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Author: Thanh Q. Huynh

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Problem Diagram (70' span Glulam, Simply Supported FBD). Derivation by Thanh Huynh

y

F
Assuming the neutral axis is
in the center of the height of
the beam at all points.

h

ho
x

a

F1

F2

L
Find MOR
Solving Reactions F1 , F2
Sum of moments about roller connection
therefore
F1 ⋅ L = F⋅ a
Sum of forces in the y direction
F⋅ a
therefore
+ F2 − F = 0
L

F⋅ a
Fa =
L
F⋅ a
a
F2 = F −
= F⋅  1 − 
L
L

Shear (Top) and Moment Diagram (Bottom)
Fa/L

(Fa(L-a)/L)
Moment is simply integral of Shear

M peak =

F⋅ a
L

⋅ ( L − a)

F(1-a/L)

Now Solve M(x), moment as a function of x
M 1 ( x) = m⋅ x + b
Note, b here is the intercept

b =0
m=
M 1 ( x) =

F⋅ a
L
F⋅ a
L

FOR 0 &lt; x &lt; L-a

⋅x

M 2 ( x) = m⋅ x + b
b = F⋅ L⋅  1 −

m = −F⋅  1 −

M 2 ( x) = −F⋅  1 −

a

L
a

L

 ⋅ x + F⋅ L⋅  1 −
L

a

a

FOR x &gt; L-a

L

h ( x) = m⋅ x + b
b = ho
m=

h ( x) =

h − ho
L

h − ho
L

⋅ x + ho

 h − ho

b⋅ 
⋅ x + h o
3
b ⋅ h ( x)
 L

I( x) =
=
12

y( x) =

h ( x)

3

Note, b here is now the width of the beam

12

=

h − ho

2

2L

⋅x +

ho
2

Extreme Fiber Stress as a function of x
Since there are many constant terms lets simply
h − ho
F⋅ a
A=
B=
L
L
F⋅ a
σ1 ( x) =

M 1 ( x) ⋅ y( x)
I( x)

=

L

 h − ho

⋅ x⋅ 

 2L

 h− ho

b⋅ 
⋅ x+ho
 L

12

ho 

ho 
B
 A⋅ x⋅  ⋅ x + 
2 
2 
2
=

⋅x +
3

(

b⋅ B⋅ x+ ho
12

)3

2

σ1 ( x) =

(

)

b ⋅ B⋅ x + h o

6  A⋅ B⋅ x + A⋅ h o ⋅ x

2

A⋅ B⋅ x + A⋅ h o ⋅ x

=

3

(

b⋅ B⋅ x + ho

)

FOR 0 &lt; x &lt; L-a

3

6

σ2 ( x) =

−F⋅  1 −
 
 
=

M 2 ( x) ⋅ y( x)

a

 ⋅ x + F⋅ L⋅  1 −
L

I( x)

ho 
 h − ho
⋅x +

 ⋅ 
L   2L
2 
a 

 h− ho
b⋅ 
⋅ x+ ho
 L

3

12

h
 F⋅ a − F ⋅ x + ( F⋅ L − F⋅ a) ⋅  h − h o ⋅ x + o 
 L

2 


  2L
σ2 ( x) =

 h− ho

b⋅ 
⋅ x+ ho
 L

3

12

C=

F⋅ a
L

D = F⋅ L − F⋅ a

−F

ho 
B
⋅x +
 6 B⋅ C⋅ x2 + C⋅ h + B⋅ D ⋅ x + D⋅ h 
( o
)
2 
o
2

=

( C⋅ x + D) ⋅ 
σ2 ( x) =

(

)

b⋅ B⋅ x+ ho

3

(

b⋅ B⋅ x + h o

12

derivatives of σ1 and σ2 set to the solution of 0
dσ1 ( x)

=

dx
dσ1 ( x)
dx
dσ2 ( x)

(

(

b ⋅ B⋅ x + ho
=0

=

dx

x=

)

3

ho
B

6 ⋅ C⋅ h o − B⋅ ( C⋅ x + 2 ⋅ D)

dx
dσ2 ( x)

)

6 ⋅ A⋅ h o − B⋅ x

(

b ⋅ B⋅ x + h o
=0

x=

ho
B

)

2D
C

3

)

3

FOR x &gt; L-a

Parameters
h := 27in
h o := 12in
L := 70ft
a := 9ft
x :=

x :=

L⋅ h o

x = 56⋅ ft

h − ho
L⋅ h o
h − ho

2 ( L − a)
a
L

x = 196 ⋅ ft

−1

Only x = 56ft makes sense, once we know the force acting at "a", you can plug in x into σ1(x)
and evaluate the stress (MOR) at that point. If I'm missing anything or did something wrong, just
give me a heads up! I beleive this is the same solution obtained in class. Note, you also need to
find MOE but you should be able to compute that with the information we obtain here.

Mathcad - Lab7 - Derivation.pdf (PDF, 40.65 KB)

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