THEORY OF
STRUCTURES 3
(INTRODUCTION)
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
The method of analyzing beams using the Cañete procedure known as Theory III was named after
Engr. Alberto C. Cañete, 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:
FIGURE 1-1
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
boundary condition. If the beam is fixed at both ends, the negative moment is
1
1
1
wL2
8
1
wL2
12
whatever the
and the positive
moment is the height 8wL2 minus the negative moment which gives 24wL2 as shown in Fig 1-2.
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
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FIGURE 1-2
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.
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
For a simply supported beam with concentrated load at the midspan, the moment diagram will
be as shown below. Its height will be
𝑃𝐿
.
4
In a fixed-fixed structure with a concentrated load at the midspan, it is known that the moment
is
𝑃𝐿
.
8
The positive moment is the total height, which is
𝑃𝐿
, minus
4
the negative moment
𝑃𝐿
.
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
resulting to
moment of
3𝑃𝐿
3𝑃𝐿
. By similar triangle the height 32 is determined. Subtracting
16
5𝑃𝐿
.
32
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𝑃𝐿
𝑃𝐿
𝑃𝐿
plus half of 8 which is 16
8
3𝑃𝐿
𝑃𝐿
from 4 gives the positive
32
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
Applying what have been discussed earlier, the formula for MA of the structure below can be
Pab 2
Pa 2 b
determined.
will be the fixed-end moment at the right support and
will be the fixed –end
L2
L2
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.
MA 
Pab 2
1 Pa 2 b
Pab 
a


b 
2
2
2 
2 L
2
L
L 
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
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,
Pab 2 Pa 2 b
, 2 : substitute the values below and sum up by integrating to get
L2
L
the fixed-end moments. Let a equals to x and b equals to L-x:
wO
x
L
dP  w(x) dx
w(x) 
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Engr. ALBERTO C. CAÑETE
Theory III
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FIXED - END M OM ENT:
Substitute the values then sum up by integratin g.
dM A 
dPx L  x 
L2
2
L
M B   dM B
0
1
L2
1
 3
L

L
M A   dM A
0

1
L2

L
0
WO x
2
x L  x  dx
L

L
0
WO x 2
x L  x  dx
L
 x
L
0
3
L
L
W  L2 x 3 2Lx 4 x 5 
 30 

 
4
5 0
L  3
W x 4L x5 
 30 
 
5 0
L  4
W L2
 0
30
W0 L2

20
To balance the moment at the left end, add
𝑤𝐿2
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

L  x 4 dx
1
7
1
20
is equal to
1
,
40
thus 30wL2 plus 24 wL2 gives 120 wL2.
MA 
wL2 1  wL2
 
30
2  20



 1 1  1 
M A  wL2    
 30 2  20 
7wL2
MA 
120
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Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
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.
W (x) = Wo x5
W = ax5
L5
dP = W dx
when x = L, W= Wo
Wo = aL5
a = Wo
and substitute “a” to the equation of the curve to get the equation for W(x), W (x)
L5
= Wo x5 . Then using the formulas for the fixed-end
moments of a concentrically loaded beam previously
Get the value, 𝑎 =
discussed,
𝑤𝑜
𝐿5
Pab 2 Pa 2 b
, 2 : substitute the values above and sum up by integrating to get the fixed-end
L2
L
moments.
To get the fixed end moment at A, substitute the values above to the equation
Pab 2
and integrate as
L2
follows:
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
WX  
W  ax 5
O
L5
dP  W dx
when x  L, W  WO
WO  aL5  a 
WO
L5
FIXED - END M OM ENTS:
dPx L  x 
L2
2
dM A 
L
L
0
0
M A   dM A  

WO
L7
 x  L
L
6
2
0
L
L
0
0
M B   dM B  
WO x 5
x L  x 2
5
L
dx
L2


 2Lx  x 2 dx
WO
L7
W
 7O
L
L
WO x 5
x 2 L  x 
5
L
dx
L2
 x L  x  dx
L
7
8
0
L
 x 8L x 9 
 

8
9 0

W  x 7 L2 2Lx 8 x 9 
 7O 

  WO x 5
2
WO  L9 x 9 
L  7
8
9 L 0 5 x  L  x 
L

L
M
  9 dM9B  
dx L7  8  9 
2
9 B
0
0


L
W  L 2L x 
 7O  
 L 
2
W L
L  7  8WO 9 x7 L  x 8 dx
MB  O
7 0
L
72
W L2
L
MA  O
W  x 8L x 9 
252
 7O 
 
FINAL M OM ENTS
L  8
9 0
𝑤𝑜 𝐿2
2
To balance the moment
toOthe
atOthe
 11W
L2 fixed-end
L2 right, resulting
1  WO Lmoment
9
9at the left end, add 72 W
WO  L x 


