Author: Thanh Q. Huynh

This PDF 1.4 document has been generated by PScript5.dll Version 5.2.2 / GPL Ghostscript 8.71, and has been sent on pdf-archive.com on 31/05/2014 at 03:45, from IP address 50.188.x.x.
The current document download page has been viewed 592 times.

File size: 40.65 KB (4 pages).

Privacy: public file

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.

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

Download

Use the permanent link to the download page to share your document on Facebook, Twitter, LinkedIn, or directly with a contact by e-Mail, Messenger, Whatsapp, Line..

Use the short link to share your document on Twitter or by text message (SMS)

Copy the following HTML code to share your document on a Website or Blog

This file has been shared publicly by a user of

Document ID: 0000166071.