Summary
A fixed support has three reactions (axial load, shear force and bending moment), eg a column fixed firmly to a base.
A pin support has two reactions (an axial load and a shear force), eg a cable pinned to a wall.
A roller has only one reaction (a load perpendicular to the sliding direction), eg a drawer in your desk.
Plate 11 Steel frame structure showing beams, rafters and columns
Plate 12 Footbridge beams and deck details
Plate 13 Column-to-base connection
Plates 11, 12 and 13 show how structural members are connected to one another and how each type of connection affects the behaviour of the structure.
2.3 Load types
Table 2 summarises common types of loads and some practical examples that are common for beams.
Table 2 Description of load types and their practical examples
Loads are mostly shown by means of arrows. The direction of the arrow (where the arrow is pointing) shows the direction of the load, and the point where the arrow is pointing is the location where the load is applied.
Uniformly distributed loads (UDL) are shown by means of lined boxes. They are either applied over the full length of the beams (self-weight or floor-loading), or over part of the beam (eg a beam or a girder supporting a wall).
A bending moment is normally shown by means of a bent arrow. The position of the arrow shows the location where the bending moment is applied, and the direction of the arrow shows the direction of the bending moment (clockwise or anticlockwise). Bending moments normally result from loads applied to other members connected to the beam.
2.4 General beam types
Table 3 summarises the main beam types and applied loading commonly found in building structures.
A simply supported beam is a beam which sits on a pin at one end and a roller at the other end.
A pin support does not allow any movement, but allows rotation (eg a rod connected to a hook in a wall). In steel structures, a pin is also represented by a single bolt.
A roller support allows for movement of the structure in one direction (in the case of beams, in a horizontal direction). It allows the structure to breathe (like expansion joints in a bridge). This relaxes excessive axial forced caused, for example by changes in temperature, to a beam.
If you have difficulties in understanding support types and type of forces resisted (support reactions) by each support type, please go back to Section 2.2 and review Table 1 and the exercise in that section.
Table 3 presents simply supported beams with common load types. The first column gives a brief description of the beam type and the applied loading.
The second column gives a pictorial description (line diagram) of the beam and loading that structural engineers use in their calculations.
The last column presents a diagram of the applied loads (actions) and support reactions to the applied loading. Note that support reactions are shown by arrows (vectors), which show their magnitude and direction.
Some examples of beams, other than simply supported, are shown in Table 4. Descriptions of table columns are the same as in Table 3.
Table 3 Description of simply supported beams with common load types
Other common beam types are cantilever beams and fixed beams. A cantilever beam is a fixed beam, which does not allow any movement or rotation. A good example of a cantilever beam is a beam under a balcony.
In a fixed beam, both ends of the beam are totally fixed (this does not allow any movement or rotation). A good example of a fixed beam is a beam welded firmly to columns or a reinforced concrete column firmly fixed to a wall at both ends.
Table 4 shows examples of cantilever and fixed beams with common load types.
Table 4 Description of beams other than simply supported
2.5 Equilibrium
Equilibrium is a very important concept. Without knowledge of equilibrium, engineers, architects and building professionals would not be able to understand the behaviour of the structures they designed. This could have disastrous consequences. It is very important that you understand this concept fully.
2.5.1 Definition of equilibrium
In simple terms, equilibrium means that an object is sitting stably in its position without falling or rolling. A stone on a floor is in equilibrium status. An object hanging by a rope is in equilibrium by its own weight and the force resisted by the rope. A seesaw is in equilibrium by balanced equal weights at each end.
For example, when you are standing on a floor, your weight is applied directly downward to the floor. This causes an upward reaction from the floor which is equal to your weight. If you place a weight on your hand, you feel the pressure of the weight in your hand and your hand pushes the weight up to keep the balance. The weight pushing your hand downwards is called the action and the upward force that your hand applies to the weight is called the reaction.
In order to keep the weight still balanced on your hand, the downward force that the weight applies to your hand must be equal to the upward force that your hand reacts on the weight. As a result of this, equilibrium status is maintained and the weight is balanced on your hand (to each action there is an equal and opposite reaction).
In structures, equilibrium means that the sum of all reactions must be equal and opposite to the sum of the applied loading.
2.5.2 Equations of equilibrium
In Section 2.1, it was mentioned that there are three types of reactions (axial force, shear force and bending moment) associated with a beam. Generally, for a structural model there are three equations of equilibrium, with one associated reacting force.
horizontal equilibrium (sum of all horizontal forces must be zero – axial load in beams)
vertical equilibrium (sum of all vertical forces must be zero – shear force in beams)
moment equilibrium (sum of moments, caused be all forces must be zero at any point on the structure).
