Simply-Supported Beams
Example
Determine  P , the displacement under the concentrated force.
P
a
b

O
R1
R2
Determine the reactions.
R1 
Pb
ab
(E.1)
R2 
Pa
ab
(E.2)
Consider the following problem. Two cantilever beams extend in opposite directions
from a “zero-thickness” wall.
a
b
R1
R2
Break a simply supported beam at a convenient location, e.g., a location at which the
deflection is to be determined, and divide the beam into two cantilever beams. Draw all the
external loads and reactions, except a concentrated force and/or concentrated moment, if one or
both exist, at the break point.
The bending-moment diagrams of the simply supported beam and the two cantilever
beams are identical. Curvature is proportional to the bending moment. Therefore, the curvature
of the simply supported beam is equal to the curvature of the two cantilever beams. In addition,
the slopes and deflections of the two cantilever beams will match at the break point, so the shape
of the deflection curve for the two cantilever beams will precisely match the shape of the
deflection curve for the simply supported beam. However, the slope of the two cantilever beams
at the break point is equal to zero, but the slope of the simply supported beam at the break point
is generally not zero. The two deflection curves will be rotated with respect to each other.
MAE3501 - 12 - 1
Sketch the deflection curves of the two cantilever beams. Draw them both upward. It
does not matter whether this is true, or which deflection is larger, the answers will be right even
if the sketch is wrong. It is the same as determining reaction forces and moments in statics,
where it is permissible to guess a direction and a minus sign identifies a wrong guess. Denote the
upward deflections at the ends of the beams by  1 and  2 , the subscripts matching the subscripts
of the reaction forces.
1
2
a
b
R1
R2
Determine 1 and 2.
R1 a 3
3EI
(E.3)
R2 b3
2 
3EI
(E.4)
1 
When cantilever beams were analyzed, all the calculated deflections and slopes were
correct, because the horizontal line used in myosotis, referred to as the myosotis-reference line,
was the same as the un-deflected line of the beam. However, when beams other than cantilever
beams are analyzed, this is not in general true. The myosotis-reference line of a simply
supported beam is the horizontal line in the sketch. It is a line tangent to the deflection curve of
the simply-supported beam at the point at which the simply-supported beam was broken into two
cantilever beams. It is necessary to construct the true un-deflected line of the beam. That is a
straight line joining the support points in the original problem.
MAE3501 - 12 - 2
P
1
a
2
b
Determine  P by linear interpolation between  1 and  2 .
 P  1 
( 2   1 ) a

ab
(E.5)
It makes no difference whether  1 or  2 is larger, the equation takes care of itself. It is
perfectly okay that  P is not measured precisely perpendicular to the un-deflected line. This is
the same approximation discussed during the last lesson, exactly the same approximation as
setting the denominator equal to 1 in the governing differential equation in the classical theory.
MAE3501 - 12 - 3
Example
Determine  P , the displacement under the concentrated force.
E I  1.0 (10) 7 kip  inch2
8 kip
40 inch
30 inch

O
R1
R2
Determine the reactions.
R1 
8 ( 30)
 3.43 kip
70
(E.1)
R 2  8  3.43  4.57 kip
(E.2)
Sketch the deflection curves of the two cantilever beams. Draw them both
upward. It does not matter whether this is true, or which deflection is larger, the answers will be
right even if the sketch is wrong. It is the same as determining reaction forces and moments in
statics, where it is permissible to guess a direction and a minus sign identifies a wrong guess.
Denote the upward deflections at the ends of the beams by  1 and  2 , the subscripts matching
the subscripts of the reaction forces.
1
2
40
30
3.43
4.57
Determine 1 and 2.
1 
2 
3.43 (40) 3
3 (10) 7
4.57 ( 30) 3
3 (10) 7
 0.00732 inch
(E.3)
 0.00411 inch
(E.4)
MAE3501 - 12 - 4
P
1
40
2
30
Determine  P by linear interpolation between  1 and  2 .
 P  0.00732 
(0.00411  0.00732) 40
 0.00549 inch 
70
(E.5)
It makes no difference whether  1 or  2 is larger, the equation takes care of itself. It is
perfectly okay that  P is not measured precisely perpendicular to the un-deflected line. This is
the same approximation discussed during the last lesson, exactly the same approximation as
setting the denominator equal to 1 in the governing differential equation in the classical theory.
MAE3501 - 12 - 5
Example
q0

L
O
R1
R2
Determine  C ,the downward displacement at the center of the uniformly loaded, simply
supported beam.
q L
(E.1)
R1  R 2  0
2
Break the beam at its center, the location at which the deflection is to be determined.
q0
q0
q0 L
2
L
2
L
2
q0 L
2
Break the simply supported beam at the location at which deflection is to be
calculated. Sketch the deflection curves of the two cantilever beams. Draw them both upward.
Denote the deflections at the ends of the beams by  1 and  2 , the subscripts matching the
subscripts of the reaction forces.
1
2
L
L
2
2
Determine the equal, upward deflections at the ends of the two cantilever beams.
3
4
 q0 L   L 
L

 
q0  
q 0 L4
2  2
2


1   2 

 0.0130
3EI
8EI
EI
(E.2)
The beam is symmetric, left-to-right, and the un-deflected line of the beam is parallel to
the myosotis-reference line.
 q L4 

