30 Method of Joints Problem Set
METHOD OF JOINTS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: In the truss system shown below, determine the forces in member AB and BC using the method of joints.
A. π΄π΅: 20 π ; π΅πΆ: 60 (πΆ) B. π΄π΅: 20 πΆ ; π΅πΆ: 60 (πΆ) C. π΄π΅: 60 πΆ ; π΅πΆ: 20 (π) D. π΄π΅: 60 π ; π΅πΆ: 20 (πΆ)
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SOLUTION 1: The process known as the METHOD OF JOINTS can be referenced under the section denoted as STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are given a structural system as:
With FORCES applied at JOINT C and JOINT D, as illustrated, we are asked to determine the FORCES that are DISTRIBUTED in to and through MEMBER AB and MEMBER BC. The first step, as will be the case in most all of our structural problems, is to define the coordinate system. We are working with a TRUSS, so with that, we will deploy your STANDARD CARTESIAN COORDINATE SYSTEM to define the X and Y AXIS that we will develop our EQUATIONS OF EQUILIBRIUM around.
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Letβs highlight the MEMBERS we are interested in, which are:
By visual inspection, we see that MEMBER AB and MEMBER BC share a connection at JOINT B, which is a PINNED SUPPORT constraint.
In order to solve for the FORCE in MEMBER AB and MEMBER BC, we will need to first determine the SUPPORT REACTIONS that are generated at the PINNED CONSTRAINT located at JOINT B.
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To start our analysis, we will draw a FREE BODY DIAGRAM for the OVERALL TRUSS SYSTEM, which will resolve as:
With a PINNED SUPPORT at JOINT B, there will be a SUPPORT REACTION that RESTRAINS against TRANSLATION VERTICALLY and as well as a SUPPORT REACTION that RESTRAINS against TRANSLATION HORIZONTALLYβ¦as we have illustrated. At JOINT A, having a simple ROLLER SUPPORT, there will only be a SUPPORT REACTION that RESTRAINS against TRANSLATION HORIZONTALLY, as that is the orientation of the ROLLER relative to the TRUSS we are analyzing. With our FREE BODY DIAGRAM now established, we can focus in on determining the what each of the SUPPORT REACTION will be acting on JOINT A and JOINT B. It is important to remember that with THREE UNKNOWNS, we will need to establish at least THREE EQUATIONS in order to solve for all of the unknowns.
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The GENERAL FORMULAS for EQUILIBRIUM REQUIREMENTS that can be used in the METHOD OF JOINTS can be referenced under the topic of STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. These GENERAL FORMULAS read as:
πΉ. = 0 πΉ0 = 0 π=0
Letβs start with SUMMING the FORCES of the TRUSS SYSTEM along the Y-AXIS, doing so we get:
πΉ0 = 0 = π΅0 β 10 ππππ β 10 ππππ
With only a single unknown, π΅0 , we can rearrange and solve, giving us: π΅0 = 20 ππππ Letβs now turn to SUMMING the FORCES of the TRUSS SYSTEM along the X-AXIS, doing so we get:
πΉ. = 0 = π΄. + π΅9
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As there are NO EXTERNAL FORCES acting on the HORIZONTAL PLANE, the only FORCES to consider along the X-AXIS are those of our SUPPORT REACTIONS. Therefore, the SUPPORT REACTIONS are EQUAL and OPPOSITE to one another, or rather: π΄. = βπ΅. We will now turn to SUMMING the MOMENTS around a SUPPORT POINT within our TRUSS SYSTEM. A solid tip here is to always SUM about a POINT where there is at least ONE UNKNOWN. Doing so will eliminate that UNKNOWN and allow for us to resolve another UNKNOWN elsewhere. With that being said, we will FOCUS on the CONSTRAINT at POINT B. Writing our EQUATION OF EQUILIBRIUM with the assumption that a COUNTERCLOCKWISE MOMENT will be represented as a POSITIVE MOMENT, and visa versa: π: = 0 = β 10 ππππ 10 ππ‘ β 10 ππππ 20 ππ‘ β (π΄. )(5 ππ‘) Both EXTERNAL FORCES at JOINTS C and D will create a CLOCKWISE MOMENT about JOINT B, therefore, they are expressed as NEGATIVE VALUES. The HORIZONTAL SUPPORT REACTION at JOINT B will create a COUNTERCLOCKWISE MOMENT about JOINT B, and for that reason, it is expressed as a POSITIVE VALUE.
