Theory of Structures

REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE

Theory III Copyright 2013

THEORY OF

STRUCTURES 3

(INTRODUCTION)



The method of analyzing beams using the Caete procedure known as Theory III was named after

Engr. Alberto C. Caete, is a shortcut method used to evaluate single span, two span, symmetrical and

antisymmetrical beams. Its objective is to prepare the students for complicated structures that will be

encountered in the board exam. Basically, it begins by assuming each span is fixed on both sides. Then, by

passing a fraction of moment, called the balancing value, from one end to another, the internal moments

at the joints are distributed and balanced.

It is useful in this chapter to remember that for a simply supported uniformly loaded beam, the

maximum moment is equal to:

In addition, load, shear, moment, slope and deflection have a common relationship. It should be

clear that, respectively, one is the integral of another. The integral of load is the shear while the integral

of shear gives the moment. Therefore, the double integral of load is the moment. The integral of moment

divided by EI is slope and the integral of slope gives the deflection. Therefore, deflection is the double

integral of moment. To be able to do this, first write its moment equation, integrate it twice and then

divide it by EI to get its deflection. With this relationship, the moment diagram and the shear diagram are

integrals of the load which is the load inside the span. It is independent of the boundary conditions.

Therefore the moment diagram and the shear diagram are simply dependent on the loading. Since the

uniform load is a zero degree curve, every time it is integrated the degree of the curve increases by one.

Thus the shear diagram is a one degree curve or a line. The moment diagram therefore is in the second

degree curve, since it is the double integral of the load, and its height will always be 1

8wL2 whatever the

boundary condition. If the beam is fixed at both ends, the negative moment is 1

12wL2 and the positive

moment is the height 1

8wL2 minus the negative moment which gives

1

24wL2 as shown in Fig 1-2.

FIGURE 1-1



SINGLE SPAN

For single span beams with propped support, half of the FEM (Fixed End Moment) on the roller

support will be transferred to the fixed support, and the roller support shall have zero moment.

FIGURE 1-2



For a simply supported beam with concentrated load at the midspan, the moment diagram will

be as shown below. Its height will be

4.

In a fixed-fixed structure with a concentrated load at the midspan, it is known that the moment

is

8. The positive moment is the total height, which is

4, minus the negative moment

8.

To show once more that the height of the moment diagram does not change, below is a propped

cantilever. As discussed earlier, the moment at the right support is equal to

8 plus half of

8 which is

16

resulting to 3

16. By similar triangle the height

3

32 is determined. Subtracting

3

32 from

4 gives the positive

moment of 5

32.



Applying what have been discussed earlier, the formula for MA of the structure below can be

determined. 2

2

L

Pab will be the fixed-end moment at the right support and

2

2

L

bPawill be the fixed end

moment at the left support, as the students known in their Theory of Structures II. Adding half of the

fixed-end moment at the left to the fixed-end moment at the right will give the final moment M at A.

2

a b

L

Pab

L

bPa

2

1

L

Pab M

22

2

2

2

A



Hence, in summary the process for a single span propped cantilever is simply as follows:

1) Compute the fixed end moments (FEM), MAB and MBA

2) Final moments:

MA= MAB (1/2 MBA)

MB= 0

Example 1 Determination of moments in a single-span beam. A single-span beam is shown in Figure.

Determine the moment Ma at the left end A.

Solution: As shown in the figure, the beam is subjected to an increasingly distributed load Wo. To start,

the fixed end moments must be computed. Assume the shaded differential strip as a concentrated load

with a differential length dx. The magnitude of the concentrated load is equal to w(x), which is the height

of the strip, multiplied by dx giving the area of the shaded strip which also happens to be the magnitude

of the concentrated load. Then using the formulas for the fixed-end moments of a concentrically loaded

beam previously discussed, 2

2

2

2

L

bPa ,

L

Pab: substitute the values below and sum up by integrating to get

the fixed-end moments. Let a equals to x and b equals to L-x:

dx w(x) dP

xL

w w(x) O



To balance the moment at the left end, add 2

20 to the fixed-end moment at the right, resulting to a zero

moment at the roller, and pass half of the balancing value to the other end. One half of 1

20 is equal to

1

40,

thus 1

30wL2 plus

1

24 wL2 gives

7

120 wL2.