Structures are naturally in a state of equilibrium. A structure or a structural element is in equilibrium if, and only if, all three equations of equilibrium are satisfied simultaneously.
2.5.3 Application of equations of equilibrium in simple beam models
Table 5 Definition of terms
Example
Apply three equations of equilibrium to the following beam
.
Figure 21 Equations of equilibrium
In Figure 21 a force of 100kN is applied at the centre of a simply supported beam spanning 10 metres. There is a pin support at the left-hand end of the beam at point A, which prevents the movement of the beam in both horizontal and vertical directions. Therefore, this support resists a horizontal force (axial load) HA, and a vertical force VA (shear force). There is a roller support at the right-hand end of the beam. This support only prevents the movement of the beam in the vertical direction. Therefore, it resists only a vertical force VB.
Apply the three equations of equilibrium:
1. Horizontal equilibrium
Sum of all horizontal forces must be zero:
There is no horizontal load applied to the beam, therefore HA = 0.
2. Vertical equilibrium
Sum of all vertical forces must be zero: VA + VB = 100
(sum of vertical reactions must be equal to vertical applied load of 100kN).
Although visually we know that the load is applied to the centre of the beam, therefore the vertical reactions at each end must be 50kN:
(VA = VB = 50kN)
Generally, the proportion of the load resisted by each support is determined by applying the third equation of equilibrium.
3. Moment equilibrium
Sum of moments, caused be all forces, must be zero at any point on the structure.
Take moment about point A:
VB x 10 (anticlockwise) = 100 x 5 (clockwise) ➔ VB = kN
Substituting this into the second equation of equilibrium:
VA + VB = 100 ➔ VA = 100kN.
2.6 Support reaction calculation
In this section we will look at the concept of determinacy. You need this to calculate support reactions.
2.6.1 Understanding determinacy
In Section 2.1 you learned that for a 2D beam we have only three types of reaction. In Section 2.4 the concept of equilibrium was introduced. You also learned that the three equations of equilibrium must be satisfied simultaneously for a 2D beam to be in equilibrium.
To recap, here are the three equilibrium equations:
- horizontal equilibrium (sum of all horizontal forces must be zero)
2 vertical equilibrium (sum of all vertical forces must be zero)
3 moment equilibrium (sum of moments, caused be all forces, must be zero at
any point on the structure).
If the number of unknown reactions is either one, two or three, then the structure is determinate. This means that you can use three equations to solve the problems.
If there are more than three unknown reactions, then the structure is indeterminate. This means that you cannot use the three equilibrium equations to solve the problems.
This learning pack deals with determinate structures, so you do not have to worry about solving problems for indeterminate structures. You need to be able to tell the difference between determinate and indeterminate structures.
Example
Show that the following structure is determinate.
There are three unknowns (HA, VA, and VB). These unknowns can be solved by three equations of equilibrium, therefore, the structure is determinate.
2.6.2 Calculation of support reactions
This section gives three examples of how to calculate support reactions.
Example 1
Calculate support reactions for the following structure.
Solution:
Apply equations of equilibrium.
-
Sum of horizontal forces = 0 ➔ HA = 0
(No applied horizontal load no horizontal reaction – OK)
- Sum of vertical forces = 0
VA + VB = 100
- Sum of bending moment about B = 0
VA * 10 = 100 * 5 ➔ VA = 50kN
substitute in (2) ➔ VB = 100 – 50 = 50kN
This makes sense as the load is applied at the centre of the beam and half the load is resisted by each support.
Example 2
Calculate support reactions for the following structure.
At point A, the cantilever beam has a fixed support. A fixed support, by definition, prevents the beam from movements horizontally and vertically, and it also prevents the beam from rotation at this point. Therefore, there are three reactions at point A: A horizontal reaction (axial load), a vertical reaction (the shear force) and a bending moment (due to bending prevention).
Solution:
Apply equations of equilibrium.
1. Sum of horizontal forces = 0 ➔ HA = 0
(no applied horizontal load no horizontal reaction – OK)
2. Sum of vertical forces = 0
VA = 100kN.
This makes sense as the applied load is resisted by the only support at A.
3. Sum of bending moment about A = 0
MA = 100 x 5 ➔ MA = 500kN-m
Example 3
Calculate support reactions for the following structure.