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The VERTICAL SUPPORT REACTION at JOINT B has no HORIZONTAL OFFSET, therefore, no MOMENT is created about JOINT B, and thus, it is disregarded in our SUMMATION of MOMENTS. This EQUATION leaves us with ONE UNKNOWN, π΅. . Rearranging and solving for this value, we get: π΄. = β60 ππππ We now have established that: π΅0 = 20 ππππ π΄. = β60 ππππ We need to determine π΅. . From our EQUATION OF EQUILIBRIUM established along the X-AXIS, we found that: π΄. = βπ΅. Or: π΅. = βπ΄. π΅. = β β60 ππππ = 60 ππππ
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Updating our FREE BODY DIAGRAM to include the SUPPORT REACTIONS that we have now defined, we have:
One quick note at this point. It is important that as we make our way through these longer calculations, that we maintain the integrity of our FREE BODY DIAGRAM so as to avoid any misinterpretations when establishing and working through our EQUATIONS OF EQUILIBRIUM. Due to the often complex nature of these TRUSS SYSTEMS, we will be working through and around multiple JOINTS creating the potential instance where LABELS and NOMENCLATURE could get muddied upβ¦focus on ORGANIZATION and cleanliness. Letβs move to solving for the FORCE in MEMBER AB and MEMBER BC. With these MEMBERS in mind, a good JOINT to EVALUATE around would be JOINT B.
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Isolating this JOINT, we have:
There will be TWO SUPPORT REACTIONS and TWO MEMBER FORCES, AB AND BC, acting at JOINT B. We can take the same approach as we did prior for the complete TRUSS SYSTEM, establishing a FREE BODY DIAGRAM and the EQUATIONS OF EQUILIBRIUM around this JOINT to solve for the UNKNOWNS. The FREE BODY DIAGRAM can be illustrated as:
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With TWO UNKNOWNS, we will need TWO EQUATIONS. These TWO EQUATIONS OF EQUILIBRIUM will be written using the HORIZONTAL and VERTICAL components of the FORCES acting at POINT B. To our advantage, all the FORCES we have are acting in unison with these AXES, so there will be no need for RESOLVING a FORCE in to COMPONENTS. Letβs start with SUMMING the FORCES along the X-AXIS, which gives us:
πΉ. = 0: 60 ππππ + πΉ:> = 0
Solving for the FORCE in MEMBER BC, we find: πΉ:> = β60 ππππ (πππππππ π πππ) Moving on to the SUMMATION of FORCES along the Y-AXIS, we get:
πΉ0 = 0: β20 ππππ + πΉE: = 0
Solving for the FORCE in MEMBER AB, we find: πΉE: = β20 ππππ (πππππππ π πππ) One point to NOTEβ¦when representing a MEMBER on a FREE BODY DIAGRAM and in an EQUATION OF EQUILIBRIUM, we will assume that itβs LINE OF ACTION is acting in some direction relative to the JOINTβ¦which will either be represented as TENSION or COMPRESSION.
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However, this is just an assumption until the EQUATIONS are fully vetted. If our ASSUMPTION is incorrect, for example if we assumed TENSION and we defined a value that was NEGATIVE, then the result would tell us that the MEMBER is actually in COMPRESSIONβ¦the OPPOSITE to what we assumed. This allows us to correct the FBD at the end of our problem to represent the reality of our analysis. The correct answer choice is B. π¨π©: ππ πͺ ; π©πͺ: ππ πͺ
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PROBLEM 2: In the truss system shown below, determine the forces in member AF using the method of joints.
