Method-of-sections.pdf
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Method of Sections ◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section.
Method of Sections ◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members.
Method of Sections ◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time.
Method of Sections ◮
◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time. Since the Method of Sections allows solving for up to three unknown forces at a time, you should choose sections that involve cutting through no more than three members at a time.
Method of Sections ◮
◮
◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time. Since the Method of Sections allows solving for up to three unknown forces at a time, you should choose sections that involve cutting through no more than three members at a time. When a member force points toward the joint it is attached to, the member is in compression. If that force points away from the joint it is attached to, the member is in tension.
Method of Sections Refer back to the end of the ”truss-initial-analysis.pdf” file to see what has been solved so far for the truss. This is what has been solved for so far: 120N B
D
F
150N
135N 3m
15N 120N A
C 4m
E 4m
G 4m
H 4m
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
150N
135N 3m
15N 120N A
C 4m
E 4m
G 4m
H 4m
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D b
15N 120N b
A
C
Let’s redraw this section enlarged.
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N 120 N b
A
4m
C
4m
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N
FBC
120 N b
A
4m
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ).
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N FAC
120 N A
4m
FBC b
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point B has both forces at A giving clockwise moments. Therefore, FAC must point to the right to provide a counter-clockwise moment.
Method of Sections - Cutting through AC, BC and BD 120 N B
Db FBD 3m
15 N FAC
120 N A
4m
FBC b
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point B has both forces at A giving clockwise moments. Therefore, FAC must point to the right to provide a counter-clockwise moment. Taking moments about point C has the 15 N force acting at A and the 120 N acting at B giving clockwise moments. Therefore, FBD must point to the left to provide a counter-clockwise moment.
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension)
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension) +
X
MB = −(120N)(3m) − (15N)(4m) + FAC (3m) = 0
FAC =
(360 + 60)Nm = 140N (tension) 3m
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension) +
X
MB = −(120N)(3m) − (15N)(4m) + FAC (3m) = 0
FAC = +
X
(360 + 60)Nm = 140N (tension) 3m
MC = −(15N)(4m) − (120N)(3m) + FBD (3m) = 0
FBD =
(60 + 360)Nm = 140N (compression) 3m
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
◮
Each member that is cut represents an unknown force. Look to see if there is a direction (horizontal or vertical) that has only one unknown. If this true, you should balance forces in that direction. In the section just looked at, this would be the forces in the vertical direction since only FBC has a vertical component.
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
◮
Each member that is cut represents an unknown force. Look to see if there is a direction (horizontal or vertical) that has only one unknown. If this true, you should balance forces in that direction. In the section just looked at, this would be the forces in the vertical direction since only FBC has a vertical component.
◮
If possible, take moments about points that two of the three unknown forces have lines of forces that pass through that point. This will result in just one unknown in that moment equation. In the section just looked at, taking moments about point B eliminates the unknowns FBC and FBD . Similarly, taking moments about point C eliminates the unknowns FBC and FAC from the equation.
Method of Sections - Cutting through BD, CD and CE 120 N B
Db
3m 15 N 120 N b
A
4m
C
4m
E
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD 3m
15 N 120 N b
A
4m
C
4m
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C.
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD 3m
FCD 15 N 120 N b
A
4m
C
4m
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C. Since FCD is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ).
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD FCD
3m
15 N FCE
120 N A
4m
C
4m
b
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C. Since FCD is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point D has the 120 N force and 15 N force acting at A giving clockwise moments. Therefore FCE must point to the right to give a counter-clockwise moment to balance this out.
Method of Sections - Cutting through BD, CD and CE Solving in the order of the previous page: ↑+
X
3 Fy = +15N − FCD = 0 5 5 FCD = (15N) = 25N (compression) 3
Method of Sections - Cutting through BD, CD and CE Solving in the order of the previous page: ↑+
+
X
X
3 Fy = +15N − FCD = 0 5 5 FCD = (15N) = 25N (compression) 3
MD = −(120N)(3m) − (15N)(8m) + FCE (3m) = 0 FCE =
(360 + 120)Nm = 160N (tension) 3m
Method of Sections - Cutting through DF, DG and EG Db
F
150 N
135 N b
E
4m
G
4m
H
3m
Method of Sections - Cutting through DF, DG and EG Db
F
150 N
FDG 135 N
3m
b
E
4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H.
