32 Method of Sections Problem Set
METHOD OF SECTIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Determine the best answer choice that represents the force in each of the following truss members: AB, AD, and CD.
A. ๐ด๐ด๐ด๐ด: 4.27 ๐ถ๐ถ ; ๐ด๐ด๐ด๐ด: 28.10 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 10 (๐ถ๐ถ)
B. ๐ด๐ด๐ด๐ด: 4.27 ๐๐ ; ๐ด๐ด๐ด๐ด: 28.10 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 10 (๐ถ๐ถ)
C. ๐ด๐ด๐ด๐ด: 3.97 ๐ถ๐ถ ; ๐ด๐ด๐ด๐ด: 15.04 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 16 (๐ถ๐ถ)
D. ๐ด๐ด๐ด๐ด: 3.97 ๐๐ ; ๐ด๐ด๐ด๐ด: 15.04 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 16 (๐ถ๐ถ)
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SOLUTION 1: We are looking to solve for the forces in members AB, AD, and CD 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 xโ and ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐.
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In order to solve for the forces in members AB, AD, and CD we will need to calculate the support reactions generated the reactions at joints A and C, particularly by the pinned constraint at joint C.
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In order to solve for the support reactions at joint C, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint C, there are support reactions to restrain against translation vertically and horizontally. At joint A, there is a roller that will only provide support against translation horizontally.
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 A and C. 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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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. We will first sum the forces of the overall truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: ๐ถ๐ถ> โ 6 ๐๐ โ 3 ๐๐ = 0 ๐ถ๐ถ> = 9 ๐๐๐๐๐๐๐๐ We then sum the forces of the overall truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: ๐ด๐ดB + ๐ถ๐ถB = 0 ๐ด๐ดB = โ๐ถ๐ถB
We will then sum the moments around the support constraint at point C to establish the third equation of equilibrium. Point C 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 C. ๐๐E = โ๐ด๐ดB 3 ๐๐๐๐)-(6 k)(4 ft โ 3 ๐๐ (8 ๐๐๐๐)
Solving for the horizontal support reaction at joint A, ๐ด๐ดB , we find:
๐ด๐ดB = โ16 ๐๐๐๐๐๐๐๐
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We can then plug in the calculated value for the horizontal support reaction at point A into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐, and solve for the horizontal support reaction at joint C. ๐ด๐ดB = โ๐ถ๐ถB ๐ถ๐ถB = โ๐ด๐ดB ๐ถ๐ถB = โ โ16 ๐๐ = 16 ๐๐๐๐๐๐๐๐
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 AB, AD, and CD, a good location to draw our imaginary section would be vertically between joints AB, and joints C and D. 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 AB, AD, and CD, except the angle that member AD makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member AD, 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 AD makes with the horizontal.
tan ๐๐ =
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด
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Plugging in the values of 3 ๐๐๐๐ for the opposite side of the triangle and 4 ๐๐๐๐ for the
adjacent side of the triangle, we are able to calculate the angle of the member AD as:
๐๐ = tanUV
3 = 36.87ยฐ 4
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 members AB, AD, and CD.
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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 members AB, AD, and CD. It is important to realize that the force in member AD acts on joint A with an angle of 36.87ยฐ, so we will need to break that force 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 A and set the equation equal to zero. When summing the moments about a joint, we do not need to include any forces that goes through the joint, as the joint will be located on the forceโs line of action. โ๐๐Y = 0: 16 ๐๐๐๐๐๐๐๐ 3 ๐๐๐๐ + ๐น๐นEZ 3 ๐๐๐๐ = 0
Solving for the force in member CD, we find: ๐น๐นEZ = โ16 ๐๐๐๐๐๐๐๐ (๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐)
We will then sum the forces of the sectioned truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ16 ๐๐ + ๐น๐น]Y + ๐น๐น]Z cos 36.87ยฐ + 16 ๐๐ + ๐น๐นEZ = 0 We will then sum the forces of the sectioned truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐:
๐น๐น> = 0: 9 ๐๐ โ ๐น๐น]Z sin 36.76ยฐ = 0 Made with
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Solving for the force in member AD we find: ๐น๐น]Z = 15.04 ๐๐๐๐๐๐๐๐
We can then plug in the calculated value for the forces in members AD and CD into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ16 ๐๐ + ๐น๐น]Y + ๐น๐น]Z cos 36.87ยฐ + 16 ๐๐ + ๐น๐นEZ = 0 โ16 ๐๐ + ๐น๐น]Y + (15.04 ๐๐)(cos 36.87ยฐ) + 16 ๐๐ + โ16 ๐๐ = 0
Solving for the force in member AB, we find: ๐น๐น]Y = 3.97 ๐๐๐๐๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ )
Therefore, the correct answer choice is D. ๐จ๐จ๐จ๐จ: ๐๐. ๐๐๐๐ ๐ป๐ป ; ๐จ๐จ๐จ๐จ: ๐๐๐๐. ๐๐๐๐ ๐ป๐ป ; ๐ช๐ช๐ช๐ช: ๐๐๐๐ ๐ช๐ช
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PROBLEM 2: In the method of sections, what is the maximum number of members in which a cut or imaginary section be passed through? A. 1 B. 2 C. 3
D. 4
SOLUTION 2: The topic of the METHOD OF SECTIONS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
The method of section also a maximum of three members to be cut. 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.