A. 25 (πΆ) B. 25 (π) C. 50 (πΆ)) D. 50 (π)
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SOLUTION 2: We are looking to solve for the forces in member AF of the truss system. The first step in this problem is to define the coordinate system. As we will be working with trusses, we will use a Cartesian Coordinate System to define the xβ and π¦ β ππ₯ππ .
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In order to solve for the forces in members BC and GH, we will need to calculate the support reactions generated the reactions at joints F and J, particularly by the pinned constraint at joint F.
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In order to solve for the support reactions at joint F, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint F, there are support reactions to restrain against translation vertically and horizontally. At joint J, there is a roller that will only provide support against translation vertically.
Now that we have a free body diagram of the overall truss system, we can solve for the restraint forces generated at the support constraints at joints F and J. It is important to remember that as we have 3 unknowns, so we need to write at least 3 equations in order to solve for all of the unknowns.
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We will first sum the forces of the overall truss system along the π¦ β ππ₯ππ :
πΉ0 = 0: πΉ0 + π½0 β 10 π β 10 π β 10 π β 10 π β 10 π = 0
πΉ0 + π½0 = 50 π We then sum the forces of the overall truss system along the π₯ β ππ₯ππ :
πΉ. = πΉ. = 0
πΉ. = 0 π We will then sum the moments around the support constraint at point F to establish the third equation of equilibrium. Point F is a good place to sum the moments about, as it will allow us to write an equation of equilibrium that has no terms associated with the support reactions generated by the pinned support constraint at point A. πP = π½0 40)-(10 k)(10 β 10 π 20)-(10 k)(30 β (10 π)(40") = 0 Solving for the vertical support reaction at point J, π½0 , we find: π½0 = 25 π
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We can then plug in the calculated value for the vertical support reaction at point F into the equation we derived by summing the forces about the π¦ β ππ₯ππ , and solve for the vertical support reaction at joint F. πΉ0 + π½0 = 50 π πΉ0 + 25 π = 50 π πΉ0 = 25 π Next, we will update our overall free body diagram to include the support reactions that we calculated. It is important that we maintain our overall free body diagram so we keep our equations organized, and do not mislabel or incorrectly write our equations of equilibrium. As we many need to evaluate various members in the truss system, it is imperative that your work stay organized as you continue on with the problem.
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The topic of METHOD OF JOINTS can be referenced under the topic of STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. As we are solving for the forces in member AF, a good joint to evaluate for the missing forces is joint F. Looking at the free body diagram for joint F below, we see that there are two support reactions and two members acting at joint F. As each joint can provide two equations of equilibrium, and we have only two unknowns, we can solve for the desired forces by simply looking at joint F.
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Looking at the free body diagram for joint F as shown above, we notice that we have all of the information we need to solve for the force in member AF. We can rewrite the free body diagram to include only the relevant information we need to solve for the force in member AF.
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We can now redraw the free-body diagram of joint F with all of the forces labeled. When representing a member on the free body diagram, we will assume it is in tension (+), and its line of action is acting away from the joint. If the member is actually in compression, then our answer at the end will have an opposite sign, such that if we assume a member is in tension to the right (+), then at the end our calculations, the member should be drawn with the line of action to the left to indicate compression (-).
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Moving on the next step of the problem, we write the equations of equilibrium for joint F to solve for the forces in member AF. We will then sum the forces of joint F along the π¦ β ππ₯ππ :
πΉ0 = 0: πΉEP + 25 ππππ = 0
Solving for the force in member AF, we find: πΉEP = β25 ππππ (πππππππ π πππ)
Therefore, the correct answer choice is C. ππ (πͺ)
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PROBLEM 3: Using the method of joints, what is the maximum number of equations we can write for a joint? A. 1 B. 2 C. 3 D. ππππ ππ π‘βπ ππππ£π
SOLUTION 3: The topic of the METHOD OF JOINTS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
As a joint can only can act along two axes, we can only write two equations for each joint. If a joint has more than 2 unknowns, then we will need to evaluate another joint or support nearby to write additional equations that we can use to substitute unknowns into.
