2 t / m
DESCRIPTION
Xb. Xa. 2m. 2m. 2m. 2m. Yb. Ya. Yc. Draw the B.M.D. & the S.F.D. 2 t. 2 t / m. 1m. Assumed reactions. Step 1: Stability Check. No. of Unknown Reactions? 5. No. of Equilibrium Equations: 3 No. of Extra Conditions: 2 (two intermediate pins). - PowerPoint PPT PresentationTRANSCRIPT
Tarek Hegazy, Univ. of Waterloo [email protected]
2 t / m
2 t
1m
2m 2m 2m 2m
Xa
Ya
Yb
Xb
Yc
Step 1: Stability Check
No. of Equilibrium Equations: 3No. of Extra Conditions: 2 (two intermediate pins)
Since the unknowns 5 = Equations (3 + 2) Then, Structure is Stable & Statically Determinate
Assumed reactions
No. of Unknown Reactions? 5
Draw the B.M.D. & the S.F.D.
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2mYc
2 t / m
2 t
1mXa
Ya
Step 2: Reactions Sign Convention
=0
=0
=0 M
+
X+
Y+
Any point
For the whole structure considering all forces
including reactions
2m 2mYb
Xbe
d
Start writing an equation with least number of unknowns:Start writing an equation with least number of unknowns:
M+
Since (e) is an intermediate pin, then
at (e) right side only = 0
Since (e) is an intermediate pin, then
at (e) right side only = 0
Tarek Hegazy, Univ. of Waterloo [email protected]
M+
at (e), right side only = 0at (e), right side only = 0
2 t
2m 2m
Yb
Xb
e
-2 . 2-2 . 2 +Xb . 0 = 0+Xb . 0 = 0+Yb . 4+Yb . 4
Solve, Yb = +1 i.e., correct directionSolve, Yb = +1 i.e., correct direction
Sign Force Distance
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2mYc
2 t / m
2 t
1mXa
Ya2m 2m
1
Xb
M+
Since (d) is an intermediate pin, then
at (d) right side only = 0
Since (d) is an intermediate pin, then
at (d) right side only = 0
Let’s write another equation:Let’s write another equation:
d
Tarek Hegazy, Univ. of Waterloo [email protected]
M+
at (d), right side only = 0at (d), right side only = 0
2 t
2m 2m 1
Xb
d
-2 . 2-2 . 2 +Xb . 1 = 0+Xb . 1 = 0+1 . 4+1 . 4
Solve, Xb = 0Solve, Xb = 0
1m
Sign Force Distance
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2mYc
2 t / m
2 t
1mXa
Ya2m 2m
1
Now, = 0 , +Xa - 0 = 0, or Xa = 0Now, = 0 , +Xa - 0 = 0, or Xa = 0 X+
0
M+
Since (d) is an intermediate pin, then
at (d) left side only = 0
Since (d) is an intermediate pin, then
at (d) left side only = 0
0
Let’s now write two equations to get the last two unknowns:Let’s now write two equations to get the last two unknowns:
d
Tarek Hegazy, Univ. of Waterloo [email protected]
- Ya.4- Ya.4
2m 2mYc
2 t / m
Ya
d
M+
at (d), left side only = 0at (d), left side only = 0
Equivalent = 2 x 4 = 8Equivalent = 2 x 4 = 8
- Yc.2- Yc.2 + 8 .2 = 0+ 8 .2 = 0
Or, 4 Ya + 2 Yc = 16 . . . . . (1)Or, 4 Ya + 2 Yc = 16 . . . . . (1)
Sign Force Distance
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2mYc
2 t / m
2 t
1m
Ya2m 2m
1
Now, then,Now, then, Y=0+
Solve (1) & (2), Ya = -1 , Yc = 10 Opposite direction Same direction
Solve (1) & (2), Ya = -1 , Yc = 10 Opposite direction Same direction
+Ya +Yc +1 -2x4 -2 = 0 +Ya +Yc +1 -2x4 -2 = 0
4 Ya + 2 Yc = 16 . . . (1)4 Ya + 2 Yc = 16 . . . (1)
Ya + Yc = 9 . . . (2)Ya + Yc = 9 . . . (2)
All Up All Down
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
Let’s first define the beam sections with changes in load or beam’s Shape.
Let’s first define the beam sections with changes in load or beam’s Shape.
We are now ready to draw the B.M.D. and the S.F.D.
We are now ready to draw the B.M.D. and the S.F.D.
1 2
3 45
67 8
9
10
Now, we analyze each section separately, considering only one side of the structure.Now, we analyze each section separately, considering only one side of the structure.
Final Reactions:Final Reactions:
Shear is Paralle to the section & Perpendicular to the beam
Axial (i.e., Normal) force is parallel to the beam
Section
2Section
3
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
67 8
9
10
B.M.D.B.M.D.
Section Analysis:Section Analysis:Analyze the right side or the left side, whichever has less calculation.Analyze the right side or the left side, whichever has less calculation.
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
67 8
9
10
B.M.D.B.M.D.
Section 1Left
Section 1Left
Section 1 B.M. = -1 . 0 = 0Section 1 B.M. = -1 . 0 = 0
0
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
3 45
67 8
9
10
B.M.D.B.M.D.
1 2
Section 2 B.M. = -1 . 2 - 4 . 1 = - 6Section 2 B.M. = -1 . 2 - 4 . 1 = - 6
0
-6
Section 2Left
Section 2Left
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
3 45
67 8
9
10
B.M.D.B.M.D.