30

LW

5

x

4

2Lx

3

xL

L

W

dx x L x L

xW

L

1

dM M

L

x LdPx dM

g.integratinby up sum then values theSubstitute

:MOMENT END-FIXED

2

0

L

0

5432

3

0

L

0

2O

2

L

0AA

2

2

A

20

LW

5

x

4

Lx

L

W

dx x Lx L

1

dx x L xL

xW

L

1

dM M

2

0

L

0

54

3

0

L

0

43

3

L

0

2O

2

L

0BB

120

7wL M

20

1

2

1

30

1 wL M

20

wL

2

1

30

wL M

2

A

2

A

22

A



Example 2 Propped cantilever with a 5-degree-curve increasingly distributed load. Determine the

moments at the supports of a propped cantilever shown in Fig. 1-11.

Fig. 1-11

Solution: In this problem it is important to recall how to write the equation of the curve. The formula here

equals a constant multiplied by x raised to the degree of the curve, W = ax5. First assume x equals to L and

W equals to wo. Then, substitute the values to the equation of the curve to come up with the equation,

wo=aL5.

Get the value, =5

and substitute a to the equation of the curve to get the equation for W(x), W (x)

= Wo x5 . Then using the formulas for the fixed-end moments of a concentrically loaded beam previously

discussed, 2

2

2

2

L

bPa ,

L

Pab: substitute the values above and sum up by integrating to get the fixed-end

moments.

To get the fixed end moment at A, substitute the values above to the equation 2

2

L

Pab and integrate as

follows:

W (x) = Wo x5 W = ax5

L5

dP = W dx when x = L, W= Wo

Wo = aL5

a = Wo

L5



To balance the moment at the left end, add

2

72 to the fixed-end moment at the right, resulting

to a zero moment at the roller, and pass half of the balancing value to the other end. One half of 1

72 is

equal to 1

144 , thus

1

252wL2 plus

1

144 wL2 gives

11

1008 wL2.

MB= 0

1008

L11W

72

LW

2

1

252

LW M

MOMENTS FINAL

72

LW M

9

x

8

L

L

W

9

x

8

Lx

L

W

dx xLx L

W

dx L

x LxL

xW

dM M

2

O

2

O

2

OA

2

OB

99

7

O

L

0

98

7

O

L

0

87

7

O

L

0

L

0 2

2

5

5

O

BB

1008

L11W

72

LW

2

1

252

LW M

MOMENTS FINAL

72

LW M

9

x

8

L

L

W

9

x

8

Lx

L

W

dx xLx L

W

dx L

x LxL

xW

dM M

2

O

2

O

2

OA

2

OB

99

7

O

L

0

98

7

O

L

0

87

7

O

L

0

L

0 2

2

5

5

O

BB

252

LW M

9

x

8

2L

7

L

L

W

9

x

8

2Lx

7

Lx

L

W

dx x2Lx L x L

W

dx L

x LxL

xW

dM M

L

x LdPx dM

:MOMENTS END-FIXED

L

W a aL W

W WL, x when dx W dP

ax WL

xW W

2

OA

999

7

O

L

0

9827

7

O

L

0

226

7

O

L

0

L

0 2

2

5

5

O

AA

2

2

A

5

O5

O

O

5

5

5

OX



MULTIPLE SPAN

TWO-SPAN WITH BOTH EXTERIOR SUPPORTS FIXED

The Caete Cross Formula is used to get the final moment easier, it is applicable for 2-span beams

or rigid frames. Considering multiple span beams and frames, Caete Cross Formula can be used if it is

converted to a 2-span beam. It is a shortcut using moment distribution method where the number of

cycles is only one.

With two-span beam, it is easier because moment distribution is use. First example is for beam

with both fixed ends.

A B C

Steps:

1. Solve for the distribution factor (DF), where:

DF= Ki

K and K=

I

L

Note: for fixed end, K=1/L and for span with roller of hinge support, K=.75/L (Sum of DF should

be equal to 1)

0.571429 DF

DF -1 0.428571

37

1

61

81

81

DF

BC

BCBA

2. Solve for the Fixed End Moment of each span

m-kN 90 63012

1 M FEM

m-kN 160 83012

1 FEM FEM

2

CBBC

2

BAAB

FE



3. Solve the unbalanced moment at joint B and then distribute to BA and BC. After getting the share of BA, half of that will be carried over to AB, since A is a fixed support it will absorb the moment and never gives back same thing with BC. Therefore the moment distribution is finished and will be the basis for computing the final moment.