Therefore, the correct answer choice is B. π
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PROBLEM 1: In the truss system shown below, determine the forces in member AB and BC using the method of joints.
A. π΄π΅: 20 π ; π΅πΆ: 60 (πΆ) B. π΄π΅: 20 πΆ ; π΅πΆ: 60 (πΆ) C. π΄π΅: 60 πΆ ; π΅πΆ: 20 (π) D. π΄π΅: 60 π ; π΅πΆ: 20 (πΆ)
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SOLUTION 1: The process known as the METHOD OF JOINTS can be referenced under the section denoted as STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are given a structural system as:
With FORCES applied at JOINT C and JOINT D, as illustrated, we are asked to determine the FORCES that are DISTRIBUTED in to and through MEMBER AB and MEMBER BC. The first step, as will be the case in most all of our structural problems, is to define the coordinate system. We are working with a TRUSS, so with that, we will deploy your STANDARD CARTESIAN COORDINATE SYSTEM to define the X and Y AXIS that we will develop our EQUATIONS OF EQUILIBRIUM around.
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Letβs highlight the MEMBERS we are interested in, which are:
By visual inspection, we see that MEMBER AB and MEMBER BC share a connection at JOINT B, which is a PINNED SUPPORT constraint.
In order to solve for the FORCE in MEMBER AB and MEMBER BC, we will need to first determine the SUPPORT REACTIONS that are generated at the PINNED CONSTRAINT located at JOINT B.
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To start our analysis, we will draw a FREE BODY DIAGRAM for the OVERALL TRUSS SYSTEM, which will resolve as:
With a PINNED SUPPORT at JOINT B, there will be a SUPPORT REACTION that RESTRAINS against TRANSLATION VERTICALLY and as well as a SUPPORT REACTION that RESTRAINS against TRANSLATION HORIZONTALLYβ¦as we have illustrated. At JOINT A, having a simple ROLLER SUPPORT, there will only be a SUPPORT REACTION that RESTRAINS against TRANSLATION HORIZONTALLY, as that is the orientation of the ROLLER relative to the TRUSS we are analyzing. With our FREE BODY DIAGRAM now established, we can focus in on determining the what each of the SUPPORT REACTION will be acting on JOINT A and JOINT B. It is important to remember that with THREE UNKNOWNS, we will need to establish at least THREE EQUATIONS in order to solve for all of the unknowns.
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The GENERAL FORMULAS for EQUILIBRIUM REQUIREMENTS that can be used in the METHOD OF JOINTS can be referenced under the topic of STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. These GENERAL FORMULAS read as:
πΉ. = 0 πΉ0 = 0 π=0
Letβs start with SUMMING the FORCES of the TRUSS SYSTEM along the Y-AXIS, doing so we get:
πΉ0 = 0 = π΅0 β 10 ππππ β 10 ππππ
With only a single unknown, π΅0 , we can rearrange and solve, giving us: π΅0 = 20 ππππ Letβs now turn to SUMMING the FORCES of the TRUSS SYSTEM along the X-AXIS, doing so we get:
πΉ. = 0 = π΄. + π΅9
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As there are NO EXTERNAL FORCES acting on the HORIZONTAL PLANE, the only FORCES to consider along the X-AXIS are those of our SUPPORT REACTIONS. Therefore, the SUPPORT REACTIONS are EQUAL and OPPOSITE to one another, or rather: π΄. = βπ΅. We will now turn to SUMMING the MOMENTS around a SUPPORT POINT within our TRUSS SYSTEM. A solid tip here is to always SUM about a POINT where there is at least ONE UNKNOWN. Doing so will eliminate that UNKNOWN and allow for us to resolve another UNKNOWN elsewhere. With that being said, we will FOCUS on the CONSTRAINT at POINT B. Writing our EQUATION OF EQUILIBRIUM with the assumption that a COUNTERCLOCKWISE MOMENT will be represented as a POSITIVE MOMENT, and visa versa: π: = 0 = β 10 ππππ 10 ππ‘ β 10 ππππ 20 ππ‘ β (π΄. )(5 ππ‘) Both EXTERNAL FORCES at JOINTS C and D will create a CLOCKWISE MOMENT about JOINT B, therefore, they are expressed as NEGATIVE VALUES. The HORIZONTAL SUPPORT REACTION at JOINT B will create a COUNTERCLOCKWISE MOMENT about JOINT B, and for that reason, it is expressed as a POSITIVE VALUE.