1 2
10
0
-6
Section 3 B.M. = - 1 . 2 - 4 . 1 +10 . 0 = -6Section 3 B.M. = - 1 . 2 - 4 . 1 +10 . 0 = -6
Section 3Left
Section 3Left
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
3 45
67 8
9
10
B.M.D.B.M.D.
1 2
10
Section 4 B.M. = -1 . 4 -4 . 3 -4 . 1 +10 . 2 = 0Section 4 B.M. = -1 . 4 -4 . 3 -4 . 1 +10 . 2 = 0
Section 4Left
Section 4Left
0
-6
0
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
3 45
67 8
9
10
B.M.D.B.M.D.
1 2
10
0
-6
0
Section 5Left
Section 5Left
Section 5 B.M. = 0 (same as section 4)Section 5 B.M. = 0 (same as section 4)
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
6
7 8
9
10
B.M.D.B.M.D.
Section 10Right
Section 10Right
Section 10 B.M. = +1 . 0 = 0Section 10 B.M. = +1 . 0 = 0
0
-6
0
0
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
6
7 8
9
10
B.M.D.B.M.D.
Section 9Right
Section 9Right
Section 9 B.M. = +1 . 2 = +2 Section 9 B.M. = +1 . 2 = +2
0
-6
0
0
+2
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
6
7 8
9
10
B.M.D.B.M.D.
Section 8Right
Section 8Right
0
-6
0
0
+2
Section 8 B.M. = +1 . 2 = +2 Section 8 B.M. = +1 . 2 = +2
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
6
7 8
9
10
B.M.D.B.M.D.
Sections 7 & 6Right
Sections 7 & 6Right
Sections 7 & 6 B.M. = 0 Sections 7 & 6 B.M. = 0
0
-6
0
0
+20
SS
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m
11 10
44 44
B.M.D.B.M.D.
0
-6
0
0
+2
0
-3-3
w.L2/8 = 1w.L2/8 = 1
Now, we connect the moment valuesNow, we connect the moment values
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
67 8
9
10
S.F.D.S.F.D.
Section 1Left
Section 1Left
Section 1 S.F. = -1Section 1 S.F. = -1 +Shear is a
force perpendicular to the beam
-1
S
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
3 45
67 8
9
10
1 2
Section 2 S.F. = -1 - 4 = - 5Section 2 S.F. = -1 - 4 = - 5
Section 2Left
Section 2Left
S.F.D.S.F.D.
-5
-1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
3
1 2
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
45
67 8
9
10
10
Section 3 S.F. = - 1 - 4 +10 = +5Section 3 S.F. = - 1 - 4 +10 = +5
Section 3Left
Section 3Left
S.F.D.S.F.D.
-5
-1
+5
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
3 456
7 8
9
10
1 2
10
Section 4 S.F. = - 1 - 4 +10 - 4 = +1Section 4 S.F. = - 1 - 4 +10 - 4 = +1
Section 4Left
Section 4Left
S.F.D.S.F.D.
-5
-1
+5+1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
10
Section 5Left
Section 5Left
Section 5 S.F. = 0 Section 5 S.F. = 0
S.F.D.S.F.D.
3 456
7 8
9
10
1 2
-5
-1
+5+1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1
44 44
10
Section 6Left
Section 6Left
Section 6 S.F. = 0 Section 6 S.F. = 0
S.F.D.S.F.D.
3 456
7 8
9
10
1 2
-5
-1
+5+1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
1 2
3 45
6
7 8
9
10
Section 10Right
Section 10Right
Section 10 S.F. = -1 Section 10 S.F. = -1
S.F.D.S.F.D.
-5
-1
+5+1
-1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
Section 9Right
Section 9Right
Section 9 S.F. = -1 Section 9 S.F. = -1
S.F.D.S.F.D.
1 2
3 45
6
7 8 9 10
-5
-1
+5+1
-1-1
S
+
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
Section 8Right
Section 8Right
S.F.D.S.F.D.
1 2
3 45
6
7 8 9 10
-5
-1
+5+1
-1-1
+1
S
+Section 8 S.F. = -1 +2 = +1Section 8 S.F. = -1 +2 = +1
Tarek Hegazy, Univ. of Waterloo [email protected]
2m 2m
2 t/m
2 t
1m
2m 2m1
1 10
44 44
Sections 7Right
Sections 7Right
Sections 7 S.F. = -1 +2 = +1 Sections 7 S.F. = -1 +2 = +1
S.F.D.S.F.D.
1 2
3 45
6
7 8 9 10
-5
-1
+5+1
-1-1
+1+1
+
S
Tarek Hegazy, Univ. of Waterloo [email protected]
S.F.D.S.F.D.
2m 2m
2 t/m
2 t
1m
2m 2m
11 10
44 44
-5
+5
+1
-1-1
+1+1
-1
Now, we connect the shear valuesNow, we connect the shear values
Tarek Hegazy, Univ. of Waterloo [email protected]
S.F.D.S.F.D.
2m 2m
2 t/m
2 t
1m
2m 2m
11 10
44 44
0
-6
0
0
+2
0
-3-3
w.L2/8 = 1w.L2/8 = 1
-5
+5
+1
-1-1
+1+1
-1
B.M.DB.M.D
Final AnswerFinal Answer