THREE – MOMENT EQUATION (Continuous Beams)
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THREE – MOMENT
EQUATION
(Continuous
Beams)
THREE-MOMENT EQUATION
This is used to determine the moments at the supports of a fully
continuous beams, in turn needed for reaction calculations.
3 – MOMENT EQUATION IN BEAMS 1 AND 2
M1L1 + 2M2 (L1+L2) + M3L2 + 6Aā/L1 + 6Aƃ/L2 = 6EI [ (h1 / L1)+(h3 / L2) ]
Where:
M1 = bending moment @ pt. 1
L1 = length of beam 1
M2 = bending moment @ pt. 2
L2 = length of beam 2
M3 = bending moment @ pt. 3
6Aā/L1 = 3-moment factor of beam 1
6AБ/L2 = 3-moment factor of beam 2
EI = flexural modulus
h1 = sag / settlement @ pt. 1
h3 = sag / settlement @ pt. 3
RULES OF SIGNS:
If the moment at any point is actually negative, sign must be
used when substituting to the equation. Similarly, if an unknown
Moment is actually negative at any point, the three moment equation
will be having a negative value for that moment.
VALUES FOR 6Aā/L AND 6Aƃ/L
Type of Loading
6Aā/L 6Aƃ/L
1.
2.
3.
4.
5.
6.
EXAMPLE 1
28Mb + 8Md + 2040 = 0 ← (Equation 1)
8Mb + 28Md + 1824 = 0 ← (Equation 2)
Multiply eqn. 1 by 8 & eqn. 2 by -28 224 Mb + 64Md + 16,320 = 0 -224 Mb - 784Md - 51,072 = 0 --------------------------------------------------------------------------------------------------------
-720Md - 34,752 = 0 ← (Equation 3)
Md = -48.27 KN∙m Substitute value of Md in eqn. 1 28Mb + 8(-48.27) +2040 = 0
Mb = -59.07 KN∙m
Solving for Reactions at beam ab
∑Mb = ∑M @ left -59.07 = 6Ra - 120(3)
Ra = 50.16 KN Solving for Reactions at beam de
Md=-48.27 Beam BCDE
∑Mb= ∑M @ left -59.07 = -40(4) - 90(10) + 21.96(14) + 8Rd
Rd = 86.69 KN
Beam BCDE
∑Md = ∑M @ right -48.27 = 50.16(14) - 120(11) - 40(4) +8Rb
Rb = 86.69 KN
SHEAR AND MOMENT DIAGRAM
Example 2
Overhang Beam ab
Mb = -15(1.5)
Mb = -22.5 KN∙m Beam bd
← (Equation 1)
Beam ce
← (Equation 2) Multiply Eqn. 1 by 2, Equate 1 by 2
56Mc + 16Md + 9570 = 0 -8Mc - 16Md - 3840 = 0 ________________________________________________________
48Mc + 5730 = 0
Mc = -119.38 KN∙m Sustitute te value of Mc in Eqn. 1
28(-119.38) + 8Md + 4785 = 0
Md = -180.30 KN∙m
SHEAR AND MOMENT DIAGRAM
BEAM
DEFLECTIONS
THREE – MOMENT
EQUATION
Example 1
Find the deflection at midspan.
DEFLECTION AT MIDSPAN:
thus;
Example 2
Find the deflection at midspan.
DEFLECTION AT MIDSPAN:
Example 3 Find the deflection at b and d.
MOMENT AT MIDSPAN OF ABC :
Mb = 12.8125 KN∙m MOMENT AT C:
MC = -5.625 KN∙m DEFLECTION AT POINT D
DEFLECTION AT POINT B
Example 4
Find the deflection at point d and f.
Example 5
Find the deflection 7m from point a.
Example 6
Find the deflection at point b and d.
DEFLECTION AT POINT B
DEFLECTION AT POINT D
Example 7
Find the deflection at point a, c, d, e and g.
E = 200,000 MPa I = 5.3333 x 10-4 m4
Example 8
Find the deflection at 2.5m from a.
Solving for Reactions
∑Fv = 0
Solving for the moment at ‘a’, ‘b’ and δ
Mb = -1.35 KN∙m
Ma = -5.6 KN∙m
Example 9
Find the deflection at point b.
DEFLECTION AT POINT B
Example 10
Find the deflection at 7m from a.
DEFLECTION AT 7m FROM A
Example 11
Find the deflection at point b.
DEFLECTION AT POINT B
Example 12
Find the deflection at 4m from a.
Solving for reactions
V 21 21
M 5.25 -5.25
DEFLECTION AT 4m FROM A
UNIT LOAD
METHOD
Where:
M = moment equation at every segment of the beam due to real load
m = moment equation at every segment of the beam due to a unit load acting at the point where deflection is desired.
E = modulus of elasticity of the beam (MPa)
I = moment of inertia of the beam at x-axis (mm4)
δ = deflection at a point and it may be vertical or horizontal
a & b = limits of integration
Where:
= slope at a point in an elastic curve
M = moment equation at every segment of the beam due to real load
m = moment equation at every segment of the beam due to a unit load applied at the point where the slope/rotation is desired
E = modulus of elasticity of the beam (MPa)
I = moment of inertia of the beam at x-axis (mm4)
a & b = limits of integration
Example 1
Find the deflection at midspan.
=
+
=
+
=
Rotation:
+
+
= 0 Example 2
Find the deflection at midspan.
DEFLECTION AT L/2
+
Example 3
Solve for the deflection at b.
Positioning unit load at b
DEFLECTION AT POINT B
+
Positioning unit load at d
DEFLECTION AT POINT B
+
+
Example 4
Solve for the deflection at point d and f.