Figure 25 Structure
Solution
Apply equations of equilibrium.
1. Sum of horizontal forces = 0 ➔ HA = 0
(No applied horizontal load no horizontal reaction – OK)
2. Sum of vertical forces = 0
VA + VB = 100 x 10 = 1000kN.
Note: For a UDL, the resulting vertical downward load from the applied UDL is = load intensity times length, over which the load is applied, in this case the whole length of the beam 10m (eg 100 x 10 = 1,000kN).
3. Sum of bending moment about B = 0
VA x 10 = (100 x 10) x 5 ➔ VA = 500kN
substitute in (2) ➔ VB = 1,000 – 500 = 500kN
This makes sense as the load is applied uniformly on the beam, and half the total applied load is shared by each support.
Activity 10 is designed for you to calculate support reactions.
Follow the guidance given in question (a) when tackling the following examples.. Note that you must apply the 3 equations of equilibrium, and make sure that the sum of all applied loading is equal and opposite to the sum of the reactions in each direction.
2.7 Shear force and bending moment diagrams
When you have learned how to do support reactions calculations, the next stage is to draw shear force and bending moment diagrams.
Use the support reactions and three equations of equilibrium at key locations on the beam to calculate shear force and bending moments.
Once you have calculated the shear force and bending moments you will be able to draw the shear force and bending moment diagrams.
2.7.1 Shear force diagram
The examples showing how to do shear force diagrams are given below.
Example 1
Draw the shear force diagram for the following structure. Ignore the self-weight of the beam.
Figure 26 Structure
- Calculate support reactions (for details refer to Section 2.5)
Apply equations of equilibrium.
1. Sum of horizontal forces = 0➔ HA = 0
(No applied horizontal load no horizontal reaction – OK)
2. Sum of vertical forces = 0
VA + VB = 100
3. Sum of bending moment about B = 0
VA x 10 = 100 x 6 ➔ VA = 60kN
substitute in (2) ➔ VB = 100 – 60 = 40kN
Note that the support closer to the applied load resists more reaction. This makes sense.
(B) Starting from the left-hand support, cut the beam at an arbitrary point at a distance X from the left-hand support, and draw all forces acting on both parts of the beam. A section of the beam, showing all forces acting on them, is called the free-body diagram for that section. For simplicity, bending moment is not shown. Note that at the cut, there is an equal and opposite vertical shear force (Vx) to maintain the equilibrium. This is an essential condition for equilibrium.
Figure 27 Structure
Figure 27 shows the free-body diagrams for the beam sections at both sides of the cut. By now it should be clear to you that all three equations of equilibrium must be satisfied, for both sections, at either sides of the cut.
(C) Apply equations of equilibrium to both sides of the cut.
There is no applied horizontal force, therefore there is no need for the first equation of equilibrium (sum of horizontal forces must be zero).
Applying the second equation of equilibrium (sum of vertical forces must be zero):
LH part: Sum of vertical forces = 0 VX = 60kN
RH part: Sum of vertical forces = 0 VX = 100 – 40 = 60kN
This means that the shear force, at any point at the LH side of the load, is 60kN, and at the RH side of the load is 40kN.
(D) Now draw the shear force diagram (SFD)
Figure 28 Structure
Explanation
- Starting from the left-hand support, there is a shear force of 60kN upwards at A (this is the vertical reaction at support A). Move upwards 60 units.
- There is no load applied between the LH support (point A), and the point of application of the load (point C). Draw a horizontal line from A to C. This line represents a constant shear force of 60kN from A to C. (Note that the self-weight of the beam is ignored for this problem).
- At C there is a 100kN point load downwards. Move 100 units downwards. (Note that the shear force to the left of C is 60kN, and the shear force at the right of C is (100 – 60 = 40kN)).
- There is no load applied between C and B, and so draw a straight line between C and B. Once again, this line represents a constant shear force of 40kN from C to B
- At B there is a 40kN upwards shear force (vertical support reaction at B). Move upwards 40 units.
- At this point, the diagram must close. If not, something is wrong and you must check your calculations.
Figure 29 Shear force diagram (SFD)
Deeper understanding of a shear force diagram
Calculate Area 1 and Area 2
Area 1 = 60 * 4 = 240kN-m
Area 2 = 40 * 6 = 240kN-m
Conclusions
Areas under the shear force diagram at top and bottom sides of the beam are equal. This is an important concept and a good check for the validity of your calculations.
Example 2
Draw the shear force diagram for the following structure.