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The VERTICAL SUPPORT REACTION at JOINT B has no HORIZONTAL OFFSET, therefore, no MOMENT is created about JOINT B, and thus, it is disregarded in our SUMMATION of MOMENTS. This EQUATION leaves us with ONE UNKNOWN, π΅. . Rearranging and solving for this value, we get: π΄. = β60 ππππ We now have established that: π΅0 = 20 ππππ π΄. = β60 ππππ We need to determine π΅. . From our EQUATION OF EQUILIBRIUM established along the X-AXIS, we found that: π΄. = βπ΅. Or: π΅. = βπ΄. π΅. = β β60 ππππ = 60 ππππ
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Updating our FREE BODY DIAGRAM to include the SUPPORT REACTIONS that we have now defined, we have:
One quick note at this point. It is important that as we make our way through these longer calculations, that we maintain the integrity of our FREE BODY DIAGRAM so as to avoid any misinterpretations when establishing and working through our EQUATIONS OF EQUILIBRIUM. Due to the often complex nature of these TRUSS SYSTEMS, we will be working through and around multiple JOINTS creating the potential instance where LABELS and NOMENCLATURE could get muddied upβ¦focus on ORGANIZATION and cleanliness. Letβs move to solving for the FORCE in MEMBER AB and MEMBER BC. With these MEMBERS in mind, a good JOINT to EVALUATE around would be JOINT B.
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Isolating this JOINT, we have:
There will be TWO SUPPORT REACTIONS and TWO MEMBER FORCES, AB AND BC, acting at JOINT B. We can take the same approach as we did prior for the complete TRUSS SYSTEM, establishing a FREE BODY DIAGRAM and the EQUATIONS OF EQUILIBRIUM around this JOINT to solve for the UNKNOWNS. The FREE BODY DIAGRAM can be illustrated as:
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With TWO UNKNOWNS, we will need TWO EQUATIONS. These TWO EQUATIONS OF EQUILIBRIUM will be written using the HORIZONTAL and VERTICAL components of the FORCES acting at POINT B. To our advantage, all the FORCES we have are acting in unison with these AXES, so there will be no need for RESOLVING a FORCE in to COMPONENTS. Letβs start with SUMMING the FORCES along the X-AXIS, which gives us:
πΉ. = 0: 60 ππππ + πΉ:> = 0
Solving for the FORCE in MEMBER BC, we find: πΉ:> = β60 ππππ (πππππππ π πππ) Moving on to the SUMMATION of FORCES along the Y-AXIS, we get:
πΉ0 = 0: β20 ππππ + πΉE: = 0
Solving for the FORCE in MEMBER AB, we find: πΉE: = β20 ππππ (πππππππ π πππ) One point to NOTEβ¦when representing a MEMBER on a FREE BODY DIAGRAM and in an EQUATION OF EQUILIBRIUM, we will assume that itβs LINE OF ACTION is acting in some direction relative to the JOINTβ¦which will either be represented as TENSION or COMPRESSION.
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However, this is just an assumption until the EQUATIONS are fully vetted. If our ASSUMPTION is incorrect, for example if we assumed TENSION and we defined a value that was NEGATIVE, then the result would tell us that the MEMBER is actually in COMPRESSIONβ¦the OPPOSITE to what we assumed. This allows us to correct the FBD at the end of our problem to represent the reality of our analysis. The correct answer choice is B. π¨π©: ππ πͺ ; π©πͺ: ππ πͺ
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PROBLEM 2: In the truss system shown below, determine the forces in member AF using the method of joints.
