Foundations of Mechanics of Materials

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Mar 19, 2018 - 2.8.3 Example of a cantilever beam with partial distributed load. ...... each little bit of force may be presented as f o.dx.x and a negative sign will ...
Foundations of Mechanics of Materials Mohammad Tawfik Academy of Knowledge http://academyofknowledge.org

DOI: 10.13140/RG.2.2.15614.48962

Draft printed on: 19. March 2018

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Table of Contents 1 Main concepts..................................................................................................................................5 1.1 Equilibrium..............................................................................................................................5 1.2 Deformation and Deflection....................................................................................................5 1.3 Compatibility of Structures......................................................................................................6 1.4 Input-Output Relations for Structures......................................................................................6 2 Supports, Reactions, and Loads.......................................................................................................8 2.1 What is a Support?...................................................................................................................8 2.2 Types of Supports....................................................................................................................8 2.3 Reactions and Free-Body Diagrams........................................................................................9 2.4 Point loads................................................................................................................................9 2.5 Examples................................................................................................................................10 2.5.1 Simply supported beam..................................................................................................10 2.5.2 Inclined Force................................................................................................................12 2.6 Distributed Loads...................................................................................................................14 2.7 Examples................................................................................................................................15 2.7.1 Constant load over a beam.............................................................................................15 2.7.2 Linear load over a beam.................................................................................................18 2.8 Bending Moment and Shear Force Diagrams........................................................................20 2.8.1 Example of a Simply supported beam...........................................................................20 2.8.2 Example of a Cantilever Beam......................................................................................24 2.8.3 Example of a cantilever beam with partial distributed load...........................................28 2.9 The Concept of Equivalent Load – A Warning!.....................................................................32 3 Frames and Trusses – The Concepts of Internal Forces................................................................33 3.1 Equilibrium of Frames...........................................................................................................33 3.1.1 Example: Two-elements frame.......................................................................................33 3.1.2 Example: Two-elements frame with pinned connection................................................34 3.2 Static Determinancy...............................................................................................................36 3.3 Internal Forces in Frames......................................................................................................38 3.3.1 Example of a simple frame............................................................................................38 3.4 Equilibrium of Trusses...........................................................................................................40 3.5 Element Forces of Trusses – The Sectioning Method...........................................................41 3.6 Equilibrium of Truss Joints....................................................................................................45 4 Stress, Strain, and Elasticity..........................................................................................................47 4.1 What is Stress?.......................................................................................................................47 4.2 Two Types of Stresses............................................................................................................47 4.3 What is Strain?.......................................................................................................................48 4.4 Types of Strain.......................................................................................................................49 4.4.1 Axial Strain....................................................................................................................49 4.4.2 Shear Strain – Engineering and Mathematical...............................................................49 4.4.3 Shear Strain Compatibility.............................................................................................50 4.5 Hooke’s Law – Constitutive Relations..................................................................................50 4.6 The Concepts of Linearity.....................................................................................................51 4.7 Poisson’s Ratio.......................................................................................................................51 4.7.1 Main Concept.................................................................................................................51 Mechanics of Materials Mohammad Tawfik

