Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Influence Lines in CE 331: Calculating Member Forces in Structural Engineering, Schemes and Mind Maps of Theory of Structures

This document from the spring 2009 ce 331 course introduces the concept of influence lines, which are used to calculate member forces in structures due to moving loads. Examples and formulas for calculating reactions, moments, and influence diagrams. Students will learn how to identify critical locations, construct influence lines, and use them to determine maximum moments and reactions.

What you will learn

  • What information can be obtained from an influence line?
  • What is an influence line and how is it used in structural engineering?
  • How do you construct an influence line for a particular member force?

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/27/2022

maya-yct
maya-yct 🇬🇧

4.8

(9)

217 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
CE 331, Spring 2009 Influence Lines 1 / 4
First of all, what is an influence line? An influence line is a plot of a particular member force
at a particular location (say moment at the middle of Span 1) due to a moving load, P(x).
ft
P
M
x
a,
Influence line for moment at “a” (Ma) due to unit load at “x”, Px
x, ft M
a, k-ft
5 2.0
14 -0.40
25 0.125
Influence lines are useful for calculating member forces due to moving loads, for example on
crane rails or bridges. A critical location is first identified (at midspan say) and the influence
line is constructed by calculating the moment at midspan due to a unit point load at each
location (x) along the bridge. The influence line can then be used to:
identify the location of the point load to cause max. moment at midspan
calculate the value of this moment
calculate the moment at midspan due to several point loads
Example: Calculate the Ma due to the truck axle loads shown below.
ftk
a
ftkftk
a
ftkftk
a
M
M
M
=
=
+=
56
460
)40.0)(10()2)(30(
a
x
Px
a
9ft
30
k
10
k
1.0
2.0
X, ft
10 20 30
pf3
pf4

Partial preview of the text

Download Influence Lines in CE 331: Calculating Member Forces in Structural Engineering and more Schemes and Mind Maps Theory of Structures in PDF only on Docsity!

First of all, what is an influence line? An influence line is a plot of a particular member force at a particular location (say moment at the middle of Span 1) due to a moving load, P(x).

ft P

M

x

a (^) ,

Influence line for moment at “a” (M (^) a) due to unit load at “x”, Px

x, ft M (^) a, k-ft

5 2.

14 -0.

25 0.

Influence lines are useful for calculating member forces due to moving loads, for example on crane rails or bridges. A critical location is first identified (at midspan say) and the influence line is constructed by calculating the moment at midspan due to a unit point load at each location (x) along the bridge. The influence line can then be used to:

  • identify the location of the point load to cause max. moment at midspan
  • calculate the value of this moment
  • calculate the moment at midspan due to several point loads

Example: Calculate the Ma due to the truck axle loads shown below.

k ft a

k ft k ft a

k ft k ft a

M

M

M

− −

=

a

x

Px

a

9 ft

30 k^10 k

X, ft

10 20 30

Influence diagrams can be constructed by applying a unit deformation associated with the type of member force.

reaction

shear

moment

1 rad

1

1

Draw the influence line for moment at E by “breaking” the beam at E and rotating the right end 1 radian^ relative to the left end, as shown. Since the beam segments are both 12 feet on either end of the break, the angles of each end are equal and equal to one half of 1.0rad^ = 0.5 rad^. The ordinate of the influence line at E is calculated from the following equation:

tan

tan

rad a 12'

a

forsmalldisplacements

a

The other ordinates are calculated using similar triangles.

The moment at E due to the concentrated load is maximum when the load is placed at E. The moment due to the 18 k^ load is then calculated from:

k ft k ft E

k M^18 = ( 18 )( 6 )= 108 −

The moment at E due to the uniform load is maximum when the load is placed at on the beam where the influence diagram is positive. The moment due to the 0.64klf^ load is then calculated from:

k ft k ft k ft E

klf ft klf ft ft k ft E

M

M

klf

− − −

ft P

M x

E (^) ,

B

5

1 rad

0.5 rad

a=

5 12’^ 12’^3