M



1

 at 7the
   pass half of the Abalancing
252 value
2  to72
1008
to a zero moment
the other
end. One half of
is

L  8roller,9 and
72

1
1
1
11
equal to 144 , thus 252wL2 2 plus 144 wL2 gives 1008 wL2.
W L
MB  O
72


FINAL M OM ENTS
MA 
WO L2 1  WO L2  11WO L2

 
252
2  72 
1008
MB= 0
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
MULTIPLE SPAN

TWO-SPAN WITH BOTH EXTERIOR SUPPORTS FIXED
The Cañete 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, Cañete 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:
Ki
DF= 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)
DFBA 
1
8  1  0.428571  1 - DF 
BC
1 1
7
8
6
3
DFBC  0.571429
2. Solve for the Fixed End Moment of each span
1
3082  160 kN - m
12
1
2
 306  90 kN - m
12
FEM AB  FEM BA 
FEM BC  FEM CB
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Engr. ALBERTO C. CAÑETE
Theory III
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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.
M
{
MB
M
BA
 160  70  0.428571  M
BC
 90  70  0.571429   130 kN - m
BC
 130 kN-m
1
700.428571  175 kN - m
2
1
M C  M CB   90  700.571429  70 kN - m
2
M c = - [FEMCB + ½ (FEMBC - FEMBA) DFBC]
M A  M AB  160 
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.
M BC  FEM BC  FEM BA  FEM BC  DFBC
M BC  FEM BA * DFBC  1  DFBC  FEM BC
M B  FEM BA  DFBC  FEM BC  DFBA
CAÑETE CROSS FORMULA
1
FEM BA  FEM BC  DFBA
2
1
 FEM BC  FEM BA  DFBC
2
M A  FEM AB 
M C  FEM CB
MBC = MBA = MB
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Theory III
Copyright © 2013
Factoring out the coefficients of the fixed end moments it will get FEM BA * DFBC  1  DFBC  FEM BC .
FEMAB
FEMBC
DFBA
DFBC
MB  FEMBA  DFBC  FEMBC  DFBA
Moments at fixed supports, moment A and moment C, from the calculations moment A is equal to
1
FEMBA  FEMBC  DFBA
2
1
FEMCB  FEM BC  FEM BA  DFBC .
2
FEM AB 
and
moment
C
is
equal
to
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”.
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Engr. ALBERTO C. CAÑETE
Theory III
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
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 ¾ 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.
DFBA 
1
5
 0.51613
1  .75
5
4
DFBC  0.48387
Note, also DFBC = 1 - DFBA = 1 – 0.51613
FIXED - END M OM ENTS:
1
3052  62.5 kN - m
12
1
2
 254  33.333
12
M AB 
M BC
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 FEM’s of span BC are equal (FEMBC=FEMCB), then this is simplified by simply multiplying the FEMBC
with a factor of 1.5
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Engr. ALBERTO C. CAÑETE
Theory III
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M BC 
1
2542  33.333  1.5  50 kN - m
12
Applying Cañete 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.
M B  62.50.48387   50 0.51613  56.0484 kN - m
M A  62.5 
1
62.5  500.516123  65.7258 kN - m
2
MC  0

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 ¾, this factor need not be included in the calculation for the distribution factors since this
factor of ¾ will be present in both the numerator and denominator and therefore will simply cancel out
.
REINFORCED CONCRETE DESIGN
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Theory III
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First, get the distribution factors of span BA and span BC.
DFBA 
1
5  0.44444
1 1
5
4
DFBC  0.55556
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 FEM’s at the end of the spans are equal.
70 5
 43.75 kNm  1.5  65.625 kN - m
8
90 4
M BC 
 45 kNm  1.5  67.5 kN - m
8
MBA 
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 Cañete 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.
MA  0
M B  65.625 0.55556  67.5 0.44444  66.458 kN - m
MC  0
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Engr. ALBERTO C. CAÑETE
Theory III
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THREE-SPAN BEAM
For beams with three-span, there is no shortcut or the Cañete 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. FEM’s (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).
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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 Cañete 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.
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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.
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Theory III
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FRAMES
Now, let’s 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 2span with both fixed end supports, use Cañete Cross Formula.
First, the stiffness factor of each span is computed on the basis of 4EI/L or by using the relativestiffness 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.
DFBC 
1
1
6