Positioning unit load at F
DEFLECTION AT POINT F
+
+
+
Example 5
Solve for the deflection at 7m from a.
Positioning unit load at 7m from a
DEFLECTION AT 7M FROM A
+
+
+
Example 6
Solve for the deflection at b and d.
Positioning Unit Load
DEFLECTION AT B
+
+
Since it is in the same beam, the values of M are the same.
Positioning unit load
DEFLECTION AT D
+
+
Example 7
Solve for the deflection at 2.5m from a.
Positioning unit load at 2.5m from a
DEFLECTION AT POINT 2.5 FROM A
+
Example 8
Find the deflection at midspan.
Positioning unit load
Rotation
Example 9
Solve for the deflection at 7m from a.
Positioning unit load at 7m from a
DEFLECTION AT 7m FROM A
Example 10
Find the deflection at free end.
Solution :
Rotation
AREA MOMENT
METHOD
Theorem 1. The change in slope between two lines tangent to the elastic
curve at point “a” and “b” is equal to the product of 1/EI and the area of the moment diagram from “a” and “b” only.
Theorem 2. The deviation of point “a” with respect to a tangent line drawn
from “b” in the elastic curve is equal to the product of 1/EI at the moment area of Aab taken at an axis of “a”.
Similarly:
Note:
Example 1 Find the deflection at midspan.
Example 2
Find the deflection at midspan.
Example 3 Solve for the deflection at b and d.
Example 4
Solve for the deflection at point d and f .
Example 5
Solve for the deflection at 7m from a.
Example 6
Solve for the deflection at point b and d.
Example 7
Find the deflection at free end.
Example 8 Find the deflection at free end
+
-
Example 9
Solve for the deflection at free end.
t b/a
Example 10
Solve for the deflection at free end.
Example 11 Solve for the deflection at 4m from a.
δ at 4m fr. a = ?
CONJUGATE BEAM
METHOD
Theorem 1:
The rotation (θ) at a point in the actual beam is equal to the
shear ( ) at that point o the fictitious (conjugate beam) of equal span
loaded with M/EI diagram.
Theorem 2:
The deflection (δ) at a point in the actual beam is equal to the
moment ( ) at that point o the fictitious (conjugate beam) of equal span loaded with M/EI diagram.
Condition of the supports
Actual beam Conjugate
Example 1
Find the deflection at midspan.
∑M @C = 0
↑+∑Fv = 0
∑M @C = 0
Example 2 Find the deflection at midspan.
∑M @C = 0
∑M @B = 0
Example 3 Find the deflection at b.
∑M @C = 0
Deflection at point b
Deflection at point d
Example 4 Find the deflection at point d and f.
∑M @ hinge= 0
Ra = 432.222 KN
∑M @ d= 0
∑M @ f= 0
Example 5 Find deflection 7m from point a.
δ at 7m fr. a = ?
∑M @ b= 0
Ra = 98.9091 KN ∑M @ 7m fr. a= 0
Example 6 Find the deflection at point b and d.
∑M @C = 0
Deflection at point b
Deflection at point d
Example 7
Find the deflection at point a, c, d, e and g.
E = 200,000 MPa I = 5.3333 x 10-4
δ at a,c,d,e,and g = ?
∑M @ f = 0
→ Equation 1 ∑M @ b = 0
→Equation 2 Equate Equation 1 to 2: Mg = 10.125 KN∙m Rg = 1.0625 KN
∑M @ e = 0
-
∑M @ d = 0
∑M @ c = 0
∑M @ a = 0
Example 8
Find the deflection at 2.5m from a.
δ at 2.5 from a = ? ∑M @ b = 0
→ Equation 1 ∑M @ a = 0
→ Equation 2 Rc = 0.4042 KN Mc = 1.82 KN∙m ∑M @ 2.5 fr.a = 0
Example 9 Find the deflection at point b.
Example 10
δ at 7m fr. a = ?
Example 11
∑M @b = 0
Example 12
δ at c = ?
Example 14 Find the deflection at point c.
E = 200,000 MPa I = 100 cm4
δ at 1m from b = ?
∑M @ b= 0
→ Equation 1
∑M @ d= 0
→ Equation 2
Equate 1 and 2
Md = -172.5 KN∙m Rc = -96.25 KN
∑M @ 1m fr. b= 0
Example 15
E = 22 GPa
I = 10,000 mm4
δ at free end = ?
∑M @ free end= 0
DOUBLE
INTEGRATION
METHOD
DOUBLE INTEGRATION METHOD
Procedure:
1.) Determine the moment equation usually at the last
segment of the beam.
2.) Integrate the moment equation to get the slope equation.
3.) Integrate the slope equation to get the deflection equation.
4.) Evaluate the constants of integration using boundary
conditions.
5.) Calculate the required slope or deflection (δ).
Mathematically:
Moment Equation
Slope Equation
Deflection Equation
Example 1
M = y” =
y’ =
y =
at x=0 , y=0 ,
at x = L , y = 0 , C =
0 =
At
Y =
δ=
Example 2
M =
y’ =
y =
at x = 0 , y = 0 , = 0
at x = L , y = 0 , C = -
0 =
At x =
y =
δ=
Example 3
M = y” = 11.375x + 21.125 < x-5> - 2.5
y’ =
y =
at x=0 , y = 0 , at x=5 , y = 0 , C = -21.3542
0 =
C=-21.3542 at x=2.5 , y = δ at b
y= =
y =
Example 4 Find the deflection at point d and f.
