31 Method of Sections Concept Overview
METHOD OF SECTIONS | CONCEPT OVERVIEW The topic of METHOD OF SECTIONS can be referenced on page 68 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
CONCEPT INTRO: The METHOD OF SECTIONS is the technique of solving for the internal forces in each truss member by passing an imaginary section through the truss, and cutting it into two parts.
The idea behind the method of section is that we can calculate forces in members by creating an imaginary cut through the structure, drawing the forces acting on the ends of the cut members, and then conducting a force and moment balance for one of the two parts of the structure. Made with
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The Method of Sections is used when we have a truss system that has multiple members, and we are only interested with forces of a few particular members. The Method of Sections is always used together with the Method of Joints to analyze a truss system. Since truss members are subject to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude as the forces at the joint. This result is based on the principle of static equilibrium and Newton’s third law.
In the Method of Joints, our analysis was limited to just solving for the external reactions and the two equations of equilibrium available for each joint. When working with the Method of Sections, we can also can also account for the equilibrium of moments, allowing us to solve for up to three unknown forces at a time. Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces, which represents the internal forces of the members. If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these unknowns to determine directly the third unknown force. Made with
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PROCEDURE FOR THE METHOD OF SECTIONS: Given a structural system represented as a truss system, we can use an imaginary section to cut through the truss system, and define two separate truss systems that are in static equilibrium. Once the systems are defined, we can use the equations of equilibrium to solve for the unknown forces, a define by the process below: 1. Define the coordinate system of the of the truss system. For the purpose of the FE Exam, we should always use a Cartesian Coordinate System when working with truss systems. As there will be varying angles in each triangular group of members, we should not incline the coordinate system in any scenario.
2. Draw a Free Body Diagram of the overall truss system and find the external support reactions, if required. This may require you to use the equations of equilibrium for the overall truss system, and sum the moments about a particular support constraint or joint.
You will typically find that a truss system is
supported by a roller and pin, so you will want to take the moment about the pin support constraint to account for the horizontal and vertical support reactions. Made with by Prepineer | Prepineer.com
3. Look at the overall free body diagram of the truss system, and identify which internal forces of members, the problem is asking you to solve for. It is important to remember that when working with the Method of Sections, the entire problem will center around you cutting the overall truss system into two separate systems in static equilibrium. Therefore, it is imperative that you pass the imagine section correctly, and following the guidelines listed below: a. Cut the member that you’re interested in to calculate the internal force of a truss member, you must first expose the internal force. The only way you can do that is by cutting the member. b. Cut the truss completely into two parts – Don’t cut partway through a truss and then stop. c. Cut a maximum of three members – you’re allowed to cut up to three members total. If you cut fewer than three that is perfectly ok. But if you cut four or more members, you may have some problems with this method as you may not be able to solve for all of the unknown on the free-body diagram. d. If you cut more than three members, all but two of the members must on the same line of action. In rare circumstance, you can actually cut more than three members, but you’ll need to make sure that all but at least two of them are collinear. In general, it’s a very rare occasion when you can actually cut more than three members at a time. 4. Make a cut along the members of interest based on your analysis in step 3. Draw an imaginary line through the members you are looking to analyze, and divide the truss system into two separate structural systems, both of which are in static equilibrium. Made with
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5. Draw a free body diagram of the particular truss system you wish to analyze, as one of the truss systems will not be used any further in the problem. For each cut member, assume the member is acting in tension and draw a tensile arrow going away from the joint. By doing this, the calculated forces will yield a positive scalar if our assumption of tension is correct, and a negative scalar if the member is actually in compression. 6. Cover up the member force you are looking for, and follow all other cut member forces until they meet at a point. Then sum moment about that point, thereby cancelling all other member forces since they now have no moment arms.
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CONCEPT EXAMPLE: The magnitude of the internal forces in members BC and GH is closes to which of the following answer choices?
A. 𝐵𝐶: 60 (𝐶); 𝐺𝐻: 60 (𝑇) B. 𝐵𝐶: 28 (𝐶); 𝐺𝐻: 60 (𝑇) C. 𝐵𝐶: 20 (𝑇); 𝐺𝐻: 15 (𝐶) D. 𝐵𝐶: 20 (𝐶); 𝐺𝐻: 15 (𝑇)
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SOLUTION: We are looking to solve for the forces in members BC and BH 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 x− and 𝑦 − 𝑎𝑥𝑒𝑠.
