Introduction 14:125:208: to Biomechanics
Homework Assignment #6
Spring 2007
Due 3/27/07
Problem 7: A paraplegic patient weighing 70kg is performing a pushup in the prone
position (below). The lower legs and hands are supported by the floor. The force plate,
placed under the right hand, shows a vertical force of 200N.
The radius and ulna make a 75 degree angle with the horizontal. The triceps (thick
black line), which connects the olecranon process to the humerus, makes a 15 degree
angle with the horizontal, and is assumed to connect at the proximal end of the lower
arm. Assume that the geometry of the lower arm can be approximated as a hollow
cylinder, with inner and outer radii, and resulting cross-sectional area and moment of
inertia, as shown. You may assume that the geometry of the lower arm is uniform along
its entire length.
Conduct a biomechanical analysis of the lower arm, assuming that it acts as a beam.
This should include:
i) A free body diagram
ii) A full beam analysis, including shear and bending moment diagrams
iii) Calculation of the magnitude and location (relative to the length along the axis
and position in a cross-section) of the maximum tensile the AND maximum
compressive stresses.
You can assume that all external loads act at the center of the cross-section of the
bone.
What about a beam in 3 point bending without symmetry?
From statics we find the
reaction forces R1 and
R2
M ( x) = R1 x =
M ( x) =
Pb
x
L
0< x<a
Pb
x P(x a ) a < x < L
L
For these type of problems we have to
evaluate the moment along the length of the
beam
First lets evaluate the deflection from x=0 to x=a
M ( x) = R1 x =
M ( x) =
On the left support we find
y(0)=0 so C2=0
Pb
x
L
0< x<a
Pb
x P(x a ) a < x < L
L
d 2 y Pb
EI 2 =
x
dx
L
dy Pb x 2
=
+ C1
EI
dx
L2
Pb x 3
+ C1 x + C2
EIy =
L6
Pb x 3
+ C1 x
EIy =
L6
11
Now lets evaluate from
a<x<L
d 2 y Pb
x P(x a )
=
dx 2
L
2
dy Pb x 2 P( x a )
EI
=
+ C3
2
dx
L2
3
Pb x 3 P( x a )
EIy =
+ C3 x + C 4
6
L6
Pb a 3
EIy (a ) =
+ C3 a + C 4 C 4 = 0
L6
C1 = C3
EI
At point x=a
Pb a 3
EIy(a ) =
+ C1a
L6
Pb L3 P(L a )
+ C3 L = 0
L6
6
3
P(b ) PbL2
Pb 2
C3 L =
b L2
C1 = C3 =
6
6
6L
Pb 3
EIy ( x ) =
x L2 b 2 x
6L
Pb 3 L
3
2
2
EIy ( x ) =
x (x a ) L b x = 0
6L
b
3
EIy(L ) =
(
(
(
))
(
)
)
0< x<a
a<x<L
12