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Design of beam for bending


									Strength of Materials I
    EGCE201 กำลังวัสดุ 1

    Instructor:   ดร.วรรณสิ ริ พันธ์อุไร (อ.ปู)
        ห้องทำงำน: 6391 ภำควิชำวิศวกรรมโยธำ
      โทรศัพท์: 66(0) 2889-2138 ต่อ 6391
Design of beam for bending
Design of beam for bending
Steps in designing a beam for bending
Assume that E,G,sall, and tall
for the material selected are
1.Construct the diagrams
corresponding to the specified                     1.
loading conditions for the beam
and define |V|max and |M|max .
2.Assume that the design of the
beam is controlled by the
normal stress at +,-c in the
section and determine Smin .
3.From available tables, select
beams with S>Smin .                                2.
-small weight per unit length
-small displacement (show in the latter chapter)
              Design Example

                 1. Modeling distributed loads as equivalent concentrated force

                 2. Use the knowledge gained from lecture 4 (either by forming
                 equations or by inspection) to form moment diagram.

                                 1. the shear force is linear, the moment varies
M = -10,125                      parabolically.
                                 2. the area under the shear diagram is – so is
                                 the correspondent moment

                                  A = 0.5(9)(-2250) = -10,125
Design Example (continued)

           3. From 2, we find

           Mmax = 18,000 lb-ft =216,000 lb-in
           Vmax = 3750 lb

     Complete Beam Analysis Example

• For the beam loaded and supported as shown,
  determine the max tensile and compressive
  stresses and where they occur.
1. Determine the location of the centroid.
• The area of the cross section (A) = 18.75 in2

2. Determine the area moment of inertia w.r.t. the centroidal axis (as
shown in figure on the lower left) and use transfer formula to
calculate I of the section.
3. Begin the analysis by defining reactions at A and B using FBD,
one writes the equations of equilibrium.

                                  4. Next, the shear diagram can be
                                  constructed (see previous example
                                  or lecture 4 for more details).
5. Now the moment diagram can be

The moment is the area under the shear
force diagram and the three areas are

A1 = 0.5(3.5ft)(-17.5kip)
   = -30.63 kip-ft
A2 = 0.5(3.5ft)(52.5-35 kip)+(3.5ft)(35 kip)
   = 153.13 kip-ft
A3 = 0.5(7ft)(-35kip)
   = -122.5 kip-ft                                            A1+A2+A3=0

The moment diagram is as shown.                A1=

Positive moment, top beam is in compression
and the bottom is under tension
6. Next, we compute stresses
At x=3.5 ft, moment is negative so
the top is in tension while
the bottom is under compression.

7.At x=7 ft, moment is positive
so the top is under compression
while the bottom is in tension.
Shear Stress in Beams
A cutting plane is passed
through a beam at an arbitrary
spanwise location, the internal
reactions are required for
Equilibrium are a bending
moment and a shear force.

The moment and shear force
as shown are considered positive.

 The shear and normal stresses
 acting on an element of area are
 represented as forces by multiplying
 them by the area (dA)
3 out of 6 equations of equilibrium involve the normal force sxdA      Pure Bending
3 out of 6 equations of equilibrium involve the shearing force txydA, txzdA


  From 1.
          Vertical shearing stresses exist in a transverse section of the beam
          if a shear force exists at that section
Shear stress on a horizontal plane

    Horizontal shear derivation
• Observing that x is
  constant over the
  cross section, the
  expression for H is
  written as
   H    x
     Q is the first area moment w.r.t. to the N.A. of
  that part of the section located above the line y=y1
               Shear Flow, q
Along a horizontal plane a distance y1 above the NA,
    the horizontal shear per unit length of beam

           H PQ
                          But       P V

  The ratio of H/x is termed the shear
  flow and is denoted by “q”
• Determine the shear flow (q) of the
  following cross section
• The cross section is broken into 3 sections and
  the second area moment of inertia

• Given V = 8000 lb, one compute

• The shear flow can now be expressed as

                 q  0558Q
• Compute Q              Q  Ay
  Q is the first area moment w.r.t the N.A.
  A is the area of the cross section above the plane for which q is
  being determined.
   y is the location of that area w.r.t the N.A (+,-).
  Transverse shear stresses in beam
• The horizontal shear flow at C q 
• A shear force exists on a horizontal
  plane passing C
For a narrow rectangular beam

 b < h/4
For an I- beam   web

• A shear force acts on a cross
  section. The cross section
  shown is made by nailing
  planks together.
  1. Use shear flow to define the
  required nail spacing if each
  nail supports 700 lb shear
  before failure.
  2. Compute shear stress at
  various locations in the cross
1. Compute the moment of inertia
2. Compute the nail spacing
3. Compute stress distribution
Displacements in beams

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