]DF )FEM - (FEM +[FEM - = Mc

m-kN 70 0.571429702

1 90 M M

m-kN 175 0.428571702

1 160 M M

m-kN 130 BC

M 0.428571 70 160 BA

M

m-kN 130 0.571429 70 90 BC

MM

BCBABC CB

CBC

ABA

B{

In this example, final moment at BA is equal to moment at BA plus the unbalanced moment

multiplied to the distribution factor of BA with respect. Unbalanced moment is equal to the moment CB

subtracted from moment BA. Same with the final moment at BC, it is equal to the moment BC plus the

unbalanced moment multiplied to the distribution factor with respect to BC. As observed, moment BA is

equal to moment BC as requirement to equilibrium, they have to be equal and opposite in directions.

Considering the carry over moments, therefore moment at A and C will be derived. With moment

A it is equal to the moment AB plus half of shared moment at BC which is unbalanced moment multiplied

to distribution factor of BA likewise with moment at C.

The general formula for the above procedure in determining the final moment at joints A, B and

C can easily be derived and it is as follows.

BCBABCCBC

BABCBAABA

BABCBCBAB

BCBCBCBABC

BCBCBABCBC

DF FEM FEM2

1 FEM M

DF FEM FEM2

1 FEM M

DF FEM DF FEM M

FEM DF 1 DF*FEM M

DF FEM FEM FEM M

MBC = MBA = MB

CAETE CROSS FORMULA



Factoring out the coefficients of the fixed end moments it will get BCBCBCBA FEM DF 1 DF*FEM .

FEMAB FEMBC

DFBA DFBC

BABCBCBAB DF FEM DF FEM M

Moments at fixed supports, moment A and moment C, from the calculations moment A is equal to

BABCBAAB DF FEM FEM2

1 FEM and moment C is equal to

BCBABCCB DF FEM FEM2

1 FEM .

Please note that the above formulas will give the negative moments at the supports A, B and C. If

the computed values using the above formulas are negative, this means that the direction of the moments

are opposite, i.e. the moments are positive.



TWO-SPAN WITH ONE EXTERIOR FIXED AND THE OTHER EXTERIOR SUPPORT SIMPLE.

Another example is for beam with external fixed at one support and external simple support, it is

similar to the first example both fixed end supports but with the case of fixed at one end and simple

support at the other end, the span with external simple support the stiffness is modified because the

members becomes more flexible representing zero restrain in the moment.

From the theory of structures, 0.75 or is for the modified stiffness of spans where the exterior

support is a simple support (i.e. M=0). Once modified the stiffness next to be computed is the distribution

factors, then fixed end moments.

0.48387 DF

0.51613

4.75

51

51

DF

BC

BA

Note, also BCDF = 1 - BADF = 1 0.51613

33.333 42512

1 M

m-kN 62.5 53012

1 M

:MOMENTS END-FIXED

2

BC

2

AB

Take note, the final moment at the simple support is zero, and then it is needed to be balanced

with the same magnitude but opposite directions and carry over half of the moment to the middle

support. The factor of +1/2 that transmits moment A to moment B is known as the carry-over factor (COF).

Since the FEMs of span BC are equal (FEMBC=FEMCB), then this is simplified by simply multiplying the FEMBC

with a factor of 1.5



m-kN 50 1.5 33.333 42512

1 M

2

BC

Applying Caete Cross Formula the final moment at B is equal to the moment AB multiply to the

distribution factor BC plus the moment at BC multiplied by the distribution factor BA. Same with the final

moment at A, it is equal to moment AB plus half unbalanced moment which is 62.5 minus 50 then the

difference will be multiplied to the distribution factor BA. Also presented below is the formula for the

moment at the end support at Joint A. The moment at Joint C is zero because it is an exterior simple

support.

0 M

m-kN 65.7258 516123.050 5.622

1 62.5 M

m-kN 56.0484 0.51613 50 0.4838762.5 M

C

A

B

TWO-SPAN WITH BOTH EXTERIOR SUPPORTS SIMPLE

For beams with both external simple supports, there can be no moment at a pinned end there is

no carry over to the pinned ends if the stiffness are modified. Also, since both spans will have the

reduction factor , this factor need not be included in the calculation for the distribution factors since this

factor of will be present in both the numerator and denominator and therefore will simply cancel out

.



First, get the distribution factors of span BA and span BC.

0.55556 DF

0.44444

41

51

51

DF

BC

BA

Also, DFBC = 1-DFBA

Then get the fixed end moment of BA and BC. Note that both exterior supports are simple (i.e. M=0).