Method of Sections - Cutting through DF, DG and EG Db
FDF
F
150 N
FDG 135 N
3m
b
E
4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H. Taking moments about point G has the 135 N force at H giving a counter-clockwise moment. Therefore FDF must point to the right to give a clockwise moment about point G to balance this out.
Method of Sections - Cutting through DF, DG and EG Db
FDF
F
150 N
FDG 135 N
b
E
3m
FEG 4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H. Taking moments about point G has the 135 N force at H giving a counter-clockwise moment. Therefore FDF must point to the right to give a clockwise moment about point G to balance this out. Taking moments about point D has the 150 N force acting clockwise and the 135 N force acting counter-clockwise. The 135 N force has twice the moment arm so FEG must point left to give a clockwise moment to balance this out.
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3 X MG = +(135N)(4m) − FDF (3m) = 0 + FDF =
540Nm = 180N (compression) 3m
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3 X MG = +(135N)(4m) − FDF (3m) = 0 + FDF =
+
X
540Nm = 180N (compression) 3m
MD = −(150N)(4m) + (135N)(8m) − FEG (3m) = 0 FEG =
480Nm (−600 + 1080)Nm = = 160N (tension) 3m 3m
Method of Sections - Cutting through DF, FG and GH Db
F
150 N
135 N b
E
b
4m
G
4m
H
3m
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N b
E
3m
b
4m
G
4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result.
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N
3m
FFG b
E
b
4m
G
4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result. Since FFG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger than the 135 N force at H.
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N
3m
FFG b
b
E
G
4m
FGH 4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result. Since FFG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger than the 135 N force at H. Taking moments about point F has the 135 N force acting counter-clockwise. The means that FGH must point left to give a clockwise moment to balance this out.
Method of Sections - Cutting through DF, FG and GH Solving in the order of the previous page: X Fy = −150N + 135N + FFG = 0 ↑+ FFG = 150N − 135N = 15N (compression)
Method of Sections - Cutting through DF, FG and GH Solving in the order of the previous page: X Fy = −150N + 135N + FFG = 0 ↑+ FFG = 150N − 135N = 15N (compression) +
X
MF = +(135N)(4m) − FGH (3m) = 0
FGH =
540Nm = 180N (tension) 3m
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
◮
To solve for the force in member FH, you would cut through FH and GH. This is equivalent to applying the method of joints at joint H.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
◮
To solve for the force in member FH, you would cut through FH and GH. This is equivalent to applying the method of joints at joint H.
◮
To solve for the force in member DE, you would cut through CE, DE and EG. This is equivalent to applying the method of joints at joint E.
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section.
Method of Sections ◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members.
Method of Sections ◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time.
Method of Sections ◮
◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time. Since the Method of Sections allows solving for up to three unknown forces at a time, you should choose sections that involve cutting through no more than three members at a time.
Method of Sections ◮
◮
◮
◮
◮
The Method of Sections involves analytically cutting the truss into sections and solving for static equilibrium for each section. The sections are obtained by cutting through some of the members of the truss to expose the force inside the members. In the Method of Joints, we are dealing with static equilibrium at a point. This limits the static equilibrium equations to just the two force equations. A section has finite size and this means you can also use moment equations to solve the problem. This allows solving for up to three unknown forces at a time. Since the Method of Sections allows solving for up to three unknown forces at a time, you should choose sections that involve cutting through no more than three members at a time. When a member force points toward the joint it is attached to, the member is in compression. If that force points away from the joint it is attached to, the member is in tension.