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4.7.2 Generalized Hook’s Law................................................................................................52 4.7.3 Hydro-static Problem.....................................................................................................53 4.7.4 Auxetic Materials...........................................................................................................54 4.8 Examples................................................................................................................................54 4.8.1 Metal block with two fixed directions...........................................................................54 4.9 Yield and Ultimate Stresses...................................................................................................54 4.10 Brittle and Ductile Materials................................................................................................54 4.11 Margin of Safety and Design Considerations......................................................................54 4.12 The Concepts of Homogeneity and Isotropy.......................................................................55 4.12.1 Homogeneous Material................................................................................................55 4.12.2 Isotropic Material.........................................................................................................55 5 Mechanics of Bars.........................................................................................................................56 5.1 One-Dimensional Structures..................................................................................................56 5.2 Stress and Average Stresses...................................................................................................56 5.3 Examples................................................................................................................................57 5.3.1 End load on a bar...........................................................................................................57 5.3.2 Compound bar................................................................................................................59 5.3.3 Compound bar with multiple loads................................................................................61 5.4 Strain and Deflection.............................................................................................................64 5.5 Examples................................................................................................................................65 5.5.1 Resolving example 5.3.1 using the differential equations.............................................65 5.5.2 Resolving example 5.3.2 using the differential equations.............................................67 5.5.3 Bar under its own weight...............................................................................................69 5.6 Indeterminate Structures........................................................................................................71 5.7 Examples................................................................................................................................72 5.7.1 Compound bar fixed from both sides.............................................................................72 5.8 Thermal Strain.......................................................................................................................77 5.9 Examples................................................................................................................................78 5.9.1 Resolving example 5.3.1 with thermal field..................................................................78 5.9.2 Compound bar with thermal field..................................................................................79 5.10 Extending the Concept of Thermal Strain...........................................................................84 6 Mechanics of Shafts......................................................................................................................85 6.1 Torsion and Main Assumptions.............................................................................................85 6.2 Deflection of Shafts...............................................................................................................85 6.3 Indeterminate shafts...............................................................................................................85 6.4 Closed vs. Open Section Shafts.............................................................................................85 7 Mechanics of Beams......................................................................................................................86 7.1 Assumptions of Euler-Bernoulli Beam Theory......................................................................86 7.2 Stress Distribution in a Thin Beam........................................................................................88 7.3 Equilibrium, the Neutral Axis, and Second Moment of Area................................................88 7.3.1 Horizontal Force Equilibrium........................................................................................88 7.3.2 Vertical Force Equilibrium.............................................................................................89 7.3.3 Moments’ Equilibrium...................................................................................................90 7.4 Non-Homogeneous Beams....................................................................................................91 7.5 Load Deflection Relations.....................................................................................................91 7.6 Examples................................................................................................................................94 7.6.1 Simply-Supported beam with concentrated load...........................................................94 7.6.2 Simply-supported beam with linear distributed load.....................................................97 Mechanics of Materials Mohammad Tawfik

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7.7 Indeterminate Beams...........................................................................................................100 7.7.1 Example of Fixed-Pinned with Constant distributed Load..........................................100 7.7.2 Example of Fixed-Pinned with point load...................................................................102 7.8 Axial and Transverse Loading Problems.............................................................................108 7.9 Deflection of Frames...........................................................................................................109 8 Shear Stresses in Beams..............................................................................................................110 8.1 Axial Equilibrium of Beams Under Shear...........................................................................110 8.2 Examples..............................................................................................................................112 8.2.1 Shear stresses distribution in a rectangular cross-section............................................112 8.2.2 Shear streses in a thick I-Beam....................................................................................113 8.2.3 Shear stresses in a box cross-section............................................................................116 8.2.4 Shear stress distribution in a thick I-Beam...................................................................118 8.2.5 Shear stress distribution in a box cross-section............................................................121 8.3 Shear Flow and Shear Center...............................................................................................123 9 Combined Stresses.......................................................................................................................124 9.1 Rotation of Axes..................................................................................................................124 9.2 Mohr’s Circle.......................................................................................................................124 9.3 Stresses in Thin-Walled Pressure Vessels............................................................................124 9.4 Three Dimensional Combined Stresses Case......................................................................124

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Main concepts

1 Main concepts 1.1 Equilibrium Equilibrium is the law that governs mechanics of materials! Usually, we refer to equilibrium as “the state of a body (or a system of bodies) at which the summation of all externally applied forces and moments are equal to zero”. But that is what we may call, also, static equilibrium. In this case, the body does not move at all, or, moves with a constant velocity.