Therefore, the correct answer choice is C. 3
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PROBLEM 3: Determine the force in members GE, GC, and BC for the truss system shown below.
A. ๐ต๐ต๐ต๐ต: 400 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 300 ๐๐ ; ๐บ๐บ๐บ๐บ: 1200 (๐๐)
B. ๐ต๐ต๐ต๐ต: 400 ๐๐ ; ๐บ๐บ๐บ๐บ: 300 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 1200 (๐๐) C. ๐ต๐ต๐ต๐ต: 800 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 500 ๐๐ ; ๐บ๐บ๐บ๐บ: 737 (๐ถ๐ถ)
D. ๐ต๐ต๐ต๐ต: 800 ๐๐ ; ๐บ๐บ๐บ๐บ: 500 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 737 (๐๐)
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SOLUTION 3: We are looking to solve for the forces in members GE, GC, and BC 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 xโ and ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐.
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By visual inspection, we notice that the truss system is symmetrical, and thus the forces on the members are symmetrical as well. Thus, we only need to analyze half of the truss, which we can isolate using the method of sections. In order to solve for the forces in members GE, GC, and BC, we will need to calculate the support reactions generated the reactions at joints A and D, particularly by the pinned constraint at joint A.
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In order to solve for the support reactions at joint A, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint A there are support reactions to restrain against translation vertically and horizontally. At joint D, 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 A and D. 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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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. We will first sum the forces of the overall truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: ๐ด๐ด> + ๐ท๐ท> โ 1200 ๐๐๐๐ ๐ด๐ด> + ๐ท๐ท> = 1200 ๐๐๐๐ We then sum the forces of the overall truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = ๐ด๐ดB + 400 ๐๐๐๐ = 0 ๐ด๐ดB = โ400 ๐๐๐๐
We will then sum the moments around the support constraint at point A to establish the third equation of equilibrium. Point A 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. ๐๐] = โ1200 ๐๐๐๐ (8 ๐๐) โ 400 ๐๐๐๐ (3 ๐๐) + (๐ท๐ท> )(12 ๐๐) = 0 Solving for the vertical support reaction at point D, ๐ท๐ท> , we find:
๐ท๐ท> = 900 ๐๐๐๐
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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 ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐, and solve for the vertical support reaction at joint A. ๐ด๐ด> + ๐ท๐ท> = 1200 ๐๐๐๐ ๐ด๐ด> + (900 ๐๐๐๐) = 1200 ๐๐๐๐ ๐ด๐ด> = 300 ๐๐๐๐ 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 GE, GC, and BC, a good location to draw our imaginary section would be vertically between joints GE, and joints B and C. 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 GE, GC, and BC, except the angle that member GC makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member GC, 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 GC makes with the horizontal.
tan ๐๐ =
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด Made with
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Plugging in the values of 3 ๐๐ for the opposite side of the triangle and 4 ๐๐ for the
adjacent side of the triangle, we are able to calculate the angle of the member BH as:
๐๐ = tanUV
3 = 36.87ยฐ 4
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 members GE, GC, and BC.
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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 (-).
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 GC acts on joint G with an angle of 36.87ยฐ, so we will need to break that force into components when analyzing the vertical and horizontal planes.
