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Hi! This mail service is an extremely detailed discussion on the basics of axle analysis. This is a perfect starting point for learning the theory behind analysing beams.

Beam Analysis

For beams, there are three requirements for complete analysis: (ane) finding the values of the reaction components, (2) modelling how the main stresses (shear and moment) human action on the construction, and (iii) determining the deflected shape.

Segmenting the Beam

The first thing we demand to do is identify certain positions in the beam and so we can separate it into segments. To exercise just that, let's identify the beam forth the horizontal x-axis with the left-most end at the origin (0, 0).

We label points of interest from the following criteria: (ane) in that location is a change in loading status, or (ii) in that location is a modify in cross-sectional properties. For simplicity, we will assume that the cross-section is the same throughout the axle.

The post-obit shows the change in loading atmospheric condition:

At (0, 0), the axle experiences a 27 kN•k clockwise moment, we label it equally A.
At (2, 0), a roller supports the beam, nosotros label it as bespeak B.
When 2 ≤ x ≤ 6.5, the axle experiences a 45 kN/m triangular load with its peak reaching at x = vi.5; hence, nosotros label the finish-points of this load, C = (vi.5, 0). Coincidentally, (2, 0) is bespeak B.
When 6.five ≤ x ≤ 11, the axle experiences a 36 kN/thousand rectangular distributed load; hence, we characterization the cease-points of the load, D = (eleven, 0). Also, at point D, a hinge supports the beam
At (12.five, 0), the beam experiences a xc kN point load, nosotros label information technology as E.

After labelling, you'll notice that the beam is composed of four segments: AB, BC, CD, and DE.

Determinacy and Stability

When analysing any structure, the first affair to do is finding its determinacy. It gives us an idea of how to approach the problem. For a beam, D is:

In the case, a roller and hinge support the beam; hence, r = 1 + 2 = 3. Since there are no internal connections, e_c = 0:

Just because information technology is determinate does not mean it is stable; thus, we need to analyse its stability. When a construction is unstable, it cannot resist loads imposed on it. There are 2 possibilities when that can happen: (1) D is less than cypher, or (two) the beam is geometrically unstable. Such instances will crave you to provide boosted supports or remove internal connections.

Reaction Assay

One requirement for a complete assay of a beam is to determine the reaction support components. In this case, we would similar to detect the values of the roller and the hinge (R_B, D_v, and D_h). When solving for these forces, we simply assume the direction of the components and apply the equilibrium equations. In the instance that our upshot is negative, it means that we take a wrong assumption and the truthful direction of the strength is the opposite.

We assume RBand Dv acting upward (to counteract the downward loads) and Dh interim to the left.

For a determinate beam, we just employ the equilibrium principle to the whole structural system. We'll have the summation of forces along x (or horizontal) to start:

Some other equilibrium status is that the summation of moments along whatever point of the structural organisation is zero. We tin can choose whatsoever point but its best to solve by taking the pivot point at the reaction points B or D since unknown reactions can be eliminated. Taking about point D:

Finally, we now apply summation of forces along y (or vertical):

It's important to double-check your results at this point because incorrect reactions will hateful that the subsequent analysis is incorrect. You tin can do that past taking summation of moments at some other betoken and it should comply with the equilibrium principle.

Shear and Moment Analysis

Because there are multiple load conditions on the beam, the shear and moment stresses will vary depending on the load; hence, we have to analyse it segment by segment. To reveal the internal shear and moment, nosotros (literally) cut the beam at a distance x from the origin to the segment we want to investigate:

Say we desire to investigate the shear and moment at segment AB, we place a section A at a distance x from the origin (Indicate A) to any point in the segment. If we want to analyse V and Yard at BC, then we place a department B at a distance x to any betoken between B and C. This goes for every identified segment in the axle.

Convention

When y'all place a cutting plane at a distance ten from the origin, you have cut the beam into two parts; As a consequence, you can either analyse the left or right section.

Permit's recall that if you cut the axle, y'all'll be "revealing" the internal shear (V₁₀) and moment (M_ten) as shown in the figure. The directions of 5 and M will depend on the section y'all have called whether its the left or right side. As a full general rule, whatsoever force or moment that causes the beam to concave upward is positive.

Let's expand on the sign convention – consider the left section. For Five x and Chiliad x to be positive, V x must be downwardly, and M x must be counterclockwise. These directions volition cause it to concave upwards. Try getting a piece of newspaper and curve it according to the direction shown in the figure – make sure you clamp at the support. You'll notice that doing and then volition make the section concave upwards.

