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2.001 - MECHANICS AND MATERIALS I
                                Lecture #5
                                9/20/2006
                          Prof. Carol Livermore

TOPIC: FORCES AND MOMENTS TRANSMITTED BY SLENDER MEM-
BERS?
When will a structure fail?

Recall:
    1. What can the structural elements tolerate?
    2. What forces and moments are experienced in the various parts of the
structure?

EXAMPLE: Airplane




EXAMPLE: Truss




SLENDER MEMBERS
  Long, skinny structural elements




                                     1
L >> b, L >> h

  EXAMPLE: Skis, Golf Clubs, I-Beams




Local axes follow the beam.




AXIAL FORCES:




Forces that act along the axis of the slender member.

FBD




                    Fij (x) = N (x), i: face, j: direction, x: location

    SHEAR FORCES:
      Forces that act in the plane of a face.
Summarize Internal Loads:




   Moments:




                     Mxz (x2 ) > 0x: face, z: axis, x2 : location
                                 Mxz (x) = Mz (x)
   Bending Moment


   Sign Convention




Summarize sign conventions




                                           3
N >0
N <0




V >0




V <0




M >0




M <0




       4
EXAMPLE: Calculating internal forces and moments




Step 1: Find reactions at supports.




                                        Fx = 0
                                      R Ax = 0


                                        Fy = 0
                                R Ay − P = 0
                                  R Ay = P

                                        MA = 0
                                MA − P a = 0
                                  MA = P a




                                         5
Step 2: If you have a multi-component structure, you will need to find joint
forces too.




Step 3: Find internal forces and moments.
   1. Set up local coordinates.




   2. Identify interesting points.




   3. Cut beam and do FBDs.




                                     6
For 0 < x < a


                                    Fx = 0
                                N =0

                                    Fy = 0
                              P − Vy = 0
                                P = Vy

                                 MA = 0
                          P a + Mz − P x = 0

                      Mz = P x − P a = −P (a − x)
OR (From the other end)

FBD




                                    Fx = 0
                               −N = 0
                                N =0

                                    Fy = 0
                             −P + Vy = 0
                                Vy = P

                                 M∗ = 0
                          −Mz − P (a − x) = 0




                                7
Mz = −P (a − x)
Note: This is the same as before.
For a < x < L




                                        Fx = 0
                                    −N = 0
                                     N =0

                                        Fy = 0
                                    Vy = 0


                                        M∗ = 0
                                    −Mz = 0
                                    Mz = 0
Step 4: Plot result and sanity check.




                                         8

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Lec5 MECH ENG STRucture

  • 1. 2.001 - MECHANICS AND MATERIALS I Lecture #5 9/20/2006 Prof. Carol Livermore TOPIC: FORCES AND MOMENTS TRANSMITTED BY SLENDER MEM- BERS? When will a structure fail? Recall: 1. What can the structural elements tolerate? 2. What forces and moments are experienced in the various parts of the structure? EXAMPLE: Airplane EXAMPLE: Truss SLENDER MEMBERS Long, skinny structural elements 1
  • 2. L >> b, L >> h EXAMPLE: Skis, Golf Clubs, I-Beams Local axes follow the beam. AXIAL FORCES: Forces that act along the axis of the slender member. FBD Fij (x) = N (x), i: face, j: direction, x: location SHEAR FORCES: Forces that act in the plane of a face.
  • 3. Summarize Internal Loads: Moments: Mxz (x2 ) > 0x: face, z: axis, x2 : location Mxz (x) = Mz (x) Bending Moment Sign Convention Summarize sign conventions 3
  • 4. N >0 N <0 V >0 V <0 M >0 M <0 4
  • 5. EXAMPLE: Calculating internal forces and moments Step 1: Find reactions at supports. Fx = 0 R Ax = 0 Fy = 0 R Ay − P = 0 R Ay = P MA = 0 MA − P a = 0 MA = P a 5
  • 6. Step 2: If you have a multi-component structure, you will need to find joint forces too. Step 3: Find internal forces and moments. 1. Set up local coordinates. 2. Identify interesting points. 3. Cut beam and do FBDs. 6
  • 7. For 0 < x < a Fx = 0 N =0 Fy = 0 P − Vy = 0 P = Vy MA = 0 P a + Mz − P x = 0 Mz = P x − P a = −P (a − x) OR (From the other end) FBD Fx = 0 −N = 0 N =0 Fy = 0 −P + Vy = 0 Vy = P M∗ = 0 −Mz − P (a − x) = 0 7
  • 8. Mz = −P (a − x) Note: This is the same as before. For a < x < L Fx = 0 −N = 0 N =0 Fy = 0 Vy = 0 M∗ = 0 −Mz = 0 Mz = 0 Step 4: Plot result and sanity check. 8