 C   1   2  0.0130  0
 EI 


MAE3501 - 12 - 6
(E.3)
Example
0.25 kip / inch

90 inch
O
R1
R2
Determine  C ,the downward displacement at the center of the uniformly loaded, simply
supported beam. E I  1.0 (10) 7 kip  inch2
0.25 (90)
R1  R 2 
 11.25 kip
2
(E.1)
Break the beam at its center, the location at which the deflection is to be determined.
0.25
0.25
11.25
45
45
11.25
Break the simply supported beam at the location at which deflection is to be calculated.
Sketch the deflection curves of the two cantilever beams. Draw them both upward. Denote the
deflections at the ends of the beams by  1 and  2 , the subscripts matching the subscripts of the
reaction forces.
1
2
45
45
Determine the equal, upward deflections at the ends of the two cantilever beams.
1   2 
11.25 (45) 3
3 (10) 7

0.25 (45) 4
8 (10) 7
 0.0214 inch
(E.2)
The beam is symmetric, left-to-right, and the un-deflected line of the beam is parallel to
the myosotis-reference line.
 C   1   2  0.0214 inch 
MAE3501 - 12 - 7
(E.3)
Example
Determine  , the downward displacement at the right edge of the distributed force.
q0
a
b

O
R1
R2
Determine the reactions.

(q 0 a )  b 

R1 
ab
a

2
(E.1)
a
(q 0 a )  
 2
R2 
(E.2)
ab
Break the simply supported beam at the location at which deflection is to be calculated.
Sketch the deflection curves of the two cantilever beams. Draw them both upward. Denote the
deflections at the ends of the beams by  1 and  2 , the subscripts matching the subscripts of the
reaction forces.
1
2
a
b
Determine 1 and 2.
1 
R1 a 3 q0 a4

3EI
8EI
(E.3)
R2 b3
2 
3EI
(E.4)
Determine  by linear interpolation between  1 and  2 .
  1 
( 2   1 ) a
ab
(E.5)
MAE3501 - 12 - 8
Example
Determine  , the downward displacement at the right edge of the distributed force.
E I  1.0 (10) 7 kip  inch2
0.20 kip / inch
40 inch
30 inch

O
R1
R2
Determine the reactions.
R1 
0.20 (40) (50)
 5.71 kip
70
(E.1)
R 2  0.20 (40)  5.71  2.29 kip
(E.2)
Break the simply supported beam at the location at which deflection is to be calculated.
Sketch the deflection curves of the two cantilever beams. Draw them both upward. Denote the
deflections at the ends of the beams by  1 and  2 , the subscripts matching the subscripts of the
reaction forces.
1
2
40
30
Determine 1 and 2.
1 
2 
5.71 (40) 3
3 (10) 7
2.29 ( 30) 3
3 (10) 7

0.20 (40) 4
8 (10) 7
 0.00578 inch
 0.00206 inch
(E.3)
(E.4)
Determine  by linear interpolation between  1 and  2 .
  0.00578 
(0.00206  0.00578) (40)
 0.00365 inch 
70
MAE3501 - 12 - 9
(E.5)
Example
Determine  P , the downward displacement under the concentrated force.
P
q0
a
b

O
R1
R2
Determine the reactions.
a

(q 0 a )  b  
Pb
2

R1 

ab
ab
(E.1)
a
(q 0 a )  
 2   Pa
R2 
ab
ab
(E.2)
Break the simply supported beam at the location at which deflection is to be calculated.
Sketch the deflection curves of the two cantilever beams. Draw them both upward. Denote the
deflections at the ends of the beams by  1 and  2 , the subscripts matching the subscripts of the
reaction forces.
1
2
a
b
MAE3501 - 12 - 10
Determine 1 and 2.
1 
R1 a 3 q0 a4

3EI
8EI
(E.3)
R2 b3
(E.4)
3EI
Note that the concentrated force P is not included in the deflections of the two cantilever
beams. A concentrated force and/or concentrated moment at the break point does not affect the
deflections of the cantilever beams. P is included in the problem, through its effect on the
reaction forces.
2 
Determine  P by linear interpolation between  1 and  2 .
  1 
( 2   1 ) a
ab
(E.5)
MAE3501 - 12 - 11
In general, we are not concerned with angles of rotation at various points on the beam.
The primary exception is the angles of rotation at the ends of a simple supported beam. The
following example demonstrates the method used to determine these angles.
Example
Consider the beam in the first example, a non-centered concentrated load P. Determine
the slopes at the ends of the simply supported beam. Re-draw the deflection sketch, omitting all
the extraneous material. Extend the true un-deflected line. Label the angles as shown.
2
0
1
0
2
1
a
R1 
b
Pb
ab
R2 
Pa
ab
1 
R1 a 2
2EI
(E.1)
2 
R2 b2
2EI
(E.2)
All the angles shown in this figure, including  2 , are measured relative to a line parallel
to the myosotis-reference line. All the angles have now been determined except  0 .
Determine  0 from the relationship that a small angle, measured in radian, is equal to its tangent.
0 
 2  1
ab
(E.3)
MAE3501 - 12 - 12
Determine the slopes at the ends of the simply supported beam by adding or subtracting.
1T and  2T denote the true slopes at the ends, relative to the un-deflected line.
1T  1   0
(E.4)
 2T   2   0
(E.5)
Measure 1T positive clockwise, and  2T positive counter-clockwise. Both angles will be
positive for a simply supported beam with downward loads. A simple, exaggerated sketch is
very useful in keeping signs straight.
Homework
Use the same numerical values from the previous set of homework problems.
1. Determine  , the displacement, positive down, at the left end of the distributed load.
Determine the rotations, positive clockwise at the left end of the beam , and positive counterclockwise at the right end of the beam.
q0

O
a
b
2. Determine the displacement,  P , positive down, under the concentrated force.
P
q0

O
b
a
MAE3501 - 12 - 13