Strength of Materials- Statically Indeterminate Beam- Hani Aziz Ameen

Strength of Materials Handout No.14

Statically Indeterminate Beam

Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]

www.mediafire.com/haniazizameen

mailto:E-mail:[email protected]

http://www.mediafire.com/haniazizameen

Strength of materials- Handout No.14- Statically Indeterminate Beam- Dr. Hani Aziz Ameen

14.1 Introduction When the unknowns quantities of a loaded beam can not be found by applying the three equations of statics ( 0M,0F,0F yx ) i.e. the equilibrium equations are not enough to solve these types of problems ,the beam is said to be statically indeterminate ,as shown in Fig.(14-1)

Fig(14-1) In order to determine the unknown quantities , we require in addition to the three equations of statics ,one more equation established from the deflection of the beam. 14.2 The Three Moment Equation The three moment equation is considered as a general method of finding the redundant moments at the intermediate supports of the beam

Fig(14-2) From Fig(14-2) ,

t1/2= 1121

111111

L32LM

21

3LLM

21aA

IE1

t3/2= 2222232222

L32LM

21L

31LM

21bA

IE1


For similar triangle (shaded area Fig(14-2 b)

1

2/11

2

32/3L

thL

ht

Put the value of t1/2 and t3/2 and E1I1=E2I2 also h3=h1=0(supports),gives

M1L1+2M2(L1+L2)+M3L3=2

22

1

11L

bA6L

a.A6

This is called three moment equation (3M equation) (or Clapeyroequation)with unknown M1,M2,M3. This will be added to the static equilibrium equation in order to solve the static indeterminate problems ( SIP ). 14-3 Application of 3M Equation to Find the Reaction The three moment equation applied on 2-span (span is the distance between two supports)

a- Find the value of left span L

aA6 for the left span

b- Find the value of right span L

bA6 for the right span

c- Apply the three moment equation to find the moment d- After determination of the moments the reaction can be found

such that


Separate the beam at point 2 as shown in Fig(14-3a),then find R1 from equilibrium equation - a - Separate the beam at point 3 as shown in Fig(14- 3b),then find R2 and so on

- b

Fig(14-3)


14.4 Solving Fixed Fixed Beams In case of one fixed end or two fixed ends beams , we assume a span of zero length from the fixed end as shown

14.5 Determination of L

aA6 & L

bA6 For the Span

a- Concentrated load

ab31.

LPab.b

21a

32*

LPaba

21

L6

LaA6

22 aLLPa

LaA6

& L

bA6 22 bLLPb


b- Uniformly Distributed Load

4wL

2L*

8wL*

2L*

32*2

L6

LaA6 32

4wL

LbA6 3

c- Uniformly varying loads

b51wLL

41L

31

6wLL

21

L6

LbA6 22

3wL607

LbA6

3wL608

LaA6&

or if the load as in Fig(14-4)

3wL607

LaA6

3wL608

LbA6

Fig(14-4)


d) For the load as in the Fig(14-5)

Fig(14-5)

4a

3a

2wa

4aL

3a

2wa

2L.L

2wa

L6

LaA6 222

LbA6a2L3

2wa

LaA6 2

e ) For the load as in the Fig(14-6)

Fig(14-6)

22 La3LMa

2b*b*ML

32*

2L*M

L6

LaA6

)Lb3(LM

2b*b*ML

31*L*M

21

L6

LbA6 22


14.6 Examples The following examples explain the different ideas for the statically indeterminate problems . Example(14-1) Fig.(14-7) shows a continuous beam subjected to the load indicated in the figure. Find the reaction (R1) of the beam .

Fig(14-7) Solution Apply 3-M on bar 123 12 left span 23 right span left span

33

wL608

4wL

LaA6

= 1952)2(480608

42*720 3

3 N.m2

right span

)34(4

3*600L

bA6 22

+ 22 14(4

1*800 )

=6150 N.m2

M1L1+2(L1+L2)M2+L2M3= LbA6

LaA6

M1= 1*2

3*720 1080 N.m

M3= 700 N.m 1080*2+2(2+4)*M2 4*700 = 1952 6150

M2 = 261.833 N.m


0Mc 261.8+R1*2 (1200*5/2)* (1/3*5) = 0 R1= 2.37 kN Example(14-2) Fig.(14-8 ) shows a continuous beam subjected to the load indicated in the figure. Find the reactions ( R1 & R2)of the beam .