Figure 30 Structure
(A) Calculate support reactions
Apply equations of equilibrium.
1. Sum of horizontal forces = 0➔ HA = 0
(No applied horizontal load no horizontal reaction – OK)
2. Sum of vertical forces = 0
VA + VB = 100 *10 = 1,000kN.
3. Sum of bending moment about B = 0
VA * 10 = 1(100*10) * 5 ➔ VA = 500kN
substitute in (2) ➔ VB = 1,000 – 500 = 500kN
(B) Starting from the left-hand support, cut the beam at an arbitrary point, at a distance X from the support, and draw the free body diagrams for both parts.
Figure 31 Structure
(C) Apply equations of equilibrium to the LH side of the cut.
Note that the force, due to a uniformly distributed load, on the beam is equal to load intensity (100kN/m in this example; this means 100kN for each metre of length) multiplied by the length on which the load is applied (eg the total applied for 3m of length would be 100 x 3 = 300kN and for X metres of length it would be 100 X kN).
LH part: Sum of vertical forces = 0 VX = 500 –100X
500kN is the support reaction at the LH support, and 100X is the applied for X m of the beam length segment, at the left of the cut.
VX = 500 –100X is the equation for a straight line. This means that the shear force varies linearly along the length of the beam. Let us use a table to examine this.
LH side:
Note that positive Vx means an upward force, and negative means a downward force.
Example calculations to clarify the table:
If X = 0m (cut directly at LH support)
VX = 500 –100 X VX = 500 –100 x 0 VX = 500kN
If X = 1m (cut at 1m from the LH support)
VX = 500 –100 X VX = 500 –100 x 1 VX = 500 –100
VX = 400kN.
(D) Now draw the shear force diagram (SFD); plot points shown in the above table, using distance in (m) on horizontal axis, and Vx in (kN) on vertical axis, using an appropriate scale for the horizontal and vertical axes.
Once again the areas under the curve, at the top and bottom of the beam, are the same (500 * 5/2 = 1,250kN-m).
Example 3
Draw the shear force diagram for the following structure.
Figure 33 Structure
(A) Calculate support reactions
Apply the equations of equilibrium.
1. Sum of horizontal forces = 0 ➔ HA = 0
(No applied horizontal load no horizontal reaction – OK)
2. Sum of vertical forces = 0
VA = 100kN (only one vertical reaction)
3. Sum of bending moment about B = 0
MA = 100 * 6 = 600kN
(B) By inspection, it can be seen that there is a constant shear force along the length of the beam.
This means that in a cantilever beam with a point load, there is a constant shear force between the support and the point where the load is applied.
Now attempt Activity 11.
2.7.2 Bending moment diagram
In Section 2.1, the bending moment was defined as a force multiplied by a perpendicular distance to a given point. In this section, details of calculating bending moments for different types of loads are discussed. If a beam is loaded along its length, it bends (deflects downwards and curves along its entire length). It was also mentioned that, due to this bending, the top face of the beam would compress and the bottom face stretches.
The top- face of the beam (blue/green) is in tension, the bottom face (red/yellow) is in tension.
Note that the bending moment diagram is always drawn on the tension face.
Three examples of the bending moment diagram (BMD) are given below.
Example 1
Draw the BMD for the following structure.
Figure 35 Bending moment diagram
Stage 1: Sketch reactions and calculate reactions
Figure 36 Bending moment diagram
Stage 2: Starting from the left-hand support, cut the beam at an arbitrary point, at a distance X from the support, and draw the free-body diagrams for both parts of the beam. For simplicity, the shear force, discussed in the previous section, is not shown at the cut point (only bending moment (Mx) relevant to this section is shown). Once again, due to equilibrium, Mx at both sides of the cut must be equal and opposite to maintain equilibrium.
Figure 37 Bending moment diagram
Note that at the cut, the equilibrium must be maintained (ΣM = 0).
Stage 3: Apply the third equation of equilibrium (sum of moment at any point along the beam must be zero) to the LH side of the cut, assuming the counter clockwise moment is positive:
(Sum of moments at a distance X from the LH support = 0, assuming the anticlockwise moment to be positive)
- 60 * X + MX = 0 ➔ Mx = 60X
Figure 38 Diagram
The above relation is valid only from A to C, as no load is acting between these points.
Stage 4: Cut the beam at any point to the RH side of the load, and draw free-
body diagrams, as in stage 3.