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Looking at the problem, we see that along the x-axis we have 3 unknown forces and 1 unknown force acting along the y-axis. 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 G and set the equation equal to zero. โ๐๐r = 0: โ 400 ๐๐๐๐ 3 ๐๐ โ 300 ๐๐๐๐ 4 ๐๐ + (๐น๐นYE )(3 ๐๐) = 0
Solving for the force in member BC, we find: ๐น๐นYE = 800 ๐๐๐๐๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก)
We will then sum the forces of the sectioned truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + ๐น๐นrE cos 36.87ยฐ + ๐น๐นYE = 0 We will then sum the forces of the sectioned truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: 300 ๐๐๐๐ โ ๐น๐นrE sin 36.87ยฐ = 0 Using the expression written for forces about the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐, we are now able to solve for force in member GC as:
๐น๐นrE = 500 ๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก)
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We can then plug in the calculated value for the forces in members BC and GC into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + ๐น๐นrE cos 36.87ยฐ + ๐น๐นYE = 0 ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + (500 ๐๐๐๐) cos 36.87ยฐ + (800 ๐๐๐๐) = 0 Solving for the force in member GE, we find: ๐น๐นrs = โ737.77 ๐๐๐๐ (๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐)
Therefore, the correct answer choice is C. ๐ฉ๐ฉ๐ฉ๐ฉ: ๐๐๐๐๐๐ ๐ช๐ช ; ๐ฎ๐ฎ๐ฎ๐ฎ: ๐๐๐๐๐๐ ๐ป๐ป ; ๐ฎ๐ฎ๐ฎ๐ฎ: ๐๐๐๐๐๐ (๐ช๐ช)
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PROBLEM 4: If a simple truss member carries a tensile force of F along its length, what is the internal force of the member is best described as ____________. A. Tensile with a magnitude of F/2
B. Tensile wiht a magnitude of F
C. Compressive with a magnitude of F/2
D. Compressive with a magntiude of F
SOLUTION 4: The topic of the METHOD OF SECTIONS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. In the problem, we are told a member is acting in tension with a magnitude of force F.
Therefore, the correct answer choice is B. Tensile with a magnitude of F
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PROBLEM 1: Determine the best answer choice that represents the force in each of the following truss members: AB, AD, and CD.
A. ๐ด๐ด๐ด๐ด: 4.27 ๐ถ๐ถ ; ๐ด๐ด๐ด๐ด: 28.10 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 10 (๐ถ๐ถ)
B. ๐ด๐ด๐ด๐ด: 4.27 ๐๐ ; ๐ด๐ด๐ด๐ด: 28.10 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 10 (๐ถ๐ถ)
C. ๐ด๐ด๐ด๐ด: 3.97 ๐ถ๐ถ ; ๐ด๐ด๐ด๐ด: 15.04 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 16 (๐ถ๐ถ)
D. ๐ด๐ด๐ด๐ด: 3.97 ๐๐ ; ๐ด๐ด๐ด๐ด: 15.04 ๐๐ ; ๐ถ๐ถ๐ถ๐ถ: 16 (๐ถ๐ถ)
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SOLUTION 1: We are looking to solve for the forces in members AB, AD, and CD 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 xโ and ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐.
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In order to solve for the forces in members AB, AD, and CD we will need to calculate the support reactions generated the reactions at joints A and C, particularly by the pinned constraint at joint C.
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In order to solve for the support reactions at joint C, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint C, there are support reactions to restrain against translation vertically and horizontally. At joint A, there is a roller that will only provide support against translation horizontally.
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 A and C. 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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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. We will first sum the forces of the overall truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: ๐ถ๐ถ> โ 6 ๐๐ โ 3 ๐๐ = 0 ๐ถ๐ถ> = 9 ๐๐๐๐๐๐๐๐ We then sum the forces of the overall truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: ๐ด๐ดB + ๐ถ๐ถB = 0 ๐ด๐ดB = โ๐ถ๐ถB
We will then sum the moments around the support constraint at point C to establish the third equation of equilibrium. Point C 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 C. ๐๐E = โ๐ด๐ดB 3 ๐๐๐๐)-(6 k)(4 ft โ 3 ๐๐ (8 ๐๐๐๐)
Solving for the horizontal support reaction at joint A, ๐ด๐ดB , we find:
๐ด๐ดB = โ16 ๐๐๐๐๐๐๐๐
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We can then plug in the calculated value for the horizontal support reaction at point A into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐, and solve for the horizontal support reaction at joint C. ๐ด๐ดB = โ๐ถ๐ถB ๐ถ๐ถB = โ๐ด๐ดB ๐ถ๐ถB = โ โ16 ๐๐ = 16 ๐๐๐๐๐๐๐๐
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 AB, AD, and CD, a good location to draw our imaginary section would be vertically between joints AB, and joints C and D. 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 AB, AD, and CD, except the angle that member AD makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member AD, 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 AD makes with the horizontal.
tan ๐๐ =
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด
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Plugging in the values of 3 ๐๐๐๐ for the opposite side of the triangle and 4 ๐๐๐๐ for the
adjacent side of the triangle, we are able to calculate the angle of the member AD as:
๐๐ = tanUV
3 = 36.87ยฐ 4
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 members AB, AD, and CD.