A. 25 (πΆ) B. 25 (π) C. 50 (πΆ)) D. 50 (π)
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SOLUTION 2: We are looking to solve for the forces in member AF of the truss system. The first step in this problem is to define the coordinate system. As we will be working with trusses, we will use a Cartesian Coordinate System to define the xβ and π¦ β ππ₯ππ .
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In order to solve for the forces in members BC and GH, we will need to calculate the support reactions generated the reactions at joints F and J, particularly by the pinned constraint at joint F.
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In order to solve for the support reactions at joint F, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint F, there are support reactions to restrain against translation vertically and horizontally. At joint J, there is a roller that will only provide support against translation vertically.
Now that we have a free body diagram of the overall truss system, we can solve for the restraint forces generated at the support constraints at joints F and J. It is important to remember that as we have 3 unknowns, so we need to write at least 3 equations in order to solve for all of the unknowns.
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We will first sum the forces of the overall truss system along the π¦ β ππ₯ππ :
πΉ0 = 0: πΉ0 + π½0 β 10 π β 10 π β 10 π β 10 π β 10 π = 0
πΉ0 + π½0 = 50 π We then sum the forces of the overall truss system along the π₯ β ππ₯ππ :
πΉ. = πΉ. = 0
πΉ. = 0 π We will then sum the moments around the support constraint at point F to establish the third equation of equilibrium. Point F is a good place to sum the moments about, as it will allow us to write an equation of equilibrium that has no terms associated with the support reactions generated by the pinned support constraint at point A. πP = π½0 40)-(10 k)(10 β 10 π 20)-(10 k)(30 β (10 π)(40") = 0 Solving for the vertical support reaction at point J, π½0 , we find: π½0 = 25 π
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We can then plug in the calculated value for the vertical support reaction at point F into the equation we derived by summing the forces about the π¦ β ππ₯ππ , and solve for the vertical support reaction at joint F. πΉ0 + π½0 = 50 π πΉ0 + 25 π = 50 π πΉ0 = 25 π Next, we will update our overall free body diagram to include the support reactions that we calculated. It is important that we maintain our overall free body diagram so we keep our equations organized, and do not mislabel or incorrectly write our equations of equilibrium. As we many need to evaluate various members in the truss system, it is imperative that your work stay organized as you continue on with the problem.
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The topic of METHOD OF JOINTS can be referenced under the topic of STATICALLY DETERMINATE TRUSS on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. As we are solving for the forces in member AF, a good joint to evaluate for the missing forces is joint F. Looking at the free body diagram for joint F below, we see that there are two support reactions and two members acting at joint F. As each joint can provide two equations of equilibrium, and we have only two unknowns, we can solve for the desired forces by simply looking at joint F.
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Looking at the free body diagram for joint F as shown above, we notice that we have all of the information we need to solve for the force in member AF. We can rewrite the free body diagram to include only the relevant information we need to solve for the force in member AF.
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We can now redraw the free-body diagram of joint F with all of the forces labeled. When representing a member on the free body diagram, we will assume it is in tension (+), and its line of action is acting away from the joint. If the member is actually in compression, then our answer at the end will have an opposite sign, such that if we assume a member is in tension to the right (+), then at the end our calculations, the member should be drawn with the line of action to the left to indicate compression (-).
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Moving on the next step of the problem, we write the equations of equilibrium for joint F to solve for the forces in member AF. We will then sum the forces of joint F along the π¦ β ππ₯ππ :
πΉ0 = 0: πΉEP + 25 ππππ = 0
Solving for the force in member AF, we find: πΉEP = β25 ππππ (πππππππ π πππ)
Therefore, the correct answer choice is C. ππ (πͺ)
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PROBLEM 3: Using the method of joints, what is the maximum number of equations we can write for a joint? A. 1 B. 2 C. 3 D. ππππ ππ π‘βπ ππππ£π
SOLUTION 3: The topic of the METHOD OF JOINTS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
As a joint can only can act along two axes, we can only write two equations for each joint. If a joint has more than 2 unknowns, then we will need to evaluate another joint or support nearby to write additional equations that we can use to substitute unknowns into.
Therefore, the correct answer choice is B. π
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