More generally, since we said that Equilibrium is a law, we may also consider the state where the forces applied are related to the accelerations. Such case may be called dynamic equilibrium. Notice that the static equilibrium is a special case of dynamic equilibrium at which the accelerations are equal to zero.

In more sophisticated terms, we can say that Newton’s Second Law of motion is the law that governs our work in the field of mechanics of material. In such a spirit, we need to remember that the first thing we always need to satisfy, in this field of work, is Equilibrium.

1.2 Deformation and Deflection Strictly speaking, ALL solids deform when subjected to loading. A tree deforms with air blowing; your seat deforms as you sit on it; the road deforms as a car passes on it; and the great pyramids of Giza deform if you stood on the top of them! The difference between all those solid deformations lies in their stiffness (their ability to resist deformation). That is why we may call some solids rigid (do not deform) when the loads applied on them cause extremely small deformations that will make it very hard for us to measure.

On the other hand, when we mention deflection, we are concerned, usually, with how a single point on, or inside, the solid moves. Hence, we may say that the tip of the tree branch moved 20 cm to the north, the tip of the pyramid moved down 3 nm when you stood on it, or the end of the shaft rotated 5 degrees when you twisted it. Thus, sometimes, it is convenient to present the displacements in a vector form as well as scalar forms. In contrast with rigid bodies, we call solids that deform under loading elastic bodies. Let me remind you at this point that there is no real boundary between elastic and rigid bodies, rather, “does the load create a noticeable deformation to the solid?” should be the question we ask.

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Main concepts

1.3 Compatibility of Structures The compatibility means that the structure deforms in a way that keaps it in one piece (no fracture), it does not detach from the support (zero or fixed value deflections at the supports), and its adjacent points deform in a manner that can be described by a continuous function.

1.4 Input-Output Relations for Structures If a piece of wood is just sitting with no external effects on it, nothing happens, that is not an interesting problem for us! However, if you stand on it or put some loads on it, it may deform or even break down, that is interesting.

It is not common to talk about structures in terms of inputs and outputs, though, I prefer to use this analogy to describe the load-deformation relations for the structures. The piece of wood we mentioned earlier, does not deform or break down unless a load is applied. Hence, we may consider that the load is an input to the structure. For equilibrium, the structure generates internal stresses which, in turn, causes strain. The strain is not measured directly, rather, it is manifested in deformation of the structure which may, then, be measured. From that, we may refer to the “deformations” as the output of the structure.

This concept, though considered alien to structural analysis, is quite important from my point of view. Since every problem involving structures, starts with an applied load and ends with a measured deformation, we can always see the direction in which we are progressing.

Figure 1.4.1: Input-Output relation for structures

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Supports, Reactions, and Loads

2 Supports, Reactions, and Loads 2.1 What is a Support? The word “support” may have different meaning in different contexts. As far as we are concerned, in mechanics of materials, a support is anything that prevents one or more form of motion of any part of the structure. The “prevention” does not have to be full stopping of motion (rigid support) rather, it may be also flexible support such as the ground under a building or the bed mattress under your body.

According to this definition, any part of the structure maybe considered as a support to the adjacent parts of the structure. Similarly, any structure carrying or holding another structure maybe considered a support to it.



However, in most of our discussions, we will be considering the supports to be rigid.

2.2 Types of Supports At this point we will strict our discussion to 2-D problems. There are three main types of rigid supports that are important to 2-D problems; namely, roller support, pinned support, and fixed support.

Figure 2.2.1: Types of 2-D supports The roller support is one that acts as if the structure is resting on a perfect cylinder that moves freely on a horizontal, vertical, or inclined surface. That cylinder is assumed not to create any kind of friction with the structure or the surface. However, the roller support in never allowed to move in a direction normal to the surface on which is rests. Thus, if we have a roller that is resting on a horizontal surface, it will move horizontally with no resistance, but, it is never allowed to move Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads vertically. The roller support is assumed to create a reaction force normal to the surface in order to prevent the structure from moving in that direction.

