Triangular Truss (On Floor)

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Francis Franklin

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Ready to use

7 years, 5 months ago

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Francis Franklin 4 years, 6 months ago

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Francis Franklin 7 years, 5 months ago

Gave some feedback: Ready to use

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Francis Franklin 7 years, 5 months ago

Created this as a copy of Triangular Truss.

Name	Status	Author	Last Modified
Triangular Truss	Ready to use	Francis Franklin	06/11/2022 16:43
Triangular Truss (On Floor)	Ready to use	Francis Franklin	06/11/2022 16:43
Truss (Bracket)	Ready to use	Francis Franklin	06/11/2022 16:44

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Question statement

Half of mechanics is about balancing forces. Some trusses can be treated like pin-jointed structures and decomposed into axial forces, i.e., without any bending or shear forces. The solution is therefore obtained through trigonometry.

Give any introductory information the student needs.

			Name	Type	Generated Value

Force	integer	4
theta	integer	45
Pac	number	2

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Name

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Value

xxxxxxxxxx
 
random(1..5)

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Description

Applied vertical (downward) force.

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Generated value: `integer`

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Ask the student a question, and give any hints about how they should answer this part.

A pin-jointed truss is shown in the figure below. The pivot at A is fixed, but the pivot at C is free to move horizontally. The angle at B is a right angle.

Pin-jointed truss

If the applied force, $F$ , is $\var{force}$ kN vertically down, and the angle of Bar AB to the horizontal, $\theta$ , is $\var{theta}^\circ$ , what is the tension in Bar AC?

Unnamed gap [Units: kN]

Advice

Balancing horizontal forces at B: $P_{AB}\cos(\var{theta})=P_{BC}\sin(\var{theta})$ .

Balancing vertical forces at B: $P_{AB}\sin(\var{theta})+P_{BC}\cos(\var{theta})+\var{force}=0$ ,

where units are kN. Multiplying the first by $\sin(\var{theta})$ and the second by $\cos(\var{theta})$ gives:

$P_{AB}\sin(\var{theta})\cos(\var{theta})-P_{BC}\sin^2(\var{theta})=0$

$P_{AB}\sin(\var{theta})\cos(\var{theta})+P_{BC}\cos^2(\var{theta})=-\var{force}\cos(\var{theta})$

and subtracting the first from the second gives:

$P_{BC}\sin^2(\var{theta})+P_{BC}\cos^2(\var{theta})=-\var{force}\cos(\var{theta})$ ,

or, since $\sin^2+\cos^2=1$ :

$P_{BC}=-\var{force}\cos(\var{theta})$ .

Balancing horizontal forces at C: $P_{AC}+P_{BC}\sin(\var{theta})=0$

Substituting in the expression for $P_{BC}$ from above, and rearranging:

$P_{AC}=\var{force}\sin(\var{theta})\cos(\var{theta}) = \var{siground(Pac,3)}$ [units: kN].

Give a worked solution to the whole question.

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This question is used in the following exams:

Basic Statics and Stresses in Trusses by Francis Franklin in Mechanics.
Introduction to Stresses by Francis Franklin in Mechanics.
Tom's copy of Introduction to Stresses by Tom Tomlinson in Tom's workspace.
Keith's copy of Introduction to Stresses by Keith Tarnowski in Keith's workspace.
Core Skills by Francis Franklin in Mechanics.
Mechanical Engineering Practice Questions by Francis Franklin in Mechanics.
MEC2009 Semester 1 - Special Resit by Francis Franklin in Mechanics.
Core Skills Practice Questions by Francis Franklin in Mechanics.
Take-Home Core Skills 2023-24 by Francis Franklin in Mechanics.
Core Skills 2024-25 by Francis Franklin in Mechanics.

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Francis Franklin 4 years, 6 months ago

Francis Franklin 7 years, 5 months ago

Francis Franklin 7 years, 5 months ago

Error in variable testing condition

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Generated value: `integer`

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Triangular Truss (On Floor)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Francis Franklin 4 years, 6 months ago

Francis Franklin 7 years, 5 months ago

Francis Franklin 7 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`