Figure 39 Diagram
Stage 5: Apply the third equation of equilibrium to the RH side of the cut, assuming the counter-clockwise moment is positive:
(Sum of moments, at a distance X from the RH support = 0)
–MX + 40 * ( 10 – X) = 0
Mx = 400 – 40X
Summarise the result in a table:
LH side of the load (Mx = 60X); Linear variation.
RH side of the load (Mx = 400 – 40X); Linear variation again.
Stage 6: Draw a BMD by plotting the co-ordinates of each point starting from A. The horizontal co-ordinate of a point denotes the distance from the LH support (A) and the vertical co-ordinate denotes the BM value at the distance Xm from A. Note the BM drawn on the tension face of the beam is positive.
Deeper understanding
Refer to Example 1 in Section 2.7.1. It was shown that the area under the shear force diagram above the beam (at the LH side of the load) was 240kN-m. The area below the beam (at the RH side of the load) was also 240kN-m. Both the magnitude and units of these areas match the magnitude and the unit of the BM at the point of application of the load. Actually, another way to calculate BM would be to calculate the area under the shear force diagram.
Example 2
Draw the bending moment diagram for the following structure.
Figure 41 Bending moment diagram
- Starting from the left-hand support, cut the beam at an arbitrary point, at a distance X from the support, and draw the free-body diagrams.
Figure 42 Bending moment diagram
- Apply the third equation of equilibrium to the LH side of the cut, assuming the counter-clockwise moment is positive:
(Σ M at a distance X from LH support = 0).
–500 * X + (100 X) * (X/2) + MX = 0
Mx = 500X – 50 X2
This is the equation of a parabola (ie X2).
Starting from the LH support, calculate Mx for each value of X.
Figure 43 Bending moment diagram (UDL)
For a deeper understanding, you should test that BM can be obtained from the area under the SFD.
Example C
End-of-module – self-assessment tasks
The aim of this part is to give you a chance to test your level of understanding of the pack. If you cannot answer all of the questions, do not be discouraged as some of them are not straightforward. If you have any difficulty in getting these questions right, you should do the following:
- go back to the section that the task is related to and revise that section. Carefully read the section and pay extra attention to the solved examples. Try to find clues to enable you to solve your problem
- if you are still unable to do the questions, look at the references and other materials
- if your problem is still unsolved, then discuss it with your tutor. In order to get the most out of your meeting with your tutor, make sure to take with you what you have done to solve the problem by yourself; how far you have got; and most importantly, identify specific areas that you find difficult.
Plate 15 Steel frame
Plate 16 Footbridge
APPENDIX
FEEDBACK AND SOLUTIONS
Refer to Figure 6 to identify:
1. Uniformly distributed load.
Uniformly distributed loads could be the self weight of the floor slab, crowd loading (party, meeting of large group) etc.
2. Point load.
Load of a single person on a floor, load of a single piece of furniture like washing machine, etc.
3. Axial load.
In this diagram, the load on the columns are the only obvious axial load.
If you were unable to answer any part of this question fully, you should review
Section 2.1 again.
Refer to Plates 2 and 3 and Figure 7 to answer the following questions:
Plate 4 shows part of a frame, consisting of beams and columns. The block-work wall built on the beam (infill) is a common feature in every steel structure. Your task is to study the photograph carefully (discuss your views with your group members, if you are working in groups), and carry out the following tasks.
1. The type of load created by the block-work in the beam in Plate 4 is a
uniformly distributed load.
2. The load transferred from the beam to the columns is an axial load.
- Sketch the load path on the line diagram using arrows, from the block-work
to the beam, then to the columns and finally to the ground.
Load from the block-work is a UDL on the beam. This load is transferred by the beam to the columns, at either sides of the beam, as axial loads. The load from the columns is distributed on the foundation as UDL, and this load is resisted by the ground.
Use the example shown below to sketch support reactions for the other two structures. Show the direction of these reactions clearly.
Follow the example shown in the first question on this sheet, and note that an axial load is acting along the axis of the beam, a shear force perpendicular to the beam, and the symbol of the bending moment is shown as a curved arrow. Complete the diagrams using these suggestions
Examine carefully Plate 11
1. In Plate 11, the type of connection which connect the tubes to the plates can
be classified as a PIN connection, because each tube is connected by a
single bolt.
2. In Plate 12, the connection at the ends of the beams can be classified as a
FIXED connection, because the beams are welded to the main beams and
then to the columns.
3. In Plate 14, the connection between the column and the base can be
classified as a FIXED connection, because the column is fixed to the concrete
footing.