Meanwhile, a pinned support is one that acts as if the structure attached to it is prevented from moving any linear motion. That is to say that the point at which the pinned support is connected can not move in the horizontal or vertical conditions, but, it may rotate freely around the pin. In order for the pinned support to perform its function of preventing linear motions, it may need to create up to two reaction forces (in two normal directions) to prevent the structure from moving. However, it can never create any moments around the point of fixation (the pin).

Finally, the fixed support prevents all kind of motion in the plane. At the point the structure is connected to a fixed support, the structure can not have any linear motion or rotation. Hence, the fixed support is assumed to create up to two forces to prevent the motion of the structure as well as a moment to prevent it from rotating.

2.3 Reactions and Free-Body Diagrams A reaction is a force or a moment exerted by the support to prevent the point, at which the support is connected, from moving. The amount of force or moment is that needed to ensure equilibrium of the structure. When you need to calculate the reactions in a support, you pretend that the support is removed then replace each restricted motion with a force or a moment. That is what we call the free body diagram. When you have a free body diagram you can readily write down the equilibrium equations and solve them, if feasible, for the values of the reactions.

2.4 Point loads Point loads are NOT real! The contact between any two bodies can never occur, in reality, at a perfect point (zero area). However, when the load, applied to the structure, occupies a tiny area of the structure surface, we may assume that it is a point load. That is an extremely convenient assumption that allows us to analyze many cases of loading. For example, as you plate a bottle of water on a table, the bottle occupies an area of the table surface, however, if you want to calculate the reactions created in the table legs, assuming that the bottle occupies a single point on the surface of the table will simplify your calculations and provide you with accurate results.

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Supports, Reactions, and Loads

Figure 2.4.1: Types of point loads We have two types of point forces in 2-D problems. Point forces and point moments. Point forces may be perfectly aligned with the coordinate directions or not, thus, we usually resolve any inclined force into two forces parallel to the horizontal and vertical directions. Meanwhile, the point moment is always parallel to the third direction that is normal to the plane of the two dimensions we are concerned with.

2.5 Examples 2.5.1 Simply supported beam A beam is supported by a roller at point A and a pin at point B. The beam length is L. A point vertical force is applied at point C which is a distance d from point A. Calculate the reactions at point A & B.

Figure 2.5.1: Simply supported beam with point load

Solution: First step in solving the problem will be drawing the free body diagram of the beam. To do that we assume that the supports are removed and replaced by forces and moments that correspond to the supports. Point A is supported on a roller which can produce one force normal to the surface

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Supports, Reactions, and Loads (Horizontal surface), hence, we replace the roller with a vertical force Ay. The pinned support, on the other hand, creates two forces, hence, we replace it with Bx and By.

Figure 2.5.2: Free-body diagram Now, we may write down the equations of equilibrium. Σ F x =0, Σ F y =0 , Σ M z=0

For the equilibrium in the x-direction, we add up all the horizontal forces Σ F x =B x =0 From this equation, we can easily declare that the horizontal reaction at point B is zero. For the equilibrium in the y-direction, we add up all the vertical forces Σ F y = A y + B y −F y =0 Finally, we apply the equation of the moment equilibrium NOTE: When applying the moment equilibrium equation we have to: 1- Select a point at which the equation will be applied – Any point in the plane will will suffice, however, for convenience, we usually select a point that has some forces applied to in order to reduce the number of variables in the equations 2- Select the direction which will be considered the positive direction – It is advised to stick to the convention of the Cartesian coordinate system to create consistency in your work, but if you decided not to follow that convention, the results will still be correct as long as you are careful enough in setting the positive and negative signs in the equation. Here we select the point A and the counter-clockwise direction for the moment equation. Σ M z =−F y . d+ B y . L=0 A

From this equation, we may evaluate the reaction at point B to be Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads B y =F y .

d L

Now, we can use this result into the equation of vertical force equilibrium to get A y =−B y + Fy d L−d A y =F y −F y . =F y . L L Observation: From the above results, we can see that the vertical reaction force at points A&B depend on the distance d. If d becomes too small (almost zero) the reaction at point A will equal to the force Fy, while at be it becomes zero, and vise versa. In other words, we may say that the closer the force to the support, the more the reaction force in the support becomes.

