Week 7 Challenge- Concepts of Structural Engineering
Objective:-
Solution Procedure:-
1) Develop equation for the moment in terms of E ( modulus of elasticity) and moment of inertia (I) and radius of curvature
The factors or bending equation terms as implemented in the derivation of bending equation are as follows –
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M = Bending moment.
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I = Moment of inertia exerted on the bending axis.
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σ = Stress of the fiber at a distance ‘y’ from neutral/centroidal axis.
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E = Young’s Modulus of beam material.
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R = Curvature radius of this bent beam.
However, if the distance to the remotest element c replaces y, then
Therefore
Where Z = I/c. This Z is the section modulus of this beam.
Bending stress formula derivation fundamentally computes the figure of bending stresses that develops on a loaded beam.
Therefore, the bending equation of stress includes the following steps –
Strain in fiber AB =
=
Therefore, strain =
With the presence of CD and C’D’ on the neutral axis, the stress on the neutral axis comes to be zero. Thus, this neutral axis is devoid of any strain from the applied force.
Yet,
= E (E = Young’s Modulus of elasticity)
Thus, the equation of the two strains based on the two relations is
…………………………. (i)
On the other hand, let us assume any arbitrary cross-section of the beam. The strain on the fiber is at a distance of ‘y’ from the N.A. Thus, the following expression is –
σ =
However, if the shaded strip has an area of ‘dA’, the following equation denotes force on strip–
F =
Consequently, a moment of the bending equation on the neutral axis will amount to –
F
Therefore, the total moment for the entire cross-section equals to –
Here, Σy2δA is the beam material’s property and suggests the second moment of the area of cross-section. The symbol I further denote it.
As a result,
Thus, when we combine equations (i) and (ii), we arrive at the following bending equation –
The first condition of equilibrium is that the net force in all directions must be zero.
- There are two conditions that must be met for an object to be in equilibrium.
- The first condition is that the net force on the object must be zero for the object to be in equilibrium.
- If the net force is zero, then net force along any direction is zero.
In order for a system to be in equilibrium, it must satisfy all three equations of equilibrium, Sum Fx = 0, Sum Fy = 0, and Sum M = 0.
3) Give examples for the following
a) Single Force Supports
Roller Support is an example of single force support
Roller supports only resist perpendicular forces and they cannot resist parallel or horizontal forces and moments.
b) Double Force Support
Pinned Support is an example of Double force Support
Pinned support or hinged support can resist both vertical and horizontal forces but they cannot resist a moment.
c) Rigid Support ( Can explain your examples with diagram)
Fixed support is the type of rigid support
Fixed support can generate moment, and shear and axial reaction
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4) What are the common conditions for external instability in a structure?
External stability takes place if all of its reactions are concurrent or parallel.
External Stability is achieved when the structure does not go rigid body deformations as a whole.
5) What are the common conditions for internal instability in a structure?
A structure is considered to be internally stable, or rigid if it maintains its shape and remains a rigid body when detached from the supports.
Internal stability is achieved if each joint in the truss is constraint against rigid body motion.