Learning Goal: To use the principle of superposition to relate external axial loads on a bar to axial deflections. Under certain conditions, the effect of complicated loads can be calculated by combining the effects of simple component loads. This is called the principle of superposition. This method can only be used when the internal axial force N in a body is linearly related to the displacement or external loading, and the deflections are small enough that the body's geometry does not change significantly. If a complicated load can be expressed as a linear combination of the simpler loads, then the stress or displacement at a point will be the same linear combination of the corresponding responses in the simpler loads. A horizontal bar is attached to a wall. When a force of 1 kN is applied at point A (Figure 1), the deflection at point A is 8₁40.47 mm. When a 1 kN force is applied at point B (Eigure 2), the deflection at point A is AB = 0.16 mm. (A given deformation due to a unit load is known as a flexibility factor) Part A - Larger load What is the change in the length of the bar if there is a single load of 5 kN applied to the right at point A? Use a positive value to indicate an increase in length. Express your answer in mm to three significant figures. ▸ View Available Hint(s) AL= Submit Part B - Combined loads AL= ΑΣΦ.If vec Submit What is the change in the length of the bar if a load of 3.6 kN is applied to the left at A and a load of 11.4 kN is applied to the right at B? Use a positive value to indicate an increase in length. Express your answer in mm to three significant figures. ▸ View Available Hint(s) ΑΣΦ.If vec Part C - Required load FA= ? ΑΣΦ If a load of 5.3 kN is applied to the right at point B, then what load magnitude must be applied to the left at point A to make the length of the bar remain unchanged? Express your answer in KN to three significant figures. ▸ View Available Hint(s) vec mm ? mm 3 KN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Learning Goal: To use the principle of superposition to relate external axial loads on a bar to axial deflections. Under certain conditions, the effect of complicated loads can be calculated by combining the effects of simple component loads. This is called the principle of superposition. This method can only be used when the internal axial force N in a body is linearly related to the displacement or external loading, and the deflections are small enough that the body's geometry does not change significantly. If a complicated load can be expressed as a linear combination of the simpler loads, then the stress or displacement at a point will be the same linear combination of the corresponding responses in the simpler loads. Figure B A 1 of 2 1 KN deflection at point A is deflection at point A is flexibility factor.) Part A Larger load A horizontal bar is attached to a wall. When a force of 1 kN is applied at point A (Figure 1), the AA = 0.47 mm. When a 1 kN force is applied at point B (Figure 2), the AB = 0.16 mm. (A given deformation due to a unit load is known as a AL = Submit What is the change in the length of the bar if there is a single load of 5 kN applied to the right at point A? Use a positive value to indicate an increase in length. Express your answer in mm to three significant figures.

Principle of Superposition
Learning Goal:
To use the principle of superposition to relate
external axial loads on a bar to axial deflections.
Under certain conditions, the effect of complicated
loads can be calculated by combining the effects of
simple component loads. This is called the
principle of superposition. This method can only be
used when the internal axial force N in a body is
linearly related to the displacement or external
loading, and the deflections are small enough that
the body's geometry does not change significantly.
If a complicated load can be expressed as a linear
combination of the simpler loads, then the stress or
displacement at a point will be the same linear
combination of the corresponding responses in the
simpler loads.
Figure
B
< 1 of 2 >
1 kN
deflection at point A is
deflection at point A is
flexibility factor.)
A horizontal bar is attached to a wall. When a force of 1 kN is applied at point A (Figure 1), the
AA = 0.47 mm. When a 1 kN force is applied at point B (Eigure 2), the
AB = 0.16 mm. (A given deformation due to a unit load is known as a
▾ Part A-Larger load
▼
▾
AL=
What is the change in the length of the bar if there is a single load of 5 kN applied to the right
at point A? Use a positive value to indicate an increase in length.
Express your answer in mm to three significant figures.
▸ View Available Hint(s)
Submit
Part B - Combined loads
AL=
Submit
IVE ΑΣΦ.It vec
Part C - Required load
FA=
What is the change in the length of the bar if a load of 3.6 kN is applied to the left at A and a
load of 11.4 kN is applied to the right at B? Use a positive value to indicate an increase in
length.
Express your answer in mm to three significant figures.
▸ View Available Hint(s)
Submit
137 ΑΣΦ. 11 vec
↑
Provide Feedback
9
IVE ΑΣΦ
ΑΣΦ.If vec ^
B
Ċ
?
P
3 of 5 >
?
If a load of 5.3 kN is applied to the right at point B, then what load magnitude must be
applied to the left at point A to make the length of the bar remain unchanged?
Express your answer in kN to three significant figures.
▸ View Available Hint(s)
mm
?
Review
mm
KN
Next >

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