2.5.2 Inclined Force A beam is supported by a roller at point A and a pin at point B. The beam length is L. A point inclined force is applied at point C which is a distance d from point A. Calculate the reactions at point A & B.

Figure 2.5.3: Simply-supported beam with inclined point load

Solution: First step in solving this problem will be resolving the force into its horizontal and vertical components

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Supports, Reactions, and Loads

Figure 2.5.4: Resolving the point force into its horizontal and vertical components

The values of the components may be calculated using the relation: F x =F . cos( θ), F y =−F . sin(θ) Now we are abe to draw the free body diagram of the beam. To do that we assume that the supports are removed and replaced by forces and moments that correspond to the supports. Point A is supported on a roller which can produce one force normal to the surface (Horizontal surface), hence, we replace the roller with a vertical force Ay. The pinned support, on the other hand, creates two forces, hence, we replace it with Bx and By.

Figure 2.5.5: Free-body dagram Now, we may write down the equations of equilibrium. Σ F x =0, Σ F y =0 , Σ M z=0

For the equilibrium in the x-direction, we add up all the horizontal forces Σ F x =B x + F x =B x + F . cos(θ) From this equation, we can calculate the horizontal reaction at B: B x =−F .cos (θ) Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads Note: The negative sign appearing the value of the reaction force indicates that the force is acting in the opposite direction of what we have assumed in the free body diagram. For the equilibrium in the y-direction, we add up all the vertical forces Σ F y = A y + B y + F y =0 ==>

A y + B y −F . sin(θ)=0

Finally, we apply the equation of the moment equilibrium Here we select the point A and the counter-clockwise direction for the moment equation. Σ M z =−F . sin(θ).d + B y . L=0 A

From this equation, we may evaluate the reaction at point B to be B y =F . sin(θ) .

d L

Now, we can use this result into the equation of vertical force equilibrium to get A y =−B y + F . sin(θ)=0 A y =F .sin (θ).

L−d L

2.6 Distributed Loads When the area over which the load is acting becomes large enough, relative to the structure, we can not assume point loading anymore. For example, when you lie on your bed, can select a point upon which you are applying your weight? When an aircraft wing is loaded during flight, can you select a point at which the air is applying its forces?

In those, and similar cases, we need to present the force as a distributed force, which is more realistic as noted in the previous section. In such cases, the forces are presented by a distribution function. For example, the bottom surface of a water tank can be shown as loaded by the weight of the water in the tank distributed uniformly over the area, in this case, the load will be presented as force per unit area (pressure).

The distribution function of the force does not have to be a constant function, thus to evaluate the force at any point, we can only obtain the load density using the given function. Consequently, if you want to evaluate the total load applied on the structure, you will need to integrate the loading function over the entire structure (area under the curve). This fact may complicate the problem of evaluating the support reactions. Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads

2.7 Examples 2.7.1 Constant load over a beam A beam is supported by a roller at point A and a pin at point B. The beam length is L. A constant downward distributed load f ( x)=−f o is applied on the beam covering its whole span. Calculate the reactions at point A & B and the total load carried by the beam.

Figure 2.7.1: Simply-supported beam with constant distributed load

Solution: First step in solving the problem will be drawing the free body diagram of the beam. To do that we assume that the supports are removed and replaced by forces and moments that correspond to the supports. Point A is supported on a roller which can produce one force normal to the surface (Horizontal surface), hence, we replace the roller with a vertical force Ay. The pinned support, on the other hand, creates two forces, hence, we replace it with Bx and By.