M = y” = 48.3333x - 2.5x2 – 20 <x -2> - 30 <x – 6>+ 101.6667<x-12> y’ = 24.1667x2 – 0.8333x3 – 10<x-2>2 – 15<x-6>2 + 50.8333<x-12>2 + C y = 8.0556x3 – 0.20833x4 – 3.3333<x-2>3 – 5<x-6>3 + 16.9444<x-12>3 + Cx + C2
at x = 12 , y = 0 , C = 432.2222
at x = 9 , y = δd
y = 8.0566( 9 )3-0.20833( 9 )4 – 3.3333<9-2>3 – 5<9-6>3 + 16.9444<9-12>3-432.2222(9)
At x = 20, y =
Example 5
Find the deflection 7m from point a.
M = y” = 6.54545x – 10 < x – 5 > -
y’ =
–
y =
–
at x = 0 , y = 0 , =0
at x = 11 , y = 0 , C = -98.909
0 =
–
C = - 98.909
At x = 7 , y = δ
Y =
–
Example 6 Find the deflection at point b and d.
M = y” = 58.9286x -
y’ =
-
y =
-
at x= 0 , y = 0 , at x= 7 , y = 0 ,
0 =
-
C = -254.0181 at x = 4 , y = δb
y =
-
δb=
at x=8.5 , y = δd
y =
-
δd=
Example 7 Find the deflection at point b.
M = y"=
y’ =
y =
at x=L , y’ = 0 , C =
at x=L , y = 0 , =
at x=0 , y =
0 =
Example 8 Find the deflection at 7m from a.
M = y” = -10x -
y’ = -
y = --
at x = 10 , y’ = 0 , C = 2166.6667 , 68.3333
0=
at x= 10 , y = 0
0 =
At x = 3 , y = δ
y =
Example 9 Find the deflection at point b.
M = y” =
y’ =
y =
at x=L , y’ = 0 , C =
at x=L , y = 0 , C =
at x=0 , y’ =
0 =
Example 10
Find the deflection at free end.
δ at free end:
M = y” = -300 + 45x –
y” = -300 + 45x - -
y’ = -300x +
y = -
at x=0 , y’=0 ,
at x=0 , y=0 ,
at x=10 , y=δ at free end
Substitute the value of x to equation of y
y= -
δ at free end = -
Example 11
δ at 1m from b (left) = ?
M = y’’ = 0.778<x-2> + 6.2222<x-5> -
y’’ = 0.778<x-2> + 6.2222<x-5> -
y’ =
y =
y =
δ at 1m. From b =
Example 12
Maximum δ = ?
M = y” = -15.7321<x-2> - 10<x-4> - 26.7679 <x – 9>
+
y’ =
–
y =
–
C1 = 14.6006 C2= -31.3029 Location of max δ
0 =
x = 5.8223 m
Max
Max δ =
CASTIGLIANO’S
SECOND THEOREM
Castigliano’s Second Theorem
This theorem is used to determine deflections of statically
determinate structure and it’s also used to analyze indeterminate beam
and trusses.
δ =
where:
M - bending moment equation at each segment of the beam
due to real loads with respect to the chosen origin.
– partial derivation of bending moment due to load P.
EI – Flexural modulus
Note:
1. In applying the theorems, the load at a point where the
deflection is desired is referred to as P.
2. After the operations required in the equation are
completed the numerical value of P is replaced on the
equation.
3. Should there be no at the point or in the deflection is
desired. An imaginary force P will be able to replaced there
in the direction desired. After the operation is complete, the
correct value of P which is zero will be substituted in the
expression.
4. If the slope or rotation is desired in the beam the partial
derivative is taken with respect to assumed moment P
acting at the point which the rotation is desired a positive
sign. The answer indicates that the rotation is in the
assumed direction of the moment P.
Example 1
δ at midspan = ?
Span Orig M δM/δP M δM/δP δM (δM/δP)dx/EI
a - b a Px/2 x/2 Px2/4
b - c c Px/2 x/2 Px2/4
Example 2
δ at midspan = ?
Span Orig M δM/δP M δM/δP δM (δM/δP)dx/EI
a - b a wLx/2+Px/+ wx2/2
x2/2
b - c c wLx/2+Px/2+ wx2/2
x2/2
DETERMINATE
TRUSSES
Example 1
Find the stresses and deflections of each member.
