Quiz_solution_Final by stariya

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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

								
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