tarek hegazy, univ. of waterloo 2 t / m 2 t 1m 2m xa ya yb xb yc step 1: stability check no. of...

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.


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:

M+

Since (e) is an intermediate pin, then

at (e) right side only = 0


M+

at (e), right side only = 0

2 t

2m 2m

Yb

Xb

e

-2 . 2 +Xb . 0 = 0+Yb . 4

Solve, Yb = +1 i.e., correct directionSign Force Distance


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

Let’s write another equation:

d


M+

at (d), right side only = 0

2 t

2m 2m 1

Xb

d

-2 . 2 +Xb . 1 = 0+1 . 4

Solve, Xb = 0

1m

Sign Force Distance


2m 2mYc

2 t / m

2 t

1mXa

Ya2m 2m

1

Now, = 0 , +Xa - 0 = 0, or Xa = 0 X+

0

M+

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:

d


- Ya.4

2m 2mYc

2 t / m

Ya

d

M+

at (d), left side only = 0

Equivalent = 2 x 4 = 8

- Yc.2 + 8 .2 = 0

Or, 4 Ya + 2 Yc = 16 . . . . . (1)Sign Force Distance


2m 2mYc

2 t / m

2 t

1m

Ya2m 2m

1

Now, then, Y=0+

Solve (1) & (2), Ya = -1 , Yc = 10 Opposite direction Same direction

+Ya +Yc +1 -2x4 -2 = 0

4 Ya + 2 Yc = 16 . . . (1)

Ya + Yc = 9 . . . (2)

All Up All Down


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

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.

1 2

3 45

67 8

910

Now, we analyze each section separately, considering only one side of the structure.

Final Reactions:

Shear is Paralle to the section & Perpendicular to the beam

Axial (i.e., Normal) force is parallel to the beam

Section2

Section3


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Section Analysis:Analyze the right side or the left side, whichever has less calculation.

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Section 1Left

Section 1 B.M. = -1 . 0 = 0

0

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

3 45

67 8

910

B.M.D.

1 2

Section 2 B.M. = -1 . 2 - 4 . 1 = - 6

0

-6

Section 2Left

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

3 45

67 8

910

B.M.D.

1 2

10

0

-6

Section 3 B.M. = - 1 . 2 - 4 . 1 +10 . 0 = -6

Section 3Left

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

3 45

67 8

910

B.M.D.

1 2

10

Section 4 B.M. = -1 . 4 -4 . 3 -4 . 1 +10 . 2 = 0

Section 4Left

0

-6

0

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

3 45

67 8

910

B.M.D.

1 2

10

0

-6

0

Section 5Left

Section 5 B.M. = 0 (same as section 4)

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Section 10Right

Section 10 B.M. = +1 . 0 = 0

0

-6

0

0

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Section 9Right

Section 9 B.M. = +1 . 2 = +2

0

-6

0

0

+2

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Section 8Right

0

-6

0

0

+2

Section 8 B.M. = +1 . 2 = +2

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

B.M.D.

Sections 7 & 6Right

Sections 7 & 6 B.M. = 0

0

-6

0

0

+20

SS

+


2m 2m

2 t/m

2 t

1m

2m 2m

11 10

4 4

B.M.D.

0

-6

0

0

+2

0

-3-3

w.L2/8 = 1w.L2/8 = 1

Now, we connect the moment values


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

S.F.D.

Section 1Left

Section 1 S.F. = -1 +Shear is a

force perpendicular to the beam

-1

S


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

3 45

67 8

910

1 2

Section 2 S.F. = -1 - 4 = - 5

Section 2Left

S.F.D.

-5

-1

S

+


3 1 2

2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

45

67 8

910

10

Section 3 S.F. = - 1 - 4 +10 = +5

Section 3Left

S.F.D.

-5

-1

+5

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

3 4567 8

910

1 2

10

Section 4 S.F. = - 1 - 4 +10 - 4 = +1

Section 4Left

S.F.D.

-5

-1

+5+1

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

10

Section 5Left

Section 5 S.F. = 0

S.F.D.

3 4567 8

910

1 2

-5

-1

+5+1

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11

4 4

10

Section 6Left

Section 6 S.F. = 0

S.F.D.

3 4567 8

910

1 2

-5

-1

+5+1

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

1 23 4

56

7 89

10

Section 10Right

Section 10 S.F. = -1

S.F.D.

-5

-1

+5+1

-1

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

Section 9Right

Section 9 S.F. = -1

S.F.D.

1 23 4

567 8 9 10

-5

-1

+5+1

-1-1

S

+


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

Section 8Right

S.F.D.

1 23 4

567 8 9 10

-5

-1

+5+1

-1-1

+1

S

+Section 8 S.F. = -1 +2 = +1


2m 2m

2 t/m

2 t

1m

2m 2m 11 10

4 4

Sections 7Right

Sections 7 S.F. = -1 +2 = +1

S.F.D.

1 23 4

567 8 9 10

-5

-1

+5+1

-1-1

+1+1

+

S


S.F.D.

2m 2m

2 t/m

2 t

1m

2m 2m

11 10

4 4

-5

+5+1

-1-1

+1+1

-1

Now, we connect the shear values


S.F.D.

2m 2m

2 t/m

2 t

1m

2m 2m

11 10

4 4

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.D

Final Answer

tarek hegazy, univ. of waterloo 2 t / m 2 t 1m 2m xa ya yb xb yc step 1: stability check no. of...

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