A = 300x300mm2 E = 200 GPa
θ1 = 63.4349 θ2 = 53.130 ab = 1.7569 C ac = 5.2143 C bc = 3.4286 C bd = 6.7857 C df = 9.4235 C
ef = 7.7858 C cd = 0.2858 T ce = 7.7858 C de = 0
Vertical Deflections
Joint a
ab = 0 ac = 0 bc = 0
bd = 0 cd = 0 ce = 0
de = 0 df = 0 ef = 0
Joint b
ab = 0.7986 C ac = 0.3571 T bc = 0.2857 C
bd = -0.3571 C cd = 0.3771 T ce = 0.1428 T
de = 0 df = 0.3194 C ef = 0.1428 T
Joint c
ab = 0.7986 C ac = 0.3571 T bc = 0.7143 T
bd = 0.3571 C cd = 0.3571 C ce = 0.1428 T
de = 0 df = 0.3194 C ef = 0.1428 T
Joint d
ab = 0.3194 C ac = 0.1428 T bc = 0.2858 T
bd = 0.1428 C cd = 0.3571 C ce = 0.7143 T
de = 0 df = 0.6388 C ef = 0.3571 T
Joint e
ab = 0.3194 C ac = 0.1428 T bc = 0.2858 T
bd = 0.1428 C cd = 0.3571 C ce = 0.3571 T
de = 0 df = 0.7986 C ef = 0.3571 T
Joint f
ab = 0 ac = 0 bc = 0 bd = 0
cd = 0 ce = 0 de = 0 df = 0
ef = 0
Joint b
Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 -0.7986 4.4721 0.09m2 200GPa 0.00035
ac -5.2143 0.3571 2 0.09m2 200GPa -0.00021
bc -3.4286 -0.2857 4 0.09m2 200GPa 0.00022
bd -6.7857 -0.3571 3 0.09m2 200GPa 0.0004
cd 4.2858 0.3571 5 0.09m2 200GPa 0.00043
ce -7.7858 0.1428 3 0.09m2 200GPa -0.00019
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 -0.3194 2 0.09m2 200GPa -0.00012
df -9.4235 0.1428 4.4721 0.09m2 200GPa 0.00075
δb = 0.00163 mm
Joint c
Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 -0.7986 4.4721 0.09m2 200GPa 0.00035
ac -5.2143 0.3571 2 0.09m2 200GPa -0.00021
bc -3.4286 0.7143 4 0.09m2 200GPa 0.00054
bd -6.7857 -0.3571 3 0.09m2 200GPa 0.0004
cd 4.2858 0.3571 5 0.09m2 200GPa 0.00043
ce -7.7858 0.1428 3 0.09m2 200GPa -0.00019
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 -0.3194 2 0.09m2 200GPa -0.00012
df -9.4235 0.1428 4.4721 0.09m2 200GPa 0.00075
δc = 0.00127 mm
Joint d Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 -0.3194 4.4721 0.09m2 200GPa 0.00014
ac -5.2143 0.1428 2 0.09m2 200GPa -0.00083
bc -3.4286 0.2858 4 0.09m2 200GPa -0.00022
bd -6.7857 -0.1428 3 0.09m2 200GPa 0.00016
cd 4.2858 -0.3573 5 0.09m2 200GPa -0.00043
ce -7.7858 0.3573 3 0.09m2 200GPa -0.00046
de 0 1 4 0.09m2 200GPa 0
ef -7. 7858 -0.7986 2 0.09m2 200GPa -0.000187
df -9.4235 0.3573 4.4721 0.09m2 200GPa -0.0031
δd = 0.000667 mm
Joint e Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 -0.3194 4.4721 0.09m2 200GPa 0.00014
ac -5.2143 0.1428 2 0.09m2 200GPa -0.00083
bc -3.4286 0.2858 4 0.09m2 200GPa -0.00032
bd -6.7857 -0.1428 3 0.09m2 200GPa 0.00016
cd 4.2858 -0.3573 5 0.09m2 200GPa -0.00043
ce -7.7858 0.3573 3 0.09m2 200GPa -0.00046
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 -0.7986 2 0.09m2 200GPa -0.000187
df -9.4235 0.3573 4.4721 0.09m2 200GPa -0.0031
δe = 0.000667 m
Horizontal Deflections
Joint a
ab = 0 ac = 0 bc = 0
bd = 0 cd = 0 ce = 0
de = 0 df = 0 ef = 0
Joint b
ab = 0.6388 T ac = 0.2857 C bc = 0.5714 T
bd = 0.7143 C cd = 0.7143 C
ce = 0.7143 C de = 0
df = 0.7143 C ef = 0.6388 T
Joint c
ab = 0 ac = 0 bc = 0
bd = 0 cd = 0
ce = 1 C
de = 0 df = 0
ef = 1 C
Joint d
ab = 0.6388 T ac = 0.2857 C bc = 0.5714 C
bd = 0.2857 T cd = 0.7143 T ce = 0.7143 C
de = 0 df = 0.6388 C ef = 0.7143 C
Joint e
ab = 0 ac = 0 bc = 0
bd = 0 cd = 0 ce = 0
de = 0 df = 0
ef = 1 C
Joint f
ab = 0 ac = 0 bc = 0
bd = 0 cd = 0 ce = 0
de = 0 df = 0
ef = 1 C
Joint b
Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 -0.6388 4.4721 0.09m2 200GPa -0.00028
ac -5.2143 -0.2857 2 0.09m2 200GPa 0.00017
bc -3.4286 0.5714 4 0.09m2 200GPa -0.00044
bd -6.7857 -0.7143 3 0.09m2 200GPa 0.00081
cd 4.2858 -0.7143 5 0.09m2 200GPa -0.00085
ce -7.7858 -0.7143 3 0.09m2 200GPa 0.00093
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 -0.6388 2 0.09m2 200GPa 0.00055
df -9.4235 0.7143 4.4721 0.09m2 200GPa -0.00017
δb = 0.0042 mm
Joint c
Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 0 4.4721 0.09m2 200GPa 0
ac -5.2143 0 2 0.09m2 200GPa 0
bc -3.4286 0 4 0.09m2 200GPa 0
bd -6.7857 0 3 0.09m2 200GPa 0
cd 4.2858 0 5 0.09m2 200GPa 0
ce -7.7858 -1 3 0.09m2 200GPa 0.0013
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 0 2 0.09m2 200GPa 0
df -9.4235 -1 4.4721 0.09m2 200GPa 0.00087
δc = 0.00217 mm
Joint d Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 0.6388 4.4721 0.09m2 200GPa -0.00028
ac -5.2143 -0.2857 2 0.09m2 200GPa 0.00017
bc -3.4286 -0.5714 4 0.09m2 200GPa -0.00044
bd -6.7857 0.2857 3 0.09m2 200GPa -0.00032
cd 4.2858 0.7143 5 0.09m2 200GPa -0.00085
ce -7.7858 -0.7143 3 0.09m2 200GPa 0.00093
de 0 0 4 0.09m2 200GPa 0
ef -7. 7858 -0.6388 2 0.09m2 200GPa 0.0015
df -9.4235 -0.7143 4.4721 0.09m2 200GPa 0.00062
δd = 0.00391 mm
Joint e Mem. S U L(m) A E SUL/AE (mm)
ab -1.7569 0 4.4721 0.09m2 200GPa 0
ac -5.2143 0 2 0.09m2 200GPa 0
bc -3.4286 0 4 0.09m2 200GPa 0
bd -6.7857 0 3 0.09m2 200GPa 0
cd 4.2858 0 5 0.09m2 200GPa 0
ce -7.7858 0 3 0.09m2 200GPa 0
de 0 0 4 0.09m2 200GPa 0 ef -7. 7858 0 2 0.09m2 200GPa 0
df -9.4235 -1 4.4721 0.09m2 200GPa 8.65x10-3
δe = 0.000865 mm
Example 2
Find the stresses and deflections of each member.