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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 members AB, AD, and CD. It is important to realize that the force in member AD acts on joint A with an angle of 36.87ยฐ, so we will need to break that force 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 A and set the equation equal to zero. When summing the moments about a joint, we do not need to include any forces that goes through the joint, as the joint will be located on the forceโs line of action. โ๐๐Y = 0: 16 ๐๐๐๐๐๐๐๐ 3 ๐๐๐๐ + ๐น๐นEZ 3 ๐๐๐๐ = 0
Solving for the force in member CD, we find: ๐น๐นEZ = โ16 ๐๐๐๐๐๐๐๐ (๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐)
We will then sum the forces of the sectioned truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ16 ๐๐ + ๐น๐น]Y + ๐น๐น]Z cos 36.87ยฐ + 16 ๐๐ + ๐น๐นEZ = 0 We will then sum the forces of the sectioned truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐:
๐น๐น> = 0: 9 ๐๐ โ ๐น๐น]Z sin 36.76ยฐ = 0 Made with
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Solving for the force in member AD we find: ๐น๐น]Z = 15.04 ๐๐๐๐๐๐๐๐
We can then plug in the calculated value for the forces in members AD and CD into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ16 ๐๐ + ๐น๐น]Y + ๐น๐น]Z cos 36.87ยฐ + 16 ๐๐ + ๐น๐นEZ = 0 โ16 ๐๐ + ๐น๐น]Y + (15.04 ๐๐)(cos 36.87ยฐ) + 16 ๐๐ + โ16 ๐๐ = 0
Solving for the force in member AB, we find: ๐น๐น]Y = 3.97 ๐๐๐๐๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ )
Therefore, the correct answer choice is D. ๐จ๐จ๐จ๐จ: ๐๐. ๐๐๐๐ ๐ป๐ป ; ๐จ๐จ๐จ๐จ: ๐๐๐๐. ๐๐๐๐ ๐ป๐ป ; ๐ช๐ช๐ช๐ช: ๐๐๐๐ ๐ช๐ช
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PROBLEM 2: In the method of sections, what is the maximum number of members in which a cut or imaginary section be passed through? A. 1 B. 2 C. 3
D. 4
SOLUTION 2: The topic of the METHOD OF SECTIONS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing.
The method of section also a maximum of three members to be cut. 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.
Therefore, the correct answer choice is C. 3
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PROBLEM 3: Determine the force in members GE, GC, and BC for the truss system shown below.
A. ๐ต๐ต๐ต๐ต: 400 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 300 ๐๐ ; ๐บ๐บ๐บ๐บ: 1200 (๐๐)
B. ๐ต๐ต๐ต๐ต: 400 ๐๐ ; ๐บ๐บ๐บ๐บ: 300 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 1200 (๐๐) C. ๐ต๐ต๐ต๐ต: 800 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 500 ๐๐ ; ๐บ๐บ๐บ๐บ: 737 (๐ถ๐ถ)
D. ๐ต๐ต๐ต๐ต: 800 ๐๐ ; ๐บ๐บ๐บ๐บ: 500 ๐ถ๐ถ ; ๐บ๐บ๐บ๐บ: 737 (๐๐)
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SOLUTION 3: We are looking to solve for the forces in members GE, GC, and BC 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 xโ and ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐.
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By visual inspection, we notice that the truss system is symmetrical, and thus the forces on the members are symmetrical as well. Thus, we only need to analyze half of the truss, which we can isolate using the method of sections. In order to solve for the forces in members GE, GC, and BC, we will need to calculate the support reactions generated the reactions at joints A and D, particularly by the pinned constraint at joint A.