Therefore, multiply the FEMBA and FEMBC with factor 1.5 since the FEMs at the end of the spans are equal.

m-kN 67.5 1.5 kNm 45

8

4 90 M

m-kN 65.625 1.5 kNm 43.75 8

5 70 M

BC

BA

For the final moment, since the span stiffnesses are modified, there is no carry over moment. Final

moment of A and C is then equal to zero. Using Caete cross formula, final moment at B is equal to the

moment AB multiplied to the distribution factor BC plus moment at BC multiplied by distribution factor

BA. The exterior supports are both simple. Therefore, the moments there will be zero.

0 M

m-kN 66.458 0.44444 67.5 0.55556 65.625 M

0 M

C

B

A



THREE-SPAN BEAM

For beams with three-span, there is no shortcut or the Caete cross formula cannot be used.

Moment Distribution Method (MDM) is applicable but it will need several cycles of balancing and carry

over moments before the final moments can be determined.

Example:

Steps:

1. Determine distribution factors for the beams at joints A, B, C and D. These factors will be used in steps 3, 5 and 7 whenever we distribute moments at a joint.

2. Fix all joints, to create a structure of fixed end moment beams. A false structure we can solve is the result. FEMs (AB and CD) equal 5(8)8 = 5 k-ft. FEM (BC) equals10 (8)8 = 10 k-ft. Choose clockwise moments as positive.

3. The 2-span beam with both exterior supports fixed is not the real situation, so we release each joint one at a time and put in a moment to cancel the sum of the fictitious moments at a joint and when we do so, we distribute the cancellation moments in accordance with the distribution factors. While one joint is released the others are fixed still. However, in our table, every time we do distribution, we do the distribution process simultaneously for all joints in the current row of the table above (this occurs at some point in the process in rows 3, 5 and 7 of the table).



4. While one joint is fixed and the other released, 1/2 of the cancellation moment goes to the opposite side of the beam span in accordance with a 1/2 CO factor.

5. Repeat step 3, but now the sum of carry over moments at each joint are now the false fixed moments which we cancel out.

6. Repeat steps 4 and 3 repeatedly until the errors are already insignificant.

7. Finally, sum all moments and see if internal equilibrium has been achieved at each joint.

For comparison, moments determined by an exact computer analysis are given. The percent error

between the moments determined by the moment distribution process and the computer analysis

moments is quite small and well within engineering accuracy necessary for structural design purposes.

SYMMETRICAL LOADING

4-span beams that are geometrically symmetrical consider half of the span then use Caete Cross

Formula. For the other half of the span, just change the sign convention of the moment.

As observe, the above beam is geometrically symmetrical same with the loadings. With this the

beam could be converted to 2-span beam taking note that the span will have zero deflection and zero

slope at the center support. Therefore, this can be replaced by a fixed support.

Therefore, we can now use the procedure discussed in 2-span with both exterior supports fixed.



ANTI-SYMMETRICAL LOADING

Figure A

With anti-symmetrical loading, the loading is converted symmetrically. Therefore, the 4-span

beam can be converted to 2-span with both support fixed because the middle support has zero slope and

zero deflection.

Also, figure A can be converted symmetrically with opposite direction of loadings. The converted 4-span

will have an exterior support fixed and exterior support simple. The middle support becomes simple

support because of the zero deflection and zero curvature at the middle support, this is called the point

of centraflexure.



FRAMES

Now, lets consider a multiple span frame. As observed, the frame is geometrically symmetrical

therefore half of the structure is considered for analysis.

With the frame modified, the end supports is considered fixed. Since it has been converted to 2-

span with both fixed end supports, use Caete Cross Formula.

First, the stiffness factor of each span is computed on the basis of 4EI/L or by using the relative-

stiffness factor I/L. Then, get the distribution factors of span BC and span BA. To easily get the distribution

factor of span BA, subtract the distribution factor of span BC from 1. Then vice versa if ever distribution

factor BA is the first to be computed.

0.64286 DF

0.35714

4

11.2

1

41.2

61

61

DF

BA

BC



Second, compute for the fixed end moments of the supports.

m-kN 37.5

8

650 M

0 M

BC

BA

Then, get the final moment at joint A, B and C using Caete Cross Formula.

Final moment at A is equal to the fixed end moment at BA plus half of shared moment at BC which

is unbalanced moment multiplied to distribution factor of BA likewise with moment at C. With the

moment at B, it is equals to the moment BA multiplied to the distribution factor BC plus the quantity of

moment BC multiplied to the distribution factor BA.



