433kN) k http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 3 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO i j k M o r TAB 1.6 0 2 (-1.3715kN.m) i + (2.1944kN.m) j – 0.36573 0.68574 0.91433 (1.0972kN.m) k M o (1.3715) 2 (2.1944) 2 (1.0972) 2 2.8107kN.m http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 4 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO 3) The boom AB lies in the vertical y-z plane and is supported by a ball-and-socket joint at B and by rA/B the two cables at A. Calculate the tension in each cable resulting T1 from the 20 kN force acting in the horizontal plane and applied at F the midpoint M of the boom. T2 Neglect the weight of the boom. rA/M (Hint: A ball-and-socket joint provides reaction forces in all Rx three directions, but no moment.) Ry Rz Notes on Solution (what you should spot immediately): This is a full 3D problem. You are better off using the λ formula. The support at B is a 3D pin-support. You have five unknowns: T1, T2, Rx, Ry, Rz In 3D, we always have six equations: ∑Fx=0, ∑Fy=0, ∑Fz=0, ∑Mx=0, ∑My=0, ∑Mz=0 The Good: The best point to calculate moment around is B as this will eliminate three unknowns from the equation. The bad: The worst point to take moment around is M. This will not solve any thing as you are eliminating the only given quantity. The Ugly: Taking the moment around A, will force you to write and solve five equations. Steps of solution 1. Express relevant points in 3D. In this case, A, B, C, D, and M. 2. Express relevant directions in 3D. In this case we will need AD and AC. 3. Use the direction to calculate the force. You need to find the length of the force vector (in this case, AD and AC). Express the force using the λ formula. 4. Find MB=rxF using determinants. This will give you three equations (∑Mx=0, ∑My=0, ∑Mz=0) in two unknowns (T1, T2). Solution: FBD is superimposed in the initial figure. A (0,3,10) B (0,0,0) C (4,2,0) D ( 4,2,0) M (0,1.5,5) AD (4m)i (5m) j (10m)k AC (4m)i (5m) j (10m)k AD 11.874m AC 11.874m http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 5 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO M B rA / B T1 rA / B T2 rM / B F 0 rA / B 3 j 10k AD T 1 0.33687 i 0.42109 j 0.84218 k AD AD T1 (T1 )T 1 (T1 ) 0.33687 (T1 )i 0.42109 (T1 ) j 0.84218 (T1 )k AD i j k rA / B T1 0 3 10 1.6844(T1 )i 3.3687(T1 ) j 1.0106(T1 )k 0.33687(T1 ) 0.42109(T1 ) 0.84218(T1 ) AC T 2 0.33687 i 0.42109 j 0.84218 k AC AC T2 (T2 )T 2 (T2 ) 0.33687 (T2 )i 0.42109 (T2 ) j 0.84218 (T2 )k AC i j k rA / B T2 0 3 10 1.6844(T2 )i 3.3687(T2 ) j 1.0106(T2 )k 0.33687(T2 ) 0.42109(T2 ) 0.84218(T2 ) rM / B 1.5 j 5k F ( 20 kN )(sin 20 )i ( 20 kN )(cos 20 ) j (6.8404 kN )i (18 .794 kN ) j i j k rM / B F 0 1.5 5 93.97i 34.202 j 10.261k 0 6.8404 18.794 M B rA / B T1 rA / B T2 rM / B F (1.6844(T1 T2 ) 93.97)i ( 3.3687(T1 T2 ) 34.202) j (1.0106(T1 T2 ) 10.261)k 0 1.6844(T1 T2 ) 93.97 0 then we have: 3.3687(T1 T2 ) 34.202 0 1.0106(T T ) 10.261 0 1 2 Solve the first two equations, we have T1 32.971kN T2 22.818kN http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 6 of 7 Civ101 TERM QUIZ SOLUTION Assignments CIV101- STRUCTURES, MATERIALS AND DESIGN, SECTION J (9), 2004-2005 DEPT. OF CIVIL ENGINEERING, UNIVERSITY OF TORONTO Then we can calculate three reaction forces of the ball-and-socket joint by the summating the forces in three axes to zero. This step is NOT required, but is helpful to understand the forces in space. AD T1 (T1 )T 1 (T1 ) 0.33687 (T1 )i 0.42109 (T1 ) j 0.84218 (T1 )k AD AC T2 (T2 )T 2 (T2 ) 0.33687 (T2 )i 0.42109 (T2 ) j 0.84218 (T2 )k AC F (6.8404 kN )i (18 .794 kN ) j RX RX i RY RY j RZ RZ k F X 0 0.33687T1 0.33687T2 6.8404 RX 0 RX 3.4202kN F Y 0 0.42109T1 0.42109T2 18.794 RY 0 RY 4.6982kN F Z 0 0.84218T1 0.84218T2 RZ 0 RZ 46.984kN http://ccnet.utoronto.ca/20049/civ101h1fsection9/ Page 7 of 7