 0.35714
11.2
1  1.2
6
4
4
DFBA  0.64286
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Second, compute for the fixed end moments of the supports.
M BA  0
M BC 
506
 37.5 kN - m
8
Then, get the final moment at joint A, B and C using Cañete 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.
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Solve for the Moments.
2.5m
7m
6m
6m
7m
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
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FEM’S
Since the FEM’s of span BA are equal, then this is simplified by multiplying the FEMBA with a factor of 1.5
1
MBA=12 (14.32)(72 )𝑥 1.5 = 87.710 𝑘𝑁 −m
1
(14.32)(62 )
12
MBC= MCB=
𝐷𝐹𝐵𝐶
= 42.96 𝑘𝑁 − 𝑚
1
6
=
1 0.75
6+ 7
= 0.60870
DFBA = 0.39130
FINAL M’s
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
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
WL
+
0.5 WL
0.5 WL
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.
0.5 WL
0.5 WL
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.
0.5 WL
+
0.5 WL
Figure B
0.5 WL
Figure C
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
SYMMETRICAL CASE
A
B
a
b
L
𝑏
𝑤
𝑀𝑎 =
∫ 𝑥(𝑙 − 𝑥)𝑑𝑥 = 2.247𝑤
𝑙^2
0
𝑏𝑙
𝑤
𝑀𝑏 =
∫ (𝑥^2)(𝑙 − 𝑥)𝑑𝑥 = 3.5386𝑤
𝑙^2
𝑎
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.
FEM’S for span A-B
Considering Figure B
1
MAB=12 (0.5𝑤)(72 ) + 2.2471(0.5𝑤) = 3.1652𝑤
1
12
MBA= (0.5𝑤)(72 ) + 3.5386(0.5𝑤) = 3.8110𝑤
Notice that, the moments were reduced to half because the supports are simple.
Considering span B-C
1
MBC=MCB=12 𝑤62 = 3𝑤
MBA=3.811𝑤 +
3.1652𝑤
2
= 5.3936𝑤
FINAL M’s
MB=3w (0.3913) +5.393w (0.6087) =4.4566w
0.6087
(3𝑤
2
MC=3w+
− 5.3932) = 2.2717𝑤
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
UNSYMMETRICAL CASE
A
a
B
FEM’s
𝑊 𝐿𝑎 3
3
MA=72 (
𝑊 74
−
MB=72 (12 −
𝑎4
)
4
𝑊 7∗2.53
3
= 72 (
7(4.5)3
3
+
4.54
)
4
−
2.54
)
4
= 0.54475𝑤
= 1.8362𝑤
1
DFBA= 1 7 1 = 0.46154
DFBC=0.53846
+
7 6
FINAL MOMENTS:
1
MB=(0.54475(0.5𝑤) ∗ 2 + 1.8362(0.50) ∗ 053846) = 0.56770𝑤
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
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
QUIZ 1
(Theory 3)
1. DETERMINE THE MAXIMUM POSITIVE MOMENT AND THE NEGATIVE MOMENT AT THE
FIXED SUPPORT.
√58
= 1.0879𝑚
7
6(
√58
) = 6.5278
7
Cantilever M= 8.4482 x (1.0879)2/2
MB= 5 kNm
1
FEM’s= 12 (8.4482)(6.5278)2 = 30.000 𝑘𝑁𝑚
Mc = 30 +
25
= 42.5 𝑘𝑁𝑚
2
Point of zero shear:
𝑋=
21.829
= 2.5839𝑚
8.4482
1
𝑀𝑎𝑥 𝑀𝑝𝑜𝑠 = (21.829)(2.5839) − 5 = 𝟐𝟑. 𝟐𝟎𝟐 𝒌𝑵𝒎
2
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
2. DETERMINE THE MOMENT AT POINT C.
Columns – steel pipe, outer ø= 300mm, thickness=10mm, G=200 GPa
Rafters – aluminum, I section, Flanges, 250 x 20mm, web 450 x 10 mm, G=83 GPa
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013
𝜋
(3004 − 2804 ) = 95.889𝑥106 𝑚𝑚4
64
1
(250𝑥4903 − 240𝑥4503 ) = 628.52𝑥106 𝑚𝑚4
𝐼𝑟𝑎𝑓𝑡 =
12
200 𝑥 95.889
. 75 (
)
6
𝐷𝐹𝐵𝐴 =
= 0.29130
200𝑥95.889
83𝑥628.52
. 75 (
)
+
6
4√5
𝐼𝑐𝑜𝑙 =
𝐷𝐹𝐵𝐶 = 0.7087
𝐹𝐸𝑀′ 𝑠 =
1
2
(10)(4√5) = 66.6667 𝑘𝑁𝑚
12
𝑀𝑐 = 66.667 +
0.7087
(66.667 − 0) = 𝟗𝟎. 𝟐𝟗𝟎 𝒌𝑵𝒎
2
REINFORCED CONCRETE DESIGN
Engr. ALBERTO C. CAÑETE
Theory III
Copyright © 2013