Solve for the Moments.

2.5m

Solution:

The problem is an anti-symmetrical loading. To solve the problem, first isolate the loading at the bottom.

Convert it into 2-span with the other support to be fixed. Then, use the procedure for 2-span with exterior

support fixed and the other exterior support simple.

EQUAL TO THIS

WDL=10+0.3(0.6) (24) = 14.32 kN-m

7m 6m 6m 7m



FEMS

Since the FEMs of span BA are equal, then this is simplified by multiplying the FEMBA with a factor of 1.5

MBA=1

12(14.32)(72) 1.5 = 87.710 m

MBC= MCB=1

12(14.32)(62) = 42.96

=

16

16 +

0.757

= 0.60870

DFBA = 0.39130

FINAL Ms

MB=87.71*0.6087+42.96*0.3913 = 70.199 kN-m

MC=42.96+.5(.6087) (42.96-87.71) = 29.340 kN-m



+

To solve for the final moments of an anti-symmetrical beam of more than 2-span, we must first

translate the given figure into a symmetrical one. To do this, we must first convert the given loading into

equivalent loading such that these loadings are equal in magnitude but opposite direction.

When converted to symmetrical, then the succeeding procedure will be the same for 2-span. As

reflected in the figures, we can solve for the moments by the principle of superposition. We compute for

the fixed end moment by each loading and solve for it arithmetically depending on the direction of the

loading.

Figure B Figure C

WL

0.5 WL

0.5 WL

0.5 WL

0.5 WL

0.5 WL

0.5 WL 0.5 WL +



SYMMETRICAL CASE

A B

a b

=

^2 ( ) = 2.247

0

=

^2 (^2)( ) = 3.5386

Now that we have solved for the fixed end moment for the given loading in figure D, we can now

determine the total fixed end moment for the figure B by adding the computed values with the fixed end

moments for uniformly distributed loading which is wl2/12.

FEMS for span A-B

Considering Figure B

MAB=1

12(0.5)(72) + 2.2471(0.5) = 3.1652

MBA=1

12(0.5)(72) + 3.5386(0.5) = 3.8110

Notice that, the moments were reduced to half because the supports are simple.

Considering span B-C

MBC=MCB=1

1262 = 3

MBA=3.811 +3.1652

2= 5.3936

FINAL Ms

MB=3w (0.3913) +5.393w (0.6087) =4.4566w

MC=3w+0.6087

2(3 5.3932) = 2.2717

L



UNSYMMETRICAL CASE

A a B

FEMs

MA=

72(

3

3

4

4) =

72(

72.53

3

2.54

4) = 0.54475

MB=

72(

74

12

7(4.5)3

3+

4.54

4) = 1.8362

DFBA=1

71

7+

1

6

= 0.46154 DFBC=0.53846

FINAL MOMENTS:

MB=(0.54475(0.5) 1

2+ 1.8362(0.50) 053846) = 0.56770

To get the final negative moments, we could just add the values computed from the symmetrical and

unsymmetrical cases based from the principle of superposition.

TOTAL NEGATIVE MOMENTS

MB= (4.4566+0.5677) w=5.0243WL

MC= (2.2717+0) w=2.2717WL< MB



QUIZ 1 (Theory 3)

1. DETERMINE THE MAXIMUM POSITIVE MOMENT AND THE NEGATIVE MOMENT AT THE

FIXED SUPPORT.

58

7= 1.0879 6(

58

7) = 6.5278

Cantilever M= 8.4482 x (1.0879)2/2

MB= 5 kNm

FEMs= 1

12 (8.4482)(6.5278)2 = 30.000

Mc = 30 +25

2= 42.5

Point of zero shear:

=21.829

8.4482= 2.5839

=1

2(21.829)(2.5839) 5 = .



2. DETERMINE THE MOMENT AT POINT C. Columns steel pipe, outer = 300mm, thickness=10mm, G=200 GPa

Rafters aluminum, I section, Flanges, 250 x 20mm, web 450 x 10 mm, G=83 GPa



=

64(3004 2804) = 95.8891064

=1

12(2504903 2404503) = 628.521064

=. 75 (

200 95.8896 )

. 75 (20095.889

6 ) +83628.52

45

= 0.29130

= 0.7087

=1

12(10)(45)

2= 66.6667

= 66.667 +0.7087

2(66.667 0) = .

Theory of Structures

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