Fig(14-8) Solution Apply 3-M on beam 123 12 left span 23 right span left span

)42*3(4

8000L

aA6 22

= 8000 N.m2 right span

320004

4*20004

wLL

bA6 33 N.m2

M1=0

M1L1+2(L1+L2)M2+L2M3= L

bA6L

aA6

0+2(4+4)M2+4M3= 8000 32000 16M2+4M3= 40000


Apply 3-M on beam 234 23 left span 34 right span left span

4wL

LaA6 3

=32000 N.m2

right span

LPb

LbA6 (L2-b2)

=3

2*6000 (32 22 ) = 20000 N.m2

M4 = 0

M2L2+2(L2+L3)M3+L3M4= L

bA6L

aA6

4M2+2(4+3)M3= 32000 20000 4M2+14M3= 52000 *4 16M2+4M3= 40000 *14 16M2+56M3= 208000

224M2 56M3= 560000

208 M2=352000 M2= 1692.307 N.m M3= 3230.7692 N.m To find the reactions

Mc=0 4R1+8000+1690=0 R1= 2.422 kN

Mc=0 622.5*8+8000+4R2 2000*4*2+

3230 = 0 R2 =2.437 kN


Example(14-3) Fig.(14-9 ) shows a continuous beam subjected to the load indicated in the figure. Find the moment over the supports and the reactions at C and D for the continuous beam .

Fig(14-9) Solution

M1L1+2(L1+L2)M2+L2 M3 = L

bA6L

aA6

Take ABC AB left span BC right span

6 MA+2(6+10)MB+10Mc= )bL(LPb

4wL 22

3

6 MA+2(6+10)MB+10MC = 105.7*16

410*8 3

(102 7.52)

10

5*24 (102 52)

MA= 20*2.5= 50 N.m 32 MB +10Mc= 1557 ...... ( i ) Take BCD BC left span CD right span

10MB+2(10+5)Mc+5MD = 051010

5*245.21010

5.2*16 2222

10 MB+30 Mc = 1275 ............ (ii) Solving Eq(i) & Eq(ii) give MB= 39.5 N.m Mc = 29.3 N.m


RD*5 = 29.3 RD = 29.3/ 5 = 5.86 N

0MB

RD*15 Rc*10 + 24*5+ 16*2.5 39.5 = 0 Rc = 20.84 N Example (14- 4 ) Fig.(14-10) shows a continuous beam subjected to the load indicated in the figure. Find the moment over the supports using three moment equation

Fig(14-10)


Apply the three-moment equation for ABC AB: left span BC : right span

3MA+2(3+6)MB+6MC= )26(6

2*454

3*40 223

MA=0 8MB+6MC= 750 ... (i) Take BCD BC: left span CD: right span (Zero length)

6MB + 2(6+0) MC + (0) MD = 0466

4*45 22

6MB+12MC= 600 ... (ii) Solving eq(i) & eq(ii) MB= 30 kN.m MC= 35kN.m Example(14-5 ) Fig.(14-11) shows a fixed-fixed beam subjected to the load indicated in the figure. Find the reactions .

Fig(14-11) Take A1AB A1A =left span AB =right span

(0) MA1+2(0+L.)MA+LMB=0+4wL3

2LMA+LMB = 4

wL3

2LMA + MB = 4

wL2 (i)


Take ABB1 AB : left span BB1 : right span

LMA+2(L+0)MB+(0)MB1= 4wL3

MA+ 2 MB=4wL2

..... (ii)

MA=12

wL2

MB=12

wL2

RA=RB=2

wL

Example(14-6 ) Fig.(14-12) shows a fixed beam subjected to the load indicated in the figure. Find the reactions

Fig(14-12) Solution Apply 3-M equation on beam 123

12 left span 23 right span


left span

LaA6

= 2*41*

32*24004*

31*4*2400

46

= 8400 N.m2

8400]0)416*2(*4[4*4

1200)aL2(a)bL2(b(L4

W 222222

right span

0L

bA6

M1= 600 N.m

M1L1+2(L2+L1)M2+L3.M3=

m.N750MM08400M)0(M)04(24*600

LbA6

LaA6

2

232

4R=1200*3(3.5) 750 kN96.2R Example(14-7 ) Fig.(14-13) shows a fixed beam subjected to the load indicated in the figure. Find the reactions

Fig(14-13)


Left Span

33

wL608

4wL

LaA6

= 33

)3)(12(608

43*88 = 9.72 kN.m2

Right Span

0L

bA6

M1=(0.8*2)/2)*(1/3)*2= 0.5333 kN.m

M1L1+2(L2+L1)M2+L3.M3= L

bA6L

aA6

0.533*3+(3+0)M2+ (0) M3 = 9.72 6M2 =19.902 M2 = 1.887 kN.m 3R= [(2*5)/2] (1/3)5 + 1.888 R= 2.14 kN


Example(14-8) Fig.(14-14) shows a fixed beam subjected to the load indicated in the figure. Find the reactions

Fig(14-14) Solution Apply 3-M on beam 123 12 left span 23 right span left span

23

22 m.N420004

6*600)26(6

2*900L

aA6

right span

)2)4(224)4(2(44*4

600

))cL2(a)dL2(d(L4

wL

bA6

22222

222222

=5400 N.m2 M1=0 M1L1+2(L1+L2)M2+L2M3

540042000M4M)46(2L

bA6L

aA6

32

20M2+4M3 = 47400 .................. (i) Apply 3-M on bean 234


2222

222222

m.N42000)2)4(2(24*4

600

aL2(a)bL2(bL4

wL

aA6

L2M2+2(L2+L3)M3+L3M4 = 4200 4M2+8M3= 4200 (ii) solving Eq(i) and Eq(ii) , we get 20M2+4M3= 47400 *2 4M2+8M3= 4200 *1 36 M2 = 90600 M2 = 2.5166 kN.m M3 = 733.33 N.m Example(14-9 ) Fig.(14-15) shows a fixed beam subjected to the load indicated in the figure. Find the reactions.