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In order to solve for the support reactions at joint A, we will draw a free body diagram for the overall truss system. As there is a pinned support at joint A there are support reactions to restrain against translation vertically and horizontally. At joint D, 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 A and D. 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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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. We will first sum the forces of the overall truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: ๐ด๐ด> + ๐ท๐ท> โ 1200 ๐๐๐๐ ๐ด๐ด> + ๐ท๐ท> = 1200 ๐๐๐๐ We then sum the forces of the overall truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = ๐ด๐ดB + 400 ๐๐๐๐ = 0 ๐ด๐ดB = โ400 ๐๐๐๐
We will then sum the moments around the support constraint at point A to establish the third equation of equilibrium. Point A 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. ๐๐] = โ1200 ๐๐๐๐ (8 ๐๐) โ 400 ๐๐๐๐ (3 ๐๐) + (๐ท๐ท> )(12 ๐๐) = 0 Solving for the vertical support reaction at point D, ๐ท๐ท> , we find:
๐ท๐ท> = 900 ๐๐๐๐
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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 ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐, and solve for the vertical support reaction at joint A. ๐ด๐ด> + ๐ท๐ท> = 1200 ๐๐๐๐ ๐ด๐ด> + (900 ๐๐๐๐) = 1200 ๐๐๐๐ ๐ด๐ด> = 300 ๐๐๐๐ 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 GE, GC, and BC, a good location to draw our imaginary section would be vertically between joints GE, and joints B and C. 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 GE, GC, and BC, except the angle that member GC makes with the horizontal.
We are given the vertical distance and horizontal distance that define the geometry of member GC, 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 GC makes with the horizontal.
tan ๐๐ =
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด๐ด Made with
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Plugging in the values of 3 ๐๐ for the opposite side of the triangle and 4 ๐๐ for the
adjacent side of the triangle, we are able to calculate the angle of the member BH as:
๐๐ = tanUV
3 = 36.87ยฐ 4
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 members GE, GC, and BC.
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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 (-).
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 GC acts on joint G with an angle of 36.87ยฐ, so we will need to break that force into components when analyzing the vertical and horizontal planes.
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Looking at the problem, we see that along the x-axis we have 3 unknown forces and 1 unknown force acting along the y-axis. 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 G and set the equation equal to zero. โ๐๐r = 0: โ 400 ๐๐๐๐ 3 ๐๐ โ 300 ๐๐๐๐ 4 ๐๐ + (๐น๐นYE )(3 ๐๐) = 0
Solving for the force in member BC, we find: ๐น๐นYE = 800 ๐๐๐๐๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก)
We will then sum the forces of the sectioned truss system along the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + ๐น๐นrE cos 36.87ยฐ + ๐น๐นYE = 0 We will then sum the forces of the sectioned truss system along the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐: ๐น๐น> = 0: 300 ๐๐๐๐ โ ๐น๐นrE sin 36.87ยฐ = 0 Using the expression written for forces about the ๐ฆ๐ฆ โ ๐๐๐๐๐๐๐๐, we are now able to solve for force in member GC as:
๐น๐นrE = 500 ๐๐๐๐ (๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก๐ก)
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We can then plug in the calculated value for the forces in members BC and GC into the equation we derived by summing the forces about the ๐ฅ๐ฅ โ ๐๐๐๐๐๐๐๐: ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + ๐น๐นrE cos 36.87ยฐ + ๐น๐นYE = 0 ๐น๐นB = 0: โ400 ๐๐๐๐ + ๐น๐นrs + (500 ๐๐๐๐) cos 36.87ยฐ + (800 ๐๐๐๐) = 0 Solving for the force in member GE, we find: ๐น๐นrs = โ737.77 ๐๐๐๐ (๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐)
Therefore, the correct answer choice is C. ๐ฉ๐ฉ๐ฉ๐ฉ: ๐๐๐๐๐๐ ๐ช๐ช ; ๐ฎ๐ฎ๐ฎ๐ฎ: ๐๐๐๐๐๐ ๐ป๐ป ; ๐ฎ๐ฎ๐ฎ๐ฎ: ๐๐๐๐๐๐ (๐ช๐ช)
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PROBLEM 4: If a simple truss member carries a tensile force of F along its length, what is the internal force of the member is best described as ____________. A. Tensile with a magnitude of F/2
B. Tensile wiht a magnitude of F
C. Compressive with a magnitude of F/2
D. Compressive with a magntiude of F
SOLUTION 4: The topic of the METHOD OF SECTIONS can be referenced under the topic of STATICS on page 68 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. In the problem, we are told a member is acting in tension with a magnitude of force F.
Therefore, the correct answer choice is B. Tensile with a magnitude of F
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