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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. The topic of the METHOD OF SECTIONS 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. Made with
by Prepineer | Prepineer.com
We will first sum the forces of the overall truss system along the 𝑦 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑦 = 0: 𝐹𝑦 + 𝐽𝑦 − 10 𝑘 − 10 𝑘 − 10 𝑘 − 10 𝑘 − 10 𝑘 = 0
𝐹𝑦 + 𝐽𝑦 = 50 𝑘 We then sum the forces of the overall truss system along the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 𝐹𝑥 = 0
𝐹𝑥 = 0 𝑘 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.
𝑀𝐹 = (𝐽𝑦 )(40) − (10 k)(10) − (10 𝑘)(20) − (10 k)(30) − (10 𝑘)(40") = 0 Solving for the vertical support reaction at point J, 𝐽𝑦 , we find: 𝐽𝑦 = 25 𝑘
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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 𝑦 − 𝑎𝑥𝑖𝑠, and solve for the vertical support reaction at joint F. 𝐹𝑦 + 𝐽𝑦 = 50 𝑘 𝐹𝑦 + 25 𝑘 = 50 𝑘 𝐹𝑦 = 25 𝑘 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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As we are solving for the forces in members BC and GH, a good location to draw our imaginary section would be vertically between joints B and C, and joints G and H. Looking at the free body diagram for the truss system below, we see the proposed location of the imaginary section would cut through three members to divide the truss systems into two separates truss systems in equilibrium. It is important to remember the maximum amount of members we can cut using the method of sections is three members.
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Looking below at the free body diagram for the truss system that we sectioned off from the overall truss system, we notice that we have all of the information we need to solve for the forces in members BC and GH, except the angle that member BH makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member BH, so we can use trigonometry to solve for the missing angle. We can now use the tangent trigonometric function to calculate the angle that member BH makes with the horizontal.
tan(𝜃) =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Plugging in the values of 10" for the opposite side of the triangle and 10" for the adjacent side of the triangle, we are able to calculate the angle of the member BH as:
𝜃 = tan−1 (
10 ) = 45° 10 Made with
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We can then plug in the calculated angle into the free body diagram for the truss system that we sectioned off from the overall truss system. The free body diagram now has all of the relevant information we need to solve for the forces in member BC and GH.
We can now redraw the free-body diagram for the sectioned truss system 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 the x- and y- axes for the overall truss system to determine the forces in member BC. It is important to realize that the force in member BH acts on joint B with an angle of 45°, so we will need to break that force’s magnitude into components when analyzing the vertical and horizontal planes. As we have 3 unknowns, we need to utilize all three equations of equilibrium, as we need at least three equations to solve for all of the unknowns. It is important to remember that the sectioned truss system is in static equilibrium at any point or joint. Therefore, we can sum the moments about joint B and set the equation equal to zero. ∑𝑀𝐵 = 0: (𝐹𝐺𝐻 )(10") + (10 𝑘)(10") − (25 𝑘)(10") = 0
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Solving for the force in member GH, we find: 𝐹𝐺𝐻 = 15 𝑘𝑖𝑝𝑠 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛) We will then sum the forces of the sectioned truss system along the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 0: 𝐹𝐵𝐶 + 𝐹𝐵𝐻 (cos 45°) + 𝐹𝐺𝐻 = 0
We will then sum the forces of the sectioned truss system along the 𝑦 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑦 = 0: 25 𝑘 − 10 𝑘 − 10 𝑘 − 𝐹𝐵𝐻 (sin 45°) = 0
Using the expression written for forces about the 𝑦 − 𝑎𝑥𝑖𝑠, we are now able to solve for force in member BG as: 𝐹𝐵𝐻 = 7.07 𝑘𝑖𝑝𝑠 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛) We can then plug in the calculated value for the forces in members BG and GH into the equation we derived by summing the forces about the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 0: 𝐹𝐵𝐶 + (7.07 𝑘)(cos 45°) + (15 𝑘) = 0
Solving for the force in member BC, we find: 𝐹𝐵𝐶 = −19.99 𝑘 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) Made with
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Therefore, the correct answer choice is D. 𝑩𝑪: 𝟐𝟎 (𝑪); 𝑮𝑯: 𝟏𝟓(𝑻)
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CONCEPT INTRO: The METHOD OF SECTIONS is the technique of solving for the internal forces in each truss member by passing an imaginary section through the truss, and cutting it into two parts.