Figure 2.7.2: Free-body diagram Now, we may write down the equations of equilibrium. Σ F x =0, Σ F y =0 , Σ M z=0

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Supports, Reactions, and Loads For the equilibrium in the x-direction, we add up all the horizontal forces Σ F x =B x =0

From this equation, we can easily declare that the horizontal reaction at point B is zero. For the equilibrium in the y-direction, we add up all the vertical forces. BUT, the distributed force can not be added directly, we will need to use integration L

Σ F y = A y + B y −∫ f o . dx=0 0

Which can be readily evaluated, giving: A y + B y −f o . L=0 Finally, we apply the equation of the moment equilibrium Here we select the point A and the counter-clockwise direction for the moment equation. BUT, again, we will need to use integration for the distributed load. In this case, the moment applied but each little bit of force may be presented as f o . dx . x and a negative sign will be used because it creates a clockwise moment L

Σ M z =−∫ f o . x . dx +B y . L=0 A

0

Figure 2.7.3: The effect of the destributed force at an infinisimal part of the beam Which can be evaluated to be: −f o .

L2 +B y . L=0 2

From this equation, we may evaluate the reaction at point B to be B y =f o .

L 2

Now, we can use this result into the equation of vertical force equilibrium to get Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads A y =−B y + f o . L L L A y =−f o . + f o . L=f o . 2 2 To evaluate the total force applied on the beam, we perform the integration: L

FTotal =∫ f o . dx=f o . L 0

NOTE: 1- The reactions of the supports are equal, which is what we should expect in such a symmetric problem. 2- Also, the total load turned out to be, simply, the total length of the beam multiplied by the values of the constant distributed load. 3- The value of the moment created by the distributed load is equal to the total force multiplied by half the beam length (looks like the area multiplied by the distance from the centroid).

2.7.2 Linear load over a beam A beam is supported by a roller at point A and a pin at point B. The beam length is L. A linear x downward distributed load f ( x)=−f o (1− ) is applied on the beam covering its whole span. L Calculate the reactions at point A & B and the total load carried by the beam.

Figure 2.7.4: Simply-supported beam with linearly distributed load Solution: First step in solving the problem will be drawing the free body diagram of the beam. To do that we assume that the supports are removed and replaced by forces and moments that correspond to the supports. Point A is supported on a roller which can produce one force normal to the surface (Horizontal surface), hence, we replace the roller with a vertical force Ay. The pinned support, on the other hand, creates two forces, hence, we replace it with Bx and By. Mechanics of Materials Mohammad Tawfik

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Supports, Reactions, and Loads

Figure 2.7.5: Free-body diagram Now, we may write down the equations of equilibrium. Σ F x =0, Σ F y =0 , Σ M z=0

For the equilibrium in the x-direction, we add up all the horizontal forces Σ F x =B x =0 From this equation, we can easily declare that the horizontal reaction at point B is zero. For the equilibrium in the y-direction, we add up all the vertical forces. BUT, the distributed force can not be added directly, we will need to use integration L

x Σ F y = A y + B y −∫ f o ( 1− ). dx=0 L 0 Which can be readily evaluated, giving: L A y + B y −f o . =0 2 Finally, we apply the equation of the moment equilibrium Here we select the point A and the counter-clockwise direction for the moment equation. BUT, again, we will need to use integration for the distributed load. In this case, the moment applied but each little bit of force may be presented as f ( x).dx . x and a negative sign will be used because it creates a clockwise moment L

x Σ M z =−∫ f o (1− ). x . dx +B y . L=0 L 0 A

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Supports, Reactions, and Loads

Figure 2.7.6: The effect of the distributed load Which can be evaluated to be: −f o .

L2 +B y . L=0 6

From this equation, we may evaluate the reaction at point B to be B y =f o .

L 6

Now, we can use this result into the equation of vertical force equilibrium to get A y =−B y + f o .