A = 0.0001m2
E = 10 GPa θ= 53.1301
ab = 6 C ac = 27.5 C ad = 16.5 T bc = 0 cd = 0 ce = 24 C cf = 12.5 T df = 16.5 T ef = 16 C
eg = 24 C fg = 7.5 T fh = 19.5 T gh = 0 gi = 0 gj = 32.5 C hj = 19.5 T ij = 8 C
Joint b
ab = 6 C
ac = 0 ad = 0 bc = 0 cd = 0
ce = 0 cf = 0 df = 0 ef = 0
eg = 0 fg = 0 fh = 0 gh = 0
gi = 0 gj = 0 hj = 0 ij = 0
Joint c
ab = 0 ac = 0.9375 C ad = 0.5625 T bc = 0 cd = 1 T ce = 0.375 C
cf = 0.3125 C df = 0.5625 C ef = 0 eg = 0.375 T fg = 0.3125 T fh = 0.1875 T
gh = 0 gi = 0 gj = 0.3125 C hj = 0.1875 T ij = 0
Joint d
ab = 0 ac = 0.9375 C ad = 0.5625 T bc = 0 cd = 0 ce = 0.375 C
cf = 0.3125 C df = 0.5625 T ef = 0 eg = 0.375 C fg = 0.3125 T fh = 0.1875 T
gh = 0 gi = 0 gj = 0.3125 C hj = 0.1875 T ij = 0
Joint e
ab = 0 ac = 0.625 C ad = 0.375 T bc = 0 cd = 0 ce = 0.75 C
cf = 0.625 T df = 0.375 T ef = 1 C eg = 0.75 C fg = 0.625 T fh = 0.375 T
gh = 0 gi = 0 gj = 0.625 C hj = 0.375 T ij = 0
Joint f
ab = 0 ac = 0.625 C ad = 0.375 T bc = 0 cd = 0 ce = 0.75 C
cf = 0.625 T df = 0.375 T ef = 0 eg = 0.75 C fg = 0.625 T fh = 0.375 T
gh = 0 gi = 0 gj = 0.625 C hj = 0.375 T ij = 0
Joint g
ab = 0 ac = 0.3125 C ad = 0.1875 T bc = 0 cd = 0 ce = 0.375 C cf = 0.3125 T
df = 0.1875 T ef = 0 eg = 0.375 C fg = 0.3125 C fh = 0.5625 T gh = 0 gi = 0
gj = 0.9375 C hj = 0.5625 T ij = 0
Joint h
ab = 0 ac = 0.3125 C ad = 0.1875 T bc = 0 cd = 0 ce = 0.375 C
cf = 0.3125 T df = 0.1875 T ef = 0 eg = 0.375 C fg = 0.3125 C fh = 0.5625 T
gh = 1 T gi = 0 gj = 0.9375 C hj = 0.5625 T ij = 0
Joint i
ab = 0 ac = 0
ad = 0 bc = 0
cd = 0 ce = 0
cf = 0 df = 0 ef = 0
eg = 0 fg = 0 fh = 0 gh = 0
gi = 0 gj = 0 hj = 0 ij = 1 C
Joint b
Mem. S U L(m) A E SUL/AE(mm)
ab -6 -1 20 0.0001m2 10 GPa 120
ac -27.5 0 25 0.0001m2 10 GPa 0
ad 16.5 0 15 0.0001m2 10 GPa 0
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 0 15 0.0001m2 10 GPa 0
cf 12.5 0 25 0.0001m2 10 GPa 0
df 16.5 0 15 0.0001m2 10 GPa 0
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 0 15 0.0001m2 10 GPa 0
fg 7.5 0 25 0.0001m2 10 GPa 0
fh 19.5 0 15 0.0001m2 10 GPa 0
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 0 25 0.0001m2 10 GPa 0
hj 19.5 0 15 0.0001m2 10 GPa 0
ij -8 0 20 0.0001m2 10 GPa 0
δb = 120mm
Joint c
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.9375 25 0.0001m2 10 GPa 644.53135
ad 16.5 0.5625 15 0.0001m2 10 GPa 139.21875
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 -0.375 15 0.0001m2 10 GPa 135
cf 12.5 -0.3125 25 0.0001m2 10 GPa -97.65625
df 16.5 0.5625 15 0.0001m2 10 GPa 139.21875
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 -0.375 15 0.0001m2 10 GPa 135
fg 7.5 0.3125 25 0.0001m2 10 GPa 58.59375
fh 19.5 0.1875 15 0.0001m2 10 GPa 54.84375
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.3125 25 0.0001m2 10 GPa 253.90625
hj 19.5 0.1875 15 0.0001m2 10 GPa 54.84375
ij -8 0 20 0.0001m2 10 GPa 0
δc = 1517.5mm
Joint d
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.9375 25 0.0001m2 10 GPa 644.53135
ad 16.5 0.5625 15 0.0001m2 10 GPa 139.21875
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 1 20 0.0001m2 10 GPa 0
ce -24 -0.375 15 0.0001m2 10 GPa 135
cf 12.5 -0.3125 25 0.0001m2 10 GPa -97.65625
df 16.5 0.5625 15 0.0001m2 10 GPa 139.21875
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 -0.375 15 0.0001m2 10 GPa 135
fg 7.5 0.3125 25 0.0001m2 10 GPa 58.59375
fh 19.5 0.1875 15 0.0001m2 10 GPa 54.84375
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.3125 25 0.0001m2 10 GPa 253.90625
hj 19.5 0.1875 15 0.0001m2 10 GPa 54.84375
ij -8 0 20 0.0001m2 10 GPa 0
δd = 1517.5mm
Joint e
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.625 25 0.0001m2 10 GPa 429.6875