The idea behind the method of section is that we can calculate forces in members by creating an imaginary cut through the structure, drawing the forces acting on the ends of the cut members, and then conducting a force and moment balance for one of the two parts of the structure. Made with
by Prepineer | Prepineer.com
The Method of Sections is used when we have a truss system that has multiple members, and we are only interested with forces of a few particular members. The Method of Sections is always used together with the Method of Joints to analyze a truss system. Since truss members are subject to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude as the forces at the joint. This result is based on the principle of static equilibrium and Newton’s third law.
In the Method of Joints, our analysis was limited to just solving for the external reactions and the two equations of equilibrium available for each joint. When working with the Method of Sections, we can also can also account for the equilibrium of moments, allowing us to solve for up to three unknown forces at a time. Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces, which represents the internal forces of the members. If two of the unknown forces are parallel, forces may be summed perpendicular to the direction of these unknowns to determine directly the third unknown force. Made with
by Prepineer | Prepineer.com
PROCEDURE FOR THE METHOD OF SECTIONS: Given a structural system represented as a truss system, we can use an imaginary section to cut through the truss system, and define two separate truss systems that are in static equilibrium. Once the systems are defined, we can use the equations of equilibrium to solve for the unknown forces, a define by the process below: 1. Define the coordinate system of the of the truss system. For the purpose of the FE Exam, we should always use a Cartesian Coordinate System when working with truss systems. As there will be varying angles in each triangular group of members, we should not incline the coordinate system in any scenario.
2. Draw a Free Body Diagram of the overall truss system and find the external support reactions, if required. This may require you to use the equations of equilibrium for the overall truss system, and sum the moments about a particular support constraint or joint.
You will typically find that a truss system is
supported by a roller and pin, so you will want to take the moment about the pin support constraint to account for the horizontal and vertical support reactions. Made with by Prepineer | Prepineer.com
3. Look at the overall free body diagram of the truss system, and identify which internal forces of members, the problem is asking you to solve for. It is important to remember that when working with the Method of Sections, the entire problem will center around you cutting the overall truss system into two separate systems in static equilibrium. Therefore, it is imperative that you pass the imagine section correctly, and following the guidelines listed below: a. Cut the member that you’re interested in to calculate the internal force of a truss member, you must first expose the internal force. The only way you can do that is by cutting the member. b. Cut the truss completely into two parts – Don’t cut partway through a truss and then stop. c. Cut a maximum of three members – you’re allowed to cut up to three members total. If you cut fewer than three that is perfectly ok. But if you cut four or more members, you may have some problems with this method as you may not be able to solve for all of the unknown on the free-body diagram. d. If you cut more than three members, all but two of the members must on the same line of action. In rare circumstance, you can actually cut more than three members, but you’ll need to make sure that all but at least two of them are collinear. In general, it’s a very rare occasion when you can actually cut more than three members at a time. 4. Make a cut along the members of interest based on your analysis in step 3. Draw an imaginary line through the members you are looking to analyze, and divide the truss system into two separate structural systems, both of which are in static equilibrium. Made with
by Prepineer | Prepineer.com
5. Draw a free body diagram of the particular truss system you wish to analyze, as one of the truss systems will not be used any further in the problem. For each cut member, assume the member is acting in tension and draw a tensile arrow going away from the joint. By doing this, the calculated forces will yield a positive scalar if our assumption of tension is correct, and a negative scalar if the member is actually in compression. 6. Cover up the member force you are looking for, and follow all other cut member forces until they meet at a point. Then sum moment about that point, thereby cancelling all other member forces since they now have no moment arms.
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CONCEPT EXAMPLE: The magnitude of the internal forces in members BC and GH is closes to which of the following answer choices?
A. 𝐵𝐶: 60 (𝐶); 𝐺𝐻: 60 (𝑇) B. 𝐵𝐶: 28 (𝐶); 𝐺𝐻: 60 (𝑇) C. 𝐵𝐶: 20 (𝑇); 𝐺𝐻: 15 (𝐶) D. 𝐵𝐶: 20 (𝐶); 𝐺𝐻: 15 (𝑇)
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SOLUTION: We are looking to solve for the forces in members BC and BH 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 x− and 𝑦 − 𝑎𝑥𝑒𝑠.
Made with
by Prepineer | Prepineer.com
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.