Fig(14-15) Solution Apply 3-M equation on beam 123 12 left Span 23 right span left span

1)533(

43*34*

32*

24*3

46

LaA6

= 12.525 N.m2 2(4) M2 + 0 = 12.525 M2= 1.5656 kN.m 4R 300 +1565 = 0 R = 358.75 N


Example(14-10) Fig.(14-16) shows a fixed beam subjected to the load indicated in the figure. Plot the shear force diagram .

Fig(14-16) Solution M2 = 251606 N.m M3 =+ 733.33 N.m

0M2 6R1 900*4 600*6*3 + 2516 = 0 R1=1980.66 N

0M3 1980.66*10+4R2 900*8 600*8*4 = 0 R2=1648.35 N

2x0 1 Fx1=R1 600x1 at x1= 0 Fx1=1980.66 N at x1=2 Fx1=180.66 N

4x0 2 Fx2=R1 900 2*600 600x2 At x2=0 Fx2= 119.4 N At x2=4 Fx2= 2519.4 N

23x0 Fx3 = R1 1200 900 2400 + R2 600x3 At x3=0 Fx3= 871.05 N At x3=2 Fx3= 2071N

2x0 4 Fx4 = R1 1200 900 2400+R2 1200 = 2071 N


Shear Force Diagram Example(14-11) Fig.(14-17) shows a fixed beam subjected to the load indicated in the figure. Plot the shear force diagram .

Fig(14-17) Solution Apply 3-M on beam 012 01 left span

12 right span left span

0L

aA6

right span 233 m.N630032000

607wL

607

LbA6

L0M0+2(L0+L1)M1+L1M2= 6300 (i) 6M1+3M2= 6300 Apply 3M on beam 123



Left Span

2

33

m.N 7200

)3)(2000(608wL

608

LaA6

Right Span

33

wL607

4wL

LbA6

233

Nm 19800)3)(2000(607

4)3(*2000

M1L1+2(L1+L2)M2+L2M3 = 7200 19800 3M1+12M2+3M3 = 27000 .............. (i) Apply 3M on beam 234 23 : left span 34 : right Span Left Span

20700wL608

4wL

LaA6 3

3N.m2

L2M2+ 2(L2+L3)M3+L3M4 = 20700 3M2 + 6M3 = 20700 ........ (ii) 3M1 + 12 M2 + 3 M3 = 27000 ....... (iii) 6M1 + 3M2 = 6300 ..... (iv) from Eq.(ii) M3 = 0.5 M2 3450 sub. into Eq.(iii) yield 3M1 + 12 M2 3 (0.5M2 3450) = 27000 3M1+10.5 M2 = 16650 * 2 ................. (v) 6M1 + 3M2 = 6300 * 1 .......................(vi) 6M1 + 21M2 = 33300

M1 3M2 = 6300 18M2 = 27000 M2= 1500 N.m


M3 = 2700 N.m M1 = 300 N.m Example(14-12) Fig.(14-18) shows a fixed beam subjected to the load indicated in the figure. Find the reactions .

Fig(14-18) Solution Apply 3M on beam 123 12 left span 23 right span left span

233

m.kN6444)4(

4wL

LaA6

Right Span

2

2222

m.kN18

)14(4

1*2)34(4

3*2L

bA6

M1=4*1*0.5 = kN.m M1L1+2(L1+L2)M2+ L2M3 = 82

8+2(4+4)M2+4M3 = 82 16 M2 + 4M3 = 74 ........... (i) Apply 3M on beam 234

23 left Span 34 right span


left span

22222 m.kN18)34(4

3*2)14(4

1*2L

aA6

right span

23 m.kN9.18wL607

LbA6

M2L2+2(L2+L3)M3+L3M4 = 36.9 4M2+14M3+3M4 = 36.9 ...............(ii) Apply 3- M on beam 345


left Span

23 m.kN6.21wL608

LaA6

L3M3 +2(L3+L4) M4 +L4M5 = 21.6 3M3 +6M4 = 21.6 ..........(iii) 16M2+4M3 = 74 ........(iv) 4M2+14M3+3M4 = 36.9 .......(v) M2 = 0.25 M3 4.625 14M2 + 4(0.25 M3 4.625)+3M4 = 36.9 13 M3 + 3M4 = 18.4 *2 3M3 + 6 M4 = 21.6 * 1 26M3 + 6M4 = 36.8

3M3 6M4= 21.6 23M3 = 15.2 M3 = 660.809 N.m M2 = 4960 N.m M4 = 3273.33 N.m

Strength of Materials- Statically Indeterminate Beam- Hani Aziz Ameen

Documents