1. Suggest another type of load that could be applied on a beam.
A twisting force is also called torsion.
2. Sketch a line diagram of a simple beam, showing the load identified in (1),
above on it.
Give practical examples of the following load types:
1. A uniformly distributed load
Self-weight of structural elements (floor slab, beam, etc) crowd loading: eg students sitting a classroom, a car park which is full of cars, etc.
2. A point load
Load of a single item, eg a single car on a car park floor, weight of a filing cabinet, etc.
What are the three main beam types?
1. A simply supported beam (pin at one end and roller at the other end)
2. A fixed beam (fixed supports at both ends)
3. A cantilever beam (fixed at one end and free at the other end)
Describe briefly the support reactions resisted by the following support types:
1. Roller
A roller can resist on a force perpendicular to the direction that it slides.
2. Pin
A pin can resist on a force perpendicular to the beam (the shear force), and another along the axis of the beam (the axial force).
3. Fixed
A fixed support can resist a horizontal force, a vertical force and a bending moment.
If you still have difficulties, go back to Sections 2.2 and 2.3 of the Pack. Make sure you understand these definitions fully before going any further
1. The three equations of equilibrium are:
- Sum of horizontal forces must be zero (sum of all applied horizontal forces must be equal to the sum of all horizontal reactions resisted by the beam supports).
- Sum of vertical forces must be zero (sum of all applied vertical forces must be equal to the sum of all vertical reactions resisted by the beams supports.)
- Sum of moment, caused by all forces (applied loads and support reactions), about any point must be zero.
You are expected to understand the three equations of equilibrium before going any further. If you have any difficulties, refer to Section 1.4.
2. Apply the three equations of equilibrium to the following beam
Solution
Sketch support reactions
- Sum of horizontal forces must be zero. There is no horizontal force applied on the cantilever beam, therefore there is no horizontal reaction.
HA = 0kN
- Sum of all vertical forces must be zero. There is only one downward vertical load of 100kN, applied at the tip of cantilever. The vertical support reaction must be equal to this vertical load.
VA = 100kN
- Sum of moment about A must be equal to zero:
MA = 100 * 4 ➔ MA = 100kN-m
Determine whether the following structures are determinate or indeterminate. (State your reasons clearly).
Solution:
There is only one support at the left-hand end of this cantilever beam. The support type is a fixed support. By definition, a fixed support resists three reactions:
(1) A horizontal reaction or an axial load
(2) A vertical reaction or a shear force
(3) A bending moment.
There are three unknowns and there are three equations of equilibrium, therefore the structure is determinate.
REMEMBER that a pin support resists a horizontal and a vertical reaction, and a roller resists only a single reaction, which is perpendicular to the direction in which the roller slides.
You should have identified that this is a determinate beam: ie three unknowns (an axial load and a shear force at pin support, and a shear force at roller support), and three equations of equilibrium.
From question (a), it is clear that a fixed beam has three reactions. Therefore, two fixed ends have six reactions, and there are only three equations of equilibrium, therefore the structure is indeterminate.
Calculate support reactions for the following structures.
Solution
Sketch reactions
Equations of Equilibrium:
1. Sum of horizontal forces = 0:
HA = 0 (No applied horizontal load)
2. Sum of vertical forces = 0
Applied vertical load from the UDL = 40 x 10 = 400kN
This force is applied over the entire length A-B, therefore it is equally shared
by two supports at A and B
VA = VB = 400/2 = 200kN
Follow the guidance given in question (a). Note that you must apply the 3 equations of equilibrium, and make sure that the sum of all applied loading is equal and opposite to the sum of the reactions in each direction.
Solving this question by using the three equations of equilibrium, you should get the following support reactions:
Axial load = 0kN
Shear force = 100kN
Bending moment = 100 x 4 = 400kN-m.
For this question, the vertical reaction at the LH support = 40kN and at the RH support is 60kN. Support closer to the load attracts more reaction.
Draw the shear force diagram for the following structures. Use the solution to question (c) for reference.
(a) Your final shear force diagram should look like this
Explanations:
There are three reactions at the fixed end (axial load, shear force and bending).
There is no axial load applied to the beam, therefore the horizontal reaction is zero. The applied load is a point load of 100kN, applied at a distance of 2m from the support. The shear force must be equal to 100kN. There is no applied load on the last 2m of the beam, therefore shear force along this length is zero.
(b) Your final shear force diagram should like this.
Explanations:
There are three reactions at the fixed end (an axial load, shear force at the pin support, and a shear force at the roller).