Method of Sections Refer back to the end of the ”truss-initial-analysis.pdf” file to see what has been solved so far for the truss. This is what has been solved for so far: 120N B
D
F
150N
135N 3m
15N 120N A
C 4m
E 4m
G 4m
H 4m
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
150N
135N 3m
15N 120N A
C 4m
E 4m
G 4m
H 4m
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D
F
15N 120N A
C
E
Let’s slide the rest of the truss out of the way.
G
H
Method of Sections - Cutting through AC, BC and BD Let’s create a section by cutting through members AC, BC and BD. Recall that we want to cut through at most three members. 120N B
D b
15N 120N b
A
C
Let’s redraw this section enlarged.
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N 120 N b
A
4m
C
4m
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N
FBC
120 N b
A
4m
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ).
Method of Sections - Cutting through AC, BC and BD 120 N B
Db
3m 15 N FAC
120 N A
4m
FBC b
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point B has both forces at A giving clockwise moments. Therefore, FAC must point to the right to provide a counter-clockwise moment.
Method of Sections - Cutting through AC, BC and BD 120 N B
Db FBD 3m
15 N FAC
120 N A
4m
FBC b
C
4m
Since FBC is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point B has both forces at A giving clockwise moments. Therefore, FAC must point to the right to provide a counter-clockwise moment. Taking moments about point C has the 15 N force acting at A and the 120 N acting at B giving clockwise moments. Therefore, FBD must point to the left to provide a counter-clockwise moment.
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension)
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension) +
X
MB = −(120N)(3m) − (15N)(4m) + FAC (3m) = 0
FAC =
(360 + 60)Nm = 140N (tension) 3m
Method of Sections - Cutting through AC, BC and BD Solving in the order of the previous page: X Fy = +15N − FBC = 0 ↑+ FBC = 15N (tension) +
X
MB = −(120N)(3m) − (15N)(4m) + FAC (3m) = 0
FAC = +
X
(360 + 60)Nm = 140N (tension) 3m
MC = −(15N)(4m) − (120N)(3m) + FBD (3m) = 0
FBD =
(60 + 360)Nm = 140N (compression) 3m
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
◮
Each member that is cut represents an unknown force. Look to see if there is a direction (horizontal or vertical) that has only one unknown. If this true, you should balance forces in that direction. In the section just looked at, this would be the forces in the vertical direction since only FBC has a vertical component.
Method of Sections - Important Points ◮
When drawing your sections, include the points that the cut members would have connected to if not cut. In the section just looked at, this would be points C and D.
◮
Each member that is cut represents an unknown force. Look to see if there is a direction (horizontal or vertical) that has only one unknown. If this true, you should balance forces in that direction. In the section just looked at, this would be the forces in the vertical direction since only FBC has a vertical component.
◮
If possible, take moments about points that two of the three unknown forces have lines of forces that pass through that point. This will result in just one unknown in that moment equation. In the section just looked at, taking moments about point B eliminates the unknowns FBC and FBD . Similarly, taking moments about point C eliminates the unknowns FBC and FAC from the equation.
Method of Sections - Cutting through BD, CD and CE 120 N B
Db
3m 15 N 120 N b
A
4m
C
4m
E
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD 3m
15 N 120 N b
A
4m
C
4m
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C.
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD 3m
FCD 15 N 120 N b
A
4m
C
4m
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C. Since FCD is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ).
Method of Sections - Cutting through BD, CD and CE 120 N B
Db FBD FCD
3m
15 N FCE
120 N A
4m
C
4m
b
E
Since we know (from the previous section) the direction of FBD we draw that in first. We could also reason this direction by taking moments about point C. Since FCD is the only force that has a vertical component, it must point down to balance the 15 N force (Ay ). Taking moments about point D has the 120 N force and 15 N force acting at A giving clockwise moments. Therefore FCE must point to the right to give a counter-clockwise moment to balance this out.