The same reasoning applies to the right section. For it to concave upward, Vx must be upward, and Mx must be clockwise. Endeavour angle a slice of paper again based on the figure to become the upshot.

Now, to find for V_ten and M₁₀, we employ the equilibrium principle for the whole section; Let'due south consider the left section again:

For the downward V_x, information technology is equal to the summation of forces along y-axis but with the assumption that all upwardly forces are positive to counteract the downwardly direction.
For the counterclockwise 1000_x, it is equal to the summation of moments about the cut section with the assumption that all clockwise moments are positive to counteract it.

For the right section:

For the up V_ten, it is equal to the summation of forces along y-axis only with the assumption that all downward forces are positive to annul the upward management.
For the clockwise M_x, it is equal to the summation of moments about the cutting department with the assumption that all counterclockwise moments are positive to counteract it.

To understand farther, let's model the shear and moment for the given beam. Remember, that we are interested in expressing V and M in terms of x: V_x = f(ten) and M_x = f(x).

Segment AB

Place section A at a altitude x from the origin between points A and B and consider the left department. Every bit a result of cutting the beam, nosotros can run into V_AB and M_AB:

The domain 10 or position is explicitly-defined between points A and B: 0 ≤ x ≤ 2

Segment BC

Place section B at a distance x from the origin betwixt points B and C and consider the left section. As a issue of cutting the beam, we can see 5_BC and M_BC:

The domain x or position is explicitly-defined between points B and C: 2 ≤ x ≤ 6.5

For the triangular load, you have to express it in terms of the position x by ratio and proportion.

Segment CD

Place section C at a distance x from the origin between points C and D and consider the right section. Equally a result of cutting the axle, we can come across V_CD and M_CD:

The domain x or position is explicitly-defined between points C and D: half dozen.5 ≤ 10 ≤ 11

We decided to choose the right department because it will be easier to analyse compared to the left. Also, like the triangular load at segment BC, you lot have to limited the rectangular load in terms of the position x.

Segment DE

Place section D at a altitude x from the origin between points D and E and consider the correct section. As a result of cutting the beam, we tin come across V_DE and M_DE:

The domain 10 or position is explicitly-defined between points C and D: 11 ≤ x ≤ 12.5

We decided to choose the right section because it volition exist easier to analyse compared to the left.

The following are the shear and moment equations – expressions in terms of position x that is used to model Five and M. Take note that these functions have domains that are explicitly-defined based from the coordinates of the segments.

Shear and Moment Diagrams

With the Five and M equations, we can visualise it through the shear and moment diagrams. It is a graphical perspective of the shear and moment that the beam experiences when subjected to certain loads. It is these graphs that will help u.s. understand our structure and serves as our basis for the design phase.

Beneath is the shear diagram of the axle example; At points B, D, and Due east, there are discontinuities which is due to the reactions (B and D) and point load (E). The values of these discontinuities in the graph should equal to the value of the reaction components or the point load at that specific position.

Side by side is the moment diagram of the axle; At signal A, in that location is a discontinuity which is due to the moment at A. The values of these discontinuities should equal to the value of the moments at that specific point.

At this point, nosotros accept accomplished modelling the shear and moment of the beam using equations. Sometimes though, formulating shear and moment equations can exist boring; however, there is another user-friendly way to model it and would depend on relating the loads, shear, and moment.

Deflected Shape Analysis

The terminal step to completely analyse the axle is to sketch the deflected shape. At that place are formal methods to compute for the beam deflection such as the double integration method and surface area-moment. For now, we will satisfy ourselves with an approximate shape for simplicity.

This pace is necessary if we want to analyse the serviceability requirement of the beam. Sometimes, we would limit the beam to deflect at a certain indicate for functionality and comfort reasons.

To gauge the shape, we refer to the moment diagram. Hither are the general rules for sketching the deflected shape based on the compatibility principle:

The supports act equally the clamps for beam deflection.
If the beam experiences moment that is positive, then it concaves upward. In the example, the beam experiences a positive moment when x is between 0 and 10.145 (0 ≤ x < 10.145).
On the other mitt, if the beam experiences moment that is negative, then it concaves downward. In the instance, the axle experiences a negative moment when x is between x.145 and 12.5 (x.145 < 10 ≤ 12.5).
At (10.145, 0), it is at this signal where the point of inflexion occurs. If yous look at the moment diagram, that is where the moment is equal to zero.

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