There is no axial load applied to the beam, therefore the horizontal reaction is zero. The total applied load is a 40 x 10 = 400kN. This is shared equally between the two supports. The shear force at each support is = 200kN. The applied is a UDL, which reduces the shear force gradually until it reaches the centre of the beam. At the beam centre, the shear force is zero. From the downward force, it builds up until it reaches the RH support. At the RH, the downward load it reaches 200kN, which is equal to the support reaction.
There is no applied load on the last 2m of the beam, therefore shear force along this length is zero.
c) Solution:
Sketch reaction
1) There is no horizontal force applied on this beam, therefore HA = 0kN
2) Take Moment about A:
VB x 10 = 100 x 6 ➔ VB = 600/10 = 60kN
Sum of vertical forces = 0
VA + VB = 100 ➔ VA = 100 – VB = 100 – 60 = 40kN
Shear force diagram
Check:
Area under the SFD at the top of the beam = 40 x 6 = 240kN-m
Area under the SFD at the bottom of the beam = 60 x 4 = 240kN-m
Both areas the same, OK.
Follow the details given in question (c) to solve the other two questions. Make sure to:
- Sketch arrows, showing support reactions and their directions
- Apply the equations of equilibrium
- Note that a point load only produces rectangular shapes in the SFD, and a UDL produces triangular shapes
IV. Check that the area above and below the SFD is the same.
Draw BMD for the following structures.
For a deeper understanding, you should test that BM can be obtained from the area under the SFD.
Solution
Sketch reactions
From Activity 11 VA = 40kN and VB = 60kN.
Sum of moments, at a distance X, from LH support = 0, assuming anti-clockwise moment to be positive.
40 x X = M
This a linear relation. At X = 6m M = 40 x 6 = 240kN-m which varies from 0 at A to 240 at C. From C to B, the bending moment linearly reduces from
240kN-m to zero at support B.
In Activity 11, it was shown that the area under the shear force diagram at the top and bottom of the beam was 240kN-m. This is the same as the value of the maximum bending moment at C. This shows that the bending moment can also be calculated from the area under the shear force diagram.
Now follow the procedure used in question (a), to answer questions (b) and (c).
Your solution should look like this:
This is not an easy problem. You should not be discouraged if you did not get this question right.
Note that the bending moment diagram, over the length of the beam on which the UDL is applied (6m from the LH support), is a parabola and after that, where there is no load on the beam, the shape of the BMD is a straight line.
You should follow the following steps:
Replace the UDL by an equivalent point load of (100 x 6 = 600kN) at the centre of the UDL (3m each side).
Take moment at A
VB x 10 – 600 x 3 = 0 ➔ VB = 180kN
VA = 600 -180 = 420kN
Split the beam into two sections, and draw the free body diagrams
Take at a distance X from A (use LD section only)
– 420 * X + 100 * X * X/2 + Mx = 0
Mx = 100 x2/2 – 420 X
You need to set a table, similar to those shown in Section 2.7 of the example. Your final table should look something like this:
X 0 1 2 3 4 5 6 7 8 9 10
M 0 370 640 810 880 850 720 540 360 180 0
Now sketch the bending moment diagram
Bending moment diagram
Feedback on self-assessment tasks
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The above topics are essential knowledge that should be learned before attempting any questions. Please be vigilant when visiting structures, walking around the town or during the site visits, to build up your knowledge on these topics.
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Please be vigilant when visiting structures, walking around the town or during the site visits to build up your knowledge of building elements and how they are connected to other members.
You should always test that sum of all applied loads, both in vertical and horizontal directions, must be equal to the support reactions in relevant directions. You should make sure that a beam is not turning or twisting under the action of the loads, and the beam supports are capable of resisting these actions.
Plate 15 Steel frame
Study Plate 15 with your peers or alone, and then answer the following questions:
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The cross-bracing members are connected to the plates at the beam column junction, by means of four small bolts. This could be considered as a pin connection, because it would be unlikely that these supports are capable of resisting any shear or bending forces.
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Main beams are connected to columns by means of welding. This is a rigid connection, because it is capable of resisting any bending caused by the applied loading.
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The floor slab is made of steel decking, which is laid on the beams. This could be considered as a pin connection, because steel decking, as shown, is sitting on the supporting beams.
If you answered all three questions correctly then you are very knowledgeable in this area, and you are confident in your understanding and ability to apply your knowledge of this topic.
If you answered two questions correctly then you have some knowledge in this area, but you recognise that there are significant gaps in your knowledge.