Method of Sections - Cutting through BD, CD and CE Solving in the order of the previous page: ↑+
X
3 Fy = +15N − FCD = 0 5 5 FCD = (15N) = 25N (compression) 3
Method of Sections - Cutting through BD, CD and CE Solving in the order of the previous page: ↑+
+
X
X
3 Fy = +15N − FCD = 0 5 5 FCD = (15N) = 25N (compression) 3
MD = −(120N)(3m) − (15N)(8m) + FCE (3m) = 0 FCE =
(360 + 120)Nm = 160N (tension) 3m
Method of Sections - Cutting through DF, DG and EG Db
F
150 N
135 N b
E
4m
G
4m
H
3m
Method of Sections - Cutting through DF, DG and EG Db
F
150 N
FDG 135 N
3m
b
E
4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H.
Method of Sections - Cutting through DF, DG and EG Db
FDF
F
150 N
FDG 135 N
3m
b
E
4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H. Taking moments about point G has the 135 N force at H giving a counter-clockwise moment. Therefore FDF must point to the right to give a clockwise moment about point G to balance this out.
Method of Sections - Cutting through DF, DG and EG Db
FDF
F
150 N
FDG 135 N
b
E
3m
FEG 4m
G
4m
H
Since FDG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger the 135 N force at H. Taking moments about point G has the 135 N force at H giving a counter-clockwise moment. Therefore FDF must point to the right to give a clockwise moment about point G to balance this out. Taking moments about point D has the 150 N force acting clockwise and the 135 N force acting counter-clockwise. The 135 N force has twice the moment arm so FEG must point left to give a clockwise moment to balance this out.
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3 X MG = +(135N)(4m) − FDF (3m) = 0 + FDF =
540Nm = 180N (compression) 3m
Method of Sections - Cutting through DF, DG and EG Solving in the order of the previous page: ↑+
X
3 Fy = −150N + 135N + FDG = 0 5 5 5 FDG = (150N − 135N) = (15N) = 25N (tension) 3 3 X MG = +(135N)(4m) − FDF (3m) = 0 + FDF =
+
X
540Nm = 180N (compression) 3m
MD = −(150N)(4m) + (135N)(8m) − FEG (3m) = 0 FEG =
480Nm (−600 + 1080)Nm = = 160N (tension) 3m 3m
Method of Sections - Cutting through DF, FG and GH Db
F
150 N
135 N b
E
b
4m
G
4m
H
3m
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N b
E
3m
b
4m
G
4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result.
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N
3m
FFG b
E
b
4m
G
4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result. Since FFG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger than the 135 N force at H.
Method of Sections - Cutting through DF, FG and GH Db
FDF
F
150 N
135 N
3m
FFG b
b
E
G
4m
FGH 4m
H
From the previous section, we know FDF points right. Taking moments about G would also give this result. Since FFG is the only unknown with a vertical component, it must point up since the 150 N force at F is bigger than the 135 N force at H. Taking moments about point F has the 135 N force acting counter-clockwise. The means that FGH must point left to give a clockwise moment to balance this out.
Method of Sections - Cutting through DF, FG and GH Solving in the order of the previous page: X Fy = −150N + 135N + FFG = 0 ↑+ FFG = 150N − 135N = 15N (compression)
Method of Sections - Cutting through DF, FG and GH Solving in the order of the previous page: X Fy = −150N + 135N + FFG = 0 ↑+ FFG = 150N − 135N = 15N (compression) +
X
MF = +(135N)(4m) − FGH (3m) = 0
FGH =
540Nm = 180N (tension) 3m
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
◮
To solve for the force in member FH, you would cut through FH and GH. This is equivalent to applying the method of joints at joint H.
Method of Sections - Remaining members ◮
For the rest of the members, AB, DE and FH, the only sections that would cut through them amount to applying the Method of Joints.
◮
To solve for the force in member AB, you would cut through AB and AC. This is equivalent to applying the method of joints at joint A.
◮
To solve for the force in member FH, you would cut through FH and GH. This is equivalent to applying the method of joints at joint H.
◮
To solve for the force in member DE, you would cut through CE, DE and EG. This is equivalent to applying the method of joints at joint E.
More Documents from "Abhinandan Ramkrishnan"
Method-of-sections.pdf
June 2020 0