• Shuffle
    Toggle On
    Toggle Off
  • Alphabetize
    Toggle On
    Toggle Off
  • Front First
    Toggle On
    Toggle Off
  • Both Sides
    Toggle On
    Toggle Off
  • Read
    Toggle On
    Toggle Off
Reading...
Front

Card Range To Study

through

image

Play button

image

Play button

image

Progress

1/73

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

73 Cards in this Set

  • Front
  • Back
–pinned connections at all joints
In general, a truss should have:
–external forces acting only on the joints
In general, a truss should have:
–zero moment at each joint
In general, a truss should have:
–members subjected only to axial forces
In general, a truss should have:
trusses are usually constructed in
wood or steel. Concrete is possible but generally not practical.
A rule of thumb in determining truss depth is a depth to span ratio of
1/10 –1/15
a 50’span would have a depth of approximately
3.3’–5’depending on the load
Trusses can be looked at in several categories:
planar truss:
two-way truss system:
space truss:
space frame:
two dimensional
planar truss:
planar trusses connected perpendicular to each other, forming a truss grid.
two-way truss system
three dimensional truss
space truss:
space trusses connected to form a three dimensional grid
space frame:
two-dimensional in their section and connected with pin joints.
planar trusses
framed beam structures having vertical web members rigidly connected to parallel top and bottom chords
Verendeel Trusses
Are not true trusses because their members are subject to nonaxial bending forces
verendeel trusses
connected with rigid joints
VierendeelTruss
resist bendingmomentlike the flanges in a steel beam.
top and bottom chord of the truss
top chord is in compression while the bottom chord is in tension.
like the flanges of a steel beam in a simple span condition, the truss
The web members in the truss resist
shear,the same as the web of the steel beam.
Loads on a truss should be applied at
the panel points.
By placing loads at the panel points, forces act
axially. in all the members of the truss so that members are all in tension or compression.
Loads applied to the member rather than at the panel point introduces
bending in the member making it behave like a beam.
Now the members must resist both bending and axial forces--not very efficient.
Loads applied to the member
Uniform loads applied directly to the truss members also introduce
bending moment into the members.
secondary members that span between trusses,
Purlins,
can be used to transfer uniform loads to the panel points.
Purlins,
ideally, forces are distributed axially in tension or compression
Truss Force Distribution
forces increase towards the centerwhere bending moment increases in beams.
top and bottom chord:
forces increase towards the supportswhere shear increases in beams.
web members:
b=
number of members
n =
number of joints
is stable, generally
(equation)
b = 2n-3
is stable but indeterminate
(equation)
b > 2n-3
is unstable
(equation)
b < 2n-3
In order for the truss assembly to be stable,
the truss or any part of the truss must be in equilibrium.
A joint, shown in a free-body, must also be in
equilibrium
ΣFV= 0 ΣFH= 0 ΣM = 0
laws of static equilibrium:
all joints are pinned with zero moment, in the case of
trusses,
Diagonal members have
both vertical and horizontal components.
AEach member of the truss has
axial, tensile or compressive,internal forces.
the relationship of the
rise to run on the
diagonal member
determines the
relationship between the
vertical and horizontal
components.
find the resulting diagonal force by using the
Pythagorean Theorem.
All forces moving to the joint are
compressive
All forces moving away from the joint are
tensile
A cable with a constant cross section carrying its own
deadweight will form a
catenary curve
A cable carrying a uniform load projected horizontally along its length will form a
parabolic curve.
Cables can be looked at in several categories:
–single cable
–double curved cables
–cable-stayed
Only_____ ______are developed in cable structures
tension forces
they offer no resistance to compression or bending.
cable structures
the shallower the cable slope,
the larger the force in the cable
____ load generates an upward load on the cable (suction).
wind load
(wind loads on cable systems)

The shape of the roof changes in response to _______,resulting in ________ Pressure caused by the effect of the wind on the new shape.
suction,
downward
The alternating effects of the wind on the changing shape of the roof cause
fluttering
a dynamic and cyclical process.
fluttering
-increase dead load
(1 of 4)
Stabilizing wind loadson simple cable structures to prevent fluttering.:
-use a stiffened tension element to resist the deformation
(1 of 4)
Stabilizing wind loadson simple cable structures to prevent fluttering.:
-use anchor cables to tie the system to the ground
(1 of 4)
Stabilizing wind loadson simple cable structures to prevent fluttering.:
-use double-curved cable systems
(1 of 4)
Stabilizing wind loadson simple cable structures to prevent fluttering.:
are connected by secondary members.(cables)
double concave cables, pretensioned cables
are separated by compression struts. (cables)
double convex cables,
pretensioned cables
Similar to the analysis we looked at in trusses, we can also analyze cables using a series of
free-body diagrams.
Each free-body must also be in equilibrium satisfying the laws of
static equilibrium:
In the case of cables, all supports are pinned with
zero moment.
Unlike truss analysis, cables generally have
pinned
connections, meaning there are a total of 4 unknown
reactions, not 3.
A free-body diagram is necessary in determining
the external reactions.
Determine the vertical reactions either by ________ or by ____________.
(cable problem 1)
symmetry

the ΣM at the reaction.
The relationship between the vertical and horizontal force components is the same as the
relationship between the rise and run for the slope of the line.
general stress formula,
f = P/A,
A = P/Ft
what is Ft
Ft is the allowable tensile stressfor steel cable.
For a given shape, cable structures only develop _______________________ and no ___________________
axial tensile forces
no interior bending moment.
result in corners
(cable summary)
pointloads
horizontally projected, result in parabolic curves
(cable summary)
uniform loads,
results in catenary curves
(cable summary)
self-weight