ad 16.5 0.375 15 0.0001m2 10 GPa 92.8125
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 -0.75 15 0.0001m2 10 GPa 270
cf 12.5 0.625 25 0.0001m2 10 GPa 195.3125
df 16.5 0.375 15 0.0001m2 10 GPa 92.8125
ef -16 -1 20 0.0001m2 10 GPa 320
eg -24 -0.75 15 0.0001m2 10 GPa 270
fg 7.5 0.625 25 0.0001m2 10 GPa 117.1875
fh 19.5 0.375 15 0.0001m2 10 GPa 109.6875
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.625 25 0.0001m2 10 GPa 507.8125
hj 19.5 0.375 15 0.0001m2 10 GPa 109.6875
ij -8 0 20 0.0001m2 10 GPa 0
δe = 2515 mm
Joint f
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.625 25 0.0001m2 10 GPa 429.6875
ad 16.5 0.375 15 0.0001m2 10 GPa 92.8125
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 -0.75 15 0.0001m2 10 GPa 270
cf 12.5 0.625 25 0.0001m2 10 GPa 195.3125
df 16.5 0.375 15 0.0001m2 10 GPa 92.8125
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 -0.75 15 0.0001m2 10 GPa 270
fg 7.5 0.625 25 0.0001m2 10 GPa 117.1875
fh 19.5 0.375 15 0.0001m2 10 GPa 109.6875
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.625 25 0.0001m2 10 GPa 507.8125
hj 19.5 0.375 15 0.0001m2 10 GPa 109.6875
ij -8 0 20 0.0001m2 10 GPa 0
δf = 2915 mm
Joint g
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.3125 25 0.0001m2 10 GPa 214.84375
ad 16.5 0.1875 15 0.0001m2 10 GPa 46.40625
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 -0.375 15 0.0001m2 10 GPa 135
cf 12.5 0.3125 25 0.0001m2 10 GPa 97.65625
df 16.5 0.1875 15 0.0001m2 10 GPa 46.40625
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 -0.375 15 0.0001m2 10 GPa 135
fg 7.5 -0.3125 25 0.0001m2 10 GPa -58.59375
fh 19.5 0.5625 15 0.0001m2 10 GPa 164.53125
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.9375 25 0.0001m2 10 GPa 761.71875
hj 19.5 0.5625 15 0.0001m2 10 GPa 164.53125
ij -8 0 20 0.0001m2 10 GPa 0
δg= 1707.5 mm
Joint h
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 -0.3125 25 0.0001m2 10 GPa 214.84375
ad 16.5 0.1875 15 0.0001m2 10 GPa 46.40625
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 -0.375 15 0.0001m2 10 GPa 135
cf 12.5 0.3125 25 0.0001m2 10 GPa 97.65625
df 16.5 0.1875 15 0.0001m2 10 GPa 46.40625
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 -0.375 15 0.0001m2 10 GPa 135
fg 7.5 -0.3125 25 0.0001m2 10 GPa -58.59375
fh 19.5 0.5625 15 0.0001m2 10 GPa 164.53125
gh 0 -1 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 -0.9375 25 0.0001m2 10 GPa 761.71875
hj 19.5 0.5625 15 0.0001m2 10 GPa 164.53125
ij -8 0 20 0.0001m2 10 GPa 0
δh= 1707.5 mm
Joint i
Mem. S U L(m) A E SUL/AE(mm)
ab -6 0 20 0.0001m2 10 GPa 0
ac -27.5 0 25 0.0001m2 10 GPa 0
ad 16.5 0 15 0.0001m2 10 GPa 0
bc 0 0 15 0.0001m2 10 GPa 0
cd 0 0 20 0.0001m2 10 GPa 0
ce -24 0 15 0.0001m2 10 GPa 0
cf 12.5 0 25 0.0001m2 10 GPa 0
df 16.5 0 15 0.0001m2 10 GPa 0
ef -16 0 20 0.0001m2 10 GPa 0
eg -24 0 15 0.0001m2 10 GPa 0
fg 7.5 0 25 0.0001m2 10 GPa 0
fh 19.5 0 15 0.0001m2 10 GPa 0
gh 0 0 20 0.0001m2 10 GPa 0
gi 0 0 15 0.0001m2 10 GPa 0
gj -32.5 0 25 0.0001m2 10 GPa 0
hj 19.5 0 15 0.0001m2 10 GPa 0
ij -8 -1 20 0.0001m2 10 GPa 160
δi= 160 mm
ANALYSIS OF
INDETERMINATE
TRUSS BY
CASTIGLIANO’S
THEOREM
Procedure:
1.) Determine one support as the redundant which is to be
removed from the given truss.
2.) Determine the first set of bar forces due to applied loads in
the absence of the redundant.
3.) Determine the bar stresses due to redundant only in the
absence of real load.
4.) Tabulate the bar stresses obtained using the format below.
5.) Get the summation of
6.) Solve for the unknown reaction.
Radius of Curvature (m)
Example 1
Find the reactions.
