Section 15: Introduction to
Stress and Bending
15-1
Bending
• Long bones: beams
• Compressive stress:
inner p
portion
• Tensile stress: outer
portion
p
• Max stresses near the
edges, less near the
neutral axis
T
C
i
axis
y
σx=(M
(Mb·y)/I
)/I
15-2
From: Noffal
axis
Bending Moments
• Shear stresses max
at neutral axis and
zero at the surface
• τ = (Q·V)/(I ·b)
• Q= area moment
• V= vertical shear
force
Q
y
h
b
15-3
From: Noffal
Behavior of Bone Under Bending
• Bending subjects bone to a combination of
t
tension
i and
d compression
i (t
(tension
i on one
side of neutral axis, compression on the other
side, and no stress or strain along the neutral
axis)
• Magnitude of stresses is proportional to the
distance from the neutral axis (see figure)
• Long bone subject to increased risk of
bending fractures
15-4
From: Brown
Bending
• Cantilever bending
• Compressive force
acting
g off-center from
long axis
15-5
From: Noffal
Bending Loading
15-6
From: Brown
Muscle Activity Changing Stress
Di t ib ti
Distribution
15-7
From: Brown
Various
a ous Types
ypes o
of Beam
ea Loading
oad g a
and
d Suppo
Supportt
• Beam - structural member designed to support
loads applied at various points along its length.
• Beam can be subjected to concentrated loads or
distributed loads or combination of both.
• Beam design is two-step process:
1) d
determine
t
i shearing
h i forces
f
andd bending
b di
moments produced by applied loads
2) select cross-section best suited to resist
shearing forces and bending moments
15-8
From: Rabiei
•
6.1 SHEAR AND MOMENT
DIAGRAMS
In order to design
a beam, it is necessary to
determine the maximum shear and moment in the
beam
• Express V and M as functions of arbitrary position
x along axis.
• These functions can be represented by graphs
called shear and moment diagrams
• Engineers need to know the variation of shear and
momentt along
l
the
th beam
b
tto know
k
where
h
tto
reinforce it
15-9
From: Wang
Diagrams
15-10
From: Hornsey
•
6.1 SHEAR AND MOMENT
DIAGRAMS
Shear and bending-moment
functions must be
determined for each region
g
of the beam between
any two discontinuities of loading
15-11
From: Wang
6.1 SHEAR AND MOMENT
DIAGRAMS
Beam sign convention
• Although choice of sign convention is arbitrary, in
this course, we adopt the one often used by
engineers:
15-12
From: Wang
6.1 SHEAR AND MOMENT
DIAGRAMS
Procedure for analysis
Support reactions
• Determine all reactive forces and couple moments
acting on beam
• Resolve all forces into components acting
perpendicular and parallel to beam’s axis
Shear and moment functions
• Specify separate coordinates x having an origin at
beam’s left end, and extending to regions of beam
between concentrated forces and/or couple
p
moments, or where there is no discontinuity of
distributed loading
15-13
From: Wang
6.1 SHEAR AND MOMENT
DIAGRAMS
Procedure for analysis
Shear and moment functions
• Section beam perpendicular to its axis at each
distance x
• Draw free
free-body
body diagram of one segment
• Make sure V and M are shown acting in positive
sense, according to sign convention
• Sum forces perpendicular to beam’s axis to get
shear
• Sum moments about the sectioned end of segment
to get moment
15-14
From: Wang
6.1 SHEAR AND MOMENT
DIAGRAMS
Procedure for analysis
Shear and moment diagrams
• Plot shear diagram (V vs. x) and moment diagram
(M vs. x)
• If numerical values are positive
positive, values are plotted
above axis, otherwise, negative values are plotted
below axis
• It is convenient
con enient to sho
show the shear and moment
diagrams directly below the free-body diagram
15-15
From: Wang
EXAMPLE 6.6
66
Draw the shear and moment diagrams for beam
shown below.
15-16
From: Wang
EXAMPLE 6
6.6
6 (SOLN)
Support reactions: Shown in free-body diagram.
Shear and moment functions
Since there is a discontinuity of distributed load
and a concentrated load at beam’s center,, two
regions of x must be considered.
0 ≤ x1 ≤ 5 m,,
15-17
+↑ Σ Fy = 0; ...
V = 5.75 N
+ Σ M = 0; ...
M = (5.75x1 + 80) kN·m
From: Wang
EXAMPLE 6
6.6
6 (SOLN)
Shear and moment functions
5 m ≤ x2 ≤ 10 m,
m
+↑
↑ Σ Fy = 0;; ...
V = ((15.75 − 5x2) kN
+ Σ M = 0;; ...
M = ((−5.75x22 + 15.75x2 +92.5)) kN·m
Check results by applying w = dV/dx and V = dM/dx.
15-18
From: Wang
EXAMPLE 6
6.6
6 (SOLN)
Shear and moment diagrams
15-19
From: Wang
6.2 GRAPHICAL METHOD FOR
CONSTRUCTING SHEAR AND MOMENT
DIAGRAMS
Regions of concentrated
force and moment
15-20
From: Wang
6.2 GRAPHICAL METHOD FOR
CONSTRUCTING SHEAR AND MOMENT
DIAGRAMS
Regions of concentrated
force and moment
15-21
From: Wang
Sample Problem 7
7.2
2
SOLUTION:
• Taking entire beam as a free-body,
free body,
calculate reactions at B and D.
Draw the shear and bending
g moment
diagrams for the beam and loading
shown.
15-22
• Find equivalent internal force-couple
systems
t
for
f free-bodies
f
b di formed
f
d by
b
cutting beam on either side of load
application points.
• Plot results.
From: Rabiei
Sample Problem 7.2
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at B and D.
• Find equivalent internal force-couple systems at
sections on either side of load application points.
∑ Fy = 0 : − 20 kN − V1 = 0
V1 = −20 kN
∑ M 2 = 0 : (20 kN )(0 m ) + M 1 = 0
Similarly
Similarly,
V3
V4
V5
V6
15-23
= 26 kN
= 26 kN
= 26 kN
= 26 kN
From: Rabiei
M3
M4
M5
M6
= −50 kN ⋅ m
= −50 kN ⋅ m
= −50 kN ⋅ m
= −50 kN ⋅ m
M1 = 0
Sample Problem 7.2
• Plot results.
Note that shear is of constant value
b
between
concentratedd loads
l d andd
bending moment varies linearly.
15-24
From: Rabiei