2 t / m

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: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


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


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


at (d) right side only = 0

Let’s write another equation:Let’s write another equation:

d


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


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+


at (d) left side only = 0


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


- 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


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


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


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.

+


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

+


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

+


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

+


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

+


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

+


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

+


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

+


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

+


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

+


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


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


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

+


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

+


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

+


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

+


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

+


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

+


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

+


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


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


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


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

Date post:	31-Jan-2016
Category:	Documents
Upload:	fell
View:	28 times
Download:	0 times

2 t / m

Documents