Made with
by Prepineer | Prepineer.com
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. The topic of the METHOD OF SECTIONS 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. Made with
by Prepineer | Prepineer.com
We will first sum the forces of the overall truss system along the 𝑦 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑦 = 0: 𝐹𝑦 + 𝐽𝑦 − 10 𝑘 − 10 𝑘 − 10 𝑘 − 10 𝑘 − 10 𝑘 = 0
𝐹𝑦 + 𝐽𝑦 = 50 𝑘 We then sum the forces of the overall truss system along the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 𝐹𝑥 = 0
𝐹𝑥 = 0 𝑘 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.
𝑀𝐹 = (𝐽𝑦 )(40) − (10 k)(10) − (10 𝑘)(20) − (10 k)(30) − (10 𝑘)(40") = 0 Solving for the vertical support reaction at point J, 𝐽𝑦 , we find: 𝐽𝑦 = 25 𝑘
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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 𝑦 − 𝑎𝑥𝑖𝑠, and solve for the vertical support reaction at joint F. 𝐹𝑦 + 𝐽𝑦 = 50 𝑘 𝐹𝑦 + 25 𝑘 = 50 𝑘 𝐹𝑦 = 25 𝑘 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.
Made with
by Prepineer | Prepineer.com
As we are solving for the forces in members BC and GH, a good location to draw our imaginary section would be vertically between joints B and C, and joints G and H. Looking at the free body diagram for the truss system below, we see the proposed location of the imaginary section would cut through three members to divide the truss systems into two separates truss systems in equilibrium. It is important to remember the maximum amount of members we can cut using the method of sections is three members.
Made with
by Prepineer | Prepineer.com
Looking below at the free body diagram for the truss system that we sectioned off from the overall truss system, we notice that we have all of the information we need to solve for the forces in members BC and GH, except the angle that member BH makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member BH, so we can use trigonometry to solve for the missing angle. We can now use the tangent trigonometric function to calculate the angle that member BH makes with the horizontal.
tan(𝜃) =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Plugging in the values of 10" for the opposite side of the triangle and 10" for the adjacent side of the triangle, we are able to calculate the angle of the member BH as:
𝜃 = tan−1 (
10 ) = 45° 10 Made with
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We can then plug in the calculated angle into the free body diagram for the truss system that we sectioned off from the overall truss system. The free body diagram now has all of the relevant information we need to solve for the forces in member BC and GH.
We can now redraw the free-body diagram for the sectioned truss system 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 the x- and y- axes for the overall truss system to determine the forces in member BC. It is important to realize that the force in member BH acts on joint B with an angle of 45°, so we will need to break that force’s magnitude into components when analyzing the vertical and horizontal planes. As we have 3 unknowns, we need to utilize all three equations of equilibrium, as we need at least three equations to solve for all of the unknowns. It is important to remember that the sectioned truss system is in static equilibrium at any point or joint. Therefore, we can sum the moments about joint B and set the equation equal to zero. ∑𝑀𝐵 = 0: (𝐹𝐺𝐻 )(10") + (10 𝑘)(10") − (25 𝑘)(10") = 0
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Solving for the force in member GH, we find: 𝐹𝐺𝐻 = 15 𝑘𝑖𝑝𝑠 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛) We will then sum the forces of the sectioned truss system along the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 0: 𝐹𝐵𝐶 + 𝐹𝐵𝐻 (cos 45°) + 𝐹𝐺𝐻 = 0
We will then sum the forces of the sectioned truss system along the 𝑦 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑦 = 0: 25 𝑘 − 10 𝑘 − 10 𝑘 − 𝐹𝐵𝐻 (sin 45°) = 0
Using the expression written for forces about the 𝑦 − 𝑎𝑥𝑖𝑠, we are now able to solve for force in member BG as: 𝐹𝐵𝐻 = 7.07 𝑘𝑖𝑝𝑠 (𝑡𝑒𝑛𝑠𝑖𝑜𝑛) We can then plug in the calculated value for the forces in members BG and GH into the equation we derived by summing the forces about the 𝑥 − 𝑎𝑥𝑖𝑠:
∑ 𝐹𝑥 = 0: 𝐹𝐵𝐶 + (7.07 𝑘)(cos 45°) + (15 𝑘) = 0
Solving for the force in member BC, we find: 𝐹𝐵𝐶 = −19.99 𝑘 (𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) Made with
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Therefore, the correct answer choice is D. 𝑩𝑪: 𝟐𝟎 (𝑪); 𝑮𝑯: 𝟏𝟓(𝑻)
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