Stresses
ab =5.2778 C ac = 7.1667 T bc = 3 T bd = -8.6133 C cd = 5.4083 C
ce = 11.6667 T de = 0 df = 14.0216 C ef = 11.6667 T
Horizontal Unit Load
ab =0 ac = 1 C bc = 0 bd = 0
cd = 0
ce = 1 C de = 0 df = 0 ef = 0
Member S’ U L A E
ab -5.2778 0 5m 0.0005m2 12 Gpa
ac 7.1667 -1 3m 0.0005m2 12 Gpa
bc 3 0 4m 0.0005m2 12 Gpa
S’-uRe
0 0 -5.2778
-3.5833 1.05 17.8333
0 0 3
0 0 -8.6133
0 0 -5.4083
-5.8333 0.5 221.8333
0 0 0
0 0 -14.0216
-5.8333 0.5 21.8333
Rah = 18.1667 KN
Rfh =
Rfv = 10.7778 KN
Rav = 7.2222 KN
Example 2
Find the reactions.
bd -8.6133 0 0.0005m2 12 Gpa
cd -5.4083 0 0.0005m2 12 Gpa
ce 11.6667 -1 3m 0.0005m2 12 Gpa
de 0 0 2m 0.0005m2 12 Gpa
df -14.0216 0 0.0005m2 12 Gpa
ef 11.6667 -1 3m 0.0005m2 12 Gpa
Member S’ U L A E
ab -4 0 4m 0.0005m2 12 Gpa
ac -2.8284 0 5.6567m 0.0005m2 12 Gpa
ad 6 1 4m 0.0005m2 12 Gpa
bc -4 0 4m 0.0005m2 12 Gpa
cd -6 0 4m 0.0005m2 12 Gpa
ce -6 0 4m 0.0005m2 12 Gpa
de 8.4853 0 5.6567m 0.0005m2 12 Gpa
df 0 1 4m 0.0005m2 12 Gpa
ef -10 0 4m 0.0005m2 12 Gpa
S’-uRe
0 0 -4
0 1.05 -2.8284
4 0 9
0 0 -4
0 0 -6
-5.8333 0 -6
0 0 8.4853
0 0 3
-5.8333 0 -10
Rfh = 3 KN
Rav = 6 KN
Rfv = 10 KN
Rah = 11 KN
MOMENT
DISTRIBUTION
METHOD
(MDM)
Example 1 Find the reactions of the frames below.
Column = 2I
Beam = I
↑+∑Fv = 0
Rvc = 168.5884 KN
∑Fh= 0
Rhc = 19.4532 KN
Origin A B C D E
ab ba bd cd db dc de ed
DF 1 0.7059 0.2941 0 0.2059 0.6176 0.1765 1
FEM -14.40 9.6 -82.5 0 82.5 0 -81.6667 81.6667
1st 14.40 51.4601 21.4399 0 -0.1716 -0.5146 -0.1471 -81.6667
CO 25.7301 7.2 -0.0858 -0.2573 10.72 0 -40.8334 -0.0736
2nd
-25.7301 -5.0219 -2.0923 0 6.2003 18.5980 5.3150 0.0736
CO -2.5110 -12.8651 3.1002 9.2990 -1.0462 0 0.0368 2.6575
3rd -2.5110 6.8930 2.8719 0 0.2078 0.6234 0.1782 -2.6575
0 57.2661 -57.2661 9.0417 98.4103 18.7068 -117.1172 0
Example 1
Determine the vertical deflection at point E of the simple frame
below. Use E=11370 MPa and I = 1750x106 mm4 for all members.
Unit Load
A. “M” due to real loads 1. Segment ab
2. Segment bd
3. Segment cd
4. Segment de
5. Segment ef
B. Due to vertical load downward
1. Segment ab
2. Segment bd
3. Segment cd
4. Segment de
5. Segment ef
Solution:
Horizontal Unit Load
mab = x mbd = x mcd = 0 mde = 0.6667x mef = 0.6667x Substitute:
Moment
Distribution of
Frames with
Sidesway
Procedure:
1. The joints are held against sidesway. The fixed-end moments
caused by the applied loading are distributed and the first set of
balanced end moments is obtained, call these joint as “ M’ ”.
2. The unloaded frame is then assumed to have a certain amount
of sidesway which will cause a set of fixed-end moments. These
fixed-end moment are then distributed and a second set of
balanced end moment is obtained, call these moment as “ M ”.
3. The resulting set of end moments may be obtained by adding
the first set and the product of a ratio “k” and a second set. The
value “k” may be obtained using shear conditions.
