# Mathcad Lab7 Derivation .pdf

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**Mathcad - Lab7 - Derivation.pdf**

Title: Mathcad - Lab7

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 < x < 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 > 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 < x < 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 > 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.

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