Otherwise, you have rather limited knowledge and you are not confident in your level of understanding of this topic.
The above diagram shows a plan view of a floor, consisting of surrounding walls, main beam or girder, floor joist and floor boards. Normal floor loading is applied on the floor boards, and this load is transferred to the walls by the components mentioned. Study the above figure carefully, and draw line diagrams for the following cases:
- Floor board showing correct support types and loading (assume that the floor load
is UDL, applied uniformly over the entire board).
The board is simply supported at both ends and continuously over the joists. The joists
could be considered as rollers under the board.
- A typical floor joist showing support types and loading, transferred from the boards to
the joist
- Main beam, showing support types and loading transferred from the joist to the main
beam
If you answered all three questions correctly then you are very knowledgeable on this and you are confident of your understanding and ability to apply your knowledge on this topic.
If you answered two questions correctly then you have some knowledge of this area, but you recognise there are significant gaps in your knowledge.
Otherwise you have rather limited knowledge and you are not confident of your level of understanding of this topic.
Plate 16 Footbridge
Plate 16 shows a footbridge deck, looking from the bottom of the bridge. The bridge deck is made of steel sheets placed on joists. The joists are then welded to side beams. The side beams are welded to the columns, and the columns are fixed to the concrete footing.
Assuming a crowded load on the bridge, sketch the load path, showing the direction of the loads by means of arrows. The load path should show how the crowd load is transferred:
(1) from the deck to the joists
(2) from joists to side beams
(3) from side beams to columns
(4) and finally to the ground.
You should use sketches of the line diagrams of various members to demonstrate
your understanding.
Load from slab deck to joist
Load from joint to main beam
Load from main beam to columns
Load from foundation to the ground
If you answered all four questions correctly, then you are very knowledgeable in this area, and you are confident in your understanding and ability to apply your knowledge of this topic.
If you answered up to two questions correctly, then you have some knowledge in this area, but you recognise that there are significant gaps in your knowledge.
Otherwise, you have rather limited knowledge and you are not confident in your level of understanding of this topic
Calculate the support reactions for the following beams.
Sum of horizontal forces = 0 ➔ HA = 30kN
Take moment about A VB x 10 – 40 x 3 = 0 ➔ VB = 12kN
Sum of vertical forces = 0 VA + VB = 40 ➔ VA = 28kN
Sum of horizontal forces = 0 ➔ HA = 30kN
Take moment about A VB X 7 – 40 x 10 = 0 ➔ VB = = 57kN
Sum of vertical forces = 0 VA + VB = 40 ➔ VA = – 17kN
This means that VA must be downward NOT UPWARD
If you answered both questions correctly then you are very knowledgeable in this area, and you are confident in your understanding and ability to apply your knowledge of this topic.
If you answered one question or partly answered both questions correctly, then you have some knowledge in this area, but you recognise that there are significant gaps in your knowledge.
Otherwise, you have rather limited knowledge and you are not confident in your level of understanding of this topic.
Draw the shear force and bending moment diagrams for the following beams, showing all values
Shear force diagram:
For reactions, please refer to Task 6. Now start from the LH end, go upward 28 units, draw a straight line to the load, go down 40 units, draw a straight line to the RH support and go up 12 units
Bending moment diagram
Bending moment at supports is zero, and due to point load it increases linearly to the point load and then decreases back to zero.
The equation of line representing the bending moment at the LH side of the load is:
M = VA x Distance
At 3m M = 28 x 3 = 84kN
Shear force diagram
Similar to question 1, start from the LH end, go downwards 17 units (note that the shear force at this end is 17 kN downwards); draw a straight line to the reaction, go upwards 57 kN; draw a straight line to the end of the beam and go 40 units downwards.
Bending moment diagram
Once again, the moment at the LH support is zero, and the moment at the RH support is 40 x 3 = 120kN-m, as the load acts as a cantilever over the overhang.
The equation of the line representing bending from the LH support to the RH support is:
Reaction x distance = 17 x distance
Note that the bending moment diagram is drawn at the tension face of the beam.
For the RH support, the BM linearly decreases from 120 to zero at the tip of the overhang.
If you answered both questions correctly then you are very knowledgeable in this area, and you are confident in your understanding and ability to apply your knowledge of this topic.
If you answered one question or partly answered both questions correctly, then you have some knowledge in this area, but you recognise that there are significant gaps in your knowledge.
Otherwise, you have rather limited knowledge and you are not confident in your level of understanding of this topic.
