Transcript Design of beam for bending

Strength of Materials I
EGCE201 กำลังวัสดุ 1
Instructor:
ดร.วรรณสิ ริ พันธ์อุไร (อ.ปู)
ห้องทำงำน: 6391 ภำควิชำวิศวกรรมโยธำ
E-mail: [email protected]
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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 known
1.Construct the diagrams
corresponding to the specified
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 .
Considerations:
-small weight per unit length
-small displacement (show in the latter chapter)
1.
2.
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.
M = -10,125
1. the shear force is linear, the moment varies
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
4.
Complete Beam Analysis Example
• For the beam loaded and supported as shown,
determine the max tensile and compressive
stresses and where they occur.
Given
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
constructed.
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
The moment diagram is as shown.
A1+A2=
A1+A2+A3=0
A1=
Recall
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.
Next
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)
Pure Bending
3 out of 6 equations of equilibrium involve the normal force sxdA
3 out of 6 equations of equilibrium involve the shearing force txydA, txzdA
1.
2.
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
H
Horizontal shear derivation
• Observing that x is
constant over the
cross section, the
expression for H is
written as
PQ
H
x
I
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

x
I
But
P V
where
The ratio of H/x is termed the shear
flow and is denoted by “q”
VQ
q
I
Example
• Determine the shear flow (q) of the
following cross section
Solution
• 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  0.558Q
• 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
VQ
• The horizontal shear flow at C q 
I
• A shear force exists on a horizontal
plane passing C
For a narrow rectangular beam
b < h/4
For an I- beam
web
flange
Example
• 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
section.
1. Compute the moment of inertia
2. Compute the nail spacing
3. Compute stress distribution
Displacements in beams
Next week