L 2

L L L A y =−f o . + f o . =f o . 6 2 3 To evaluate the total force applied on the beam, we perform the integration: L

x L FTotal =∫ f o (1− ) . dx=f o . L 2 0 NOTE: 1- The total load turned out to be, simply, the area of the triangle. 2- The value of the moment created by the distributed load is equal to the total force multiplied by one third the beam length (AGAIN, looks like the area multiplied by the distance from the centroid).

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Supports, Reactions, and Loads

2.8 Bending Moment and Shear Force Diagrams Remember that equilibrium is a law. The law of equilibrium has to apply to each and every part of the structure, not only for the whole structure. If you draw the free body diagram of a part of the structure, it has to have internal forces that are in equilibrium with the externally applies loads as well.

2.8.1 Example of a Simply supported beam A beam is supported by a roller at point A and a pin at point B. The beam length is L. A point vertical force is applied at point C which is a distance d from point A. Calculate the internal shear force and bending moment at an arbitrary point D.

Figure 2.8.1: Simply supported beam with point load

Solution: First step for solving this problem is to resolve example 2.5.1. The reaction forces were found to be (see Figure 2.8.2): A y =F y .

L−d L

B x =0 B y =F y .

d L

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Supports, Reactions, and Loads

Figure 2.8.2: Free-body diagram

Now, we draw the free body diagram for a part of the beam that ends at point D. Since we have a concentrated point load, we will divide the problems into two parts, first part is for xd (we will start by measuring the distance from support A).

Figure 2.8.3: Free-body diagram for the the part [A,D] for xd), we draw a similar free body diagram (Figure 2.8.4) and apply the equilibrium.

Figure 2.8.4: Free-body diagram for the the part [A,D] for x>d

∑ F y= A y −S−F y =0 S=F y

L−d d −F y =−F y L L

∑ M z=M −Sx−F y d=0 M =Sx+ F y d →M =F y d−F y

( )

d x x=F y d 1− L L

If we plot the functions obtained for the shear force and moment, we get Figure 2.8.5 and Figure 2.8.6 respectively.

Figure 2.8.5: Shear force diagram

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Supports, Reactions, and Loads

Figure 2.8.6: bending moment diagram These are what we call the Shear-Force Diagram and Bending-Moment Diagram. Thus, we may define the shear-force diagram as a graphical presentation for the internal shear force distribution and the bending-moment diagram as a graphical presentation of the internal moment distribution.

Notes: 1- The shear-force distribution has the value of the reaction forces at the supports 2- The shear-force has a discontinuous change at the point of the application of the point-force. The change is value is equal to the magnitude of the force 3- The bending moment reaches the values of zero at the roller and pin supports 4- At the point of the force application, the bending moment has equal calues from both sides while changing the rate of change from increasing to decreasing.

Observation: Integrating the functions of the shear force, you will get the bending moment functions. (see section 7.3.3)

2.8.2 Example of a Cantilever Beam A beam is fixed at point A and free at point B. The beam length is L. Two point vertical forces are applied at points C & E which are at distances d1 & d2 respectively from point A. Calculate the internal shear force and bending moment at an arbitrary point D.

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Figure 2.8.7: Cantilever beam with two point loads

Solution: First step for solving this problem is to solve for the reactionsDrawing the free-body diagram, we get Figure 2.8.8.

Figure 2.8.8: Free-body diagram Applying the equilibrium:

∑ F y= A y−F 1 + F 2=0 A y =F 1−F 2

Note that the direction of Ay will depend on the relative magnitudes of F1 and F2. For the momment equilibrium around point A, we get:

∑ M z=−M A −F 1 d 1+ F 2 d 2=0 M A=−F 1 d1 + F 2 d 2 Now, we draw the free body diagram for a part of the beam that ends at point D. Since we have two concentrated point loads, we will divide the problems into three parts, first part is for x