M = M’ + KM”
ILLUSTRATION:
Actual frame Due to loads Due to sidesway
M M’ M”
Example 1
Orig A B C
MEM ab ba bg bc cb cf cd
DF 0 0.3636 0.2728 0.3636 0.3636 0.2728 0.3636 FEM 0 0 -2.6667 -3 3 -4 -3
1ST 0 2.0604 1.5459 2.0604 1.4544 1.0912 1.4544
CO 1.0302 0 -0.3638 0.7272 1.0302 0.5456 0.2857
2ND
0 -0.1321 -0.0991 -0.1321 -0.2801 -0.2801 -0.2801
CO -0.0661 0 -0.0063 -0.1401 -0.0661 0.0372 0.0372
3RD
0 0.0532 0.0399 0.0532 -0.0432 -0.0432 -0.0432
∑ 0.9614 1.9815 -1.5501 -0.4314 5.0952 -1.5460 -1.5460
Orig D E F
MEM dc de ed ef fe fc fg
DF 0.5714 0.4286 0.4286 0.5714 0.3636 0.2727 0.3636
FEM 3 -4 4 0 0 4 0
1ST
0.5714 0.4286 -1.7144 -2.2856 1.4544 1.0912 -0.9696
CO 0.7272 -0.8572 0.2143 0.7272 1.1428 0.5456 -0.7272
2ND 0.0734 -0.0361 0.2198 0.2931 0.3934 0.2952 -0.0167
CO -0.1401 0.1099 -0.0181 0.1967 0.1466 -0.1051 0.1967
3RD
0.0173 0.0129 -0.0963 0.1283 -0.0121 -0.0090 -0.0535
∑ 4.2501 -4.2501 2.6513 -2.6513 -2.0693 3.6354 -1.5703
Orig G H MEM gf gb gh hg
DF 0.3636 0.2127 0.3636 0 FEM 0 2.6667 0 0 1ST -1.4544 -0.7275 -0.9696 0 CO -0.4840 0.7730 0 -0.4848 2ND 0.3934 -0.0125 -0.0167 0 CO -0.0084 -0.0496 0 -0.0084 3RD -0.0120 -0.0401 -0.0535 0 ∑ -1.5662 2.6100 -1.0398 -0.4932
PORTAL
METHOD
Example 1
Determine the shear moment and axial forces, and the columns,
girders of the frame below.
Solution:
1.) Column Shear
1st level
ground level
2.) Column Moments
1st level
Mcb = Mbc = 7.50 ( 1.50 ) = 11.25 KN.m
Mde = Med = 15 ( 1.50 ) = 22.5 KN.m
Mih = Mhi = 7.50 ( 1.50 ) = 11.25 KN.m
Ground level
Mba = Mab = 10 ( 2 ) = 20 KN.m
Meh = Mfe = 20 ( 2 )= 40 KN.m
Mhg = Mgh = 10 ( 2 )= 20 KN.m
3.) Girder Moments
Joint c :Mdc = Mcb = 11.25 KN.m Joint d :22.50 = 11.25 + Mdi = 11.25 KN.m
Joint i :Mdi = Mih = 11.25 KN.m Joint b :Mba + Mhg = 31.25 KN.m Joint h :Mhi + Mhg = 31.25 KN.m Joint e :Mbe + Meh = Med + Mef = 31.25 KN.m
4.) Girder Shears
Mcd = Vcd( 2 ) 11.25 = Vcd( 2 ) = 5.625 KN Mdi = Vdi( 2 ) 11.25 = Vdi( 2 ) = 5.625 KN Mbe = Vbe( 2 ) 31.25 = Vbe( 2 ) = 15.625 KN Meh = Veh( 2 ) 31.25 = Veh( 2 ) = 15.625 KN
5.) Column Axial Forces “ s “
Joint C
3.625 – Scb = 0
Scb = 5.625 KN
Joint i
Sih – 5.625 = 0
Sih = 5.625 KN
Joint b
5.625 + 15.625 – Sba= 0 Sba = 21.25 KN
Joint d
By inspection, Sde = 0
Joint e
By inspection, Sef = 0
Joint h
-15.625 – 5.625 + Shg = 0 Shg = 21.25 KN
Cantilever
Method
Cantilever Method Assumptions:
1. There is a point of inflection at midpoint of all members. 2. The intensity of axial force in each column of a storey is
proportional to the horizontal distance of that column from the center of gravity of all columns of the storey under consideration.
Example 1 Analyze the building frame below completely using the cantilever method. Assume area of each column is 100,000mm2.
1. Coulmn Axial Forces
∑ax
1st level (Above A-A)
∑M@1=0
60(1) + Sde (6) – sin (14) = 0 60 + 0.09 sin (6) – 14 sin = 0 Sin = 4.46 KN
Thus; Scb = 0.91 (4.46) ; Scb = 4.06 KN Sde = 0.09 (4.46) ; Sde = 0.40 KN
Ground level (Above B-B)
∑M@2=0
40(1.5) + 60(3.5) + 0.09 Shg (6) – Shg (14) = 0
Shg = 20.06 KN Sba = 18.25 KN
Sef = 1.81 KN
2. Girder Shear
Joint c
∑Fv = 0
Vcd – 4.06 = 0 Vcd = 4.06 KN
Joint d
∑Fv = 0 -4.06 – 0.40 + Vdi = 0 Vdi = 4.46 KN
Joint b
∑Fv = 0 -18.25 + 4.06 + Vbe = 0 Vbe = 14.19 KN
Joint e
∑Fv = 0 -14.19 – 1.81 + 0.40 + Veh = 0 Veh = 15.60 KN
3. Girder Moments
Mcd = Mde = 4.06 (3) = 12.18 KN∙m Mdi = Mid = 4.06 (4) = 17.84 KN∙m Mbe = Meb = 14.19 (3) = 42.57 KN∙m Meh = Mhe = 15.60 (4) = 62.40 KN∙m
4. Column Moments
∑M column = ∑M girder
Joint c Mcd = Mcb = 12.18 KN∙m
Joint d Mdc + Mdi = Mde Mde = 30.02 KN∙m
Joint j Mdi = Mih = 17.84 KN∙m
Joint h Mhi + Mhg = Mhe 17.84 + Mhg = 62.40 Mhg = 44.56 KN∙m
Joint b Mbc + Mba = Mbe 12.18 + Mba = 42.57 Mba = 30.39 KN∙m
Joint e Med + Mef = Meb + Meh 30.02 + Mef = 42.57 + 62.40 Mef = 74.95 KN∙m
5. Column Shear
Vbc = 2 (12.18) / 2 = 12.18 KN Vde = 2 (30.02) / 2 = 30.02 KN Vih = 2 (17.84) / 2 = 17.84 KN Vba = 2 (30.39) / 3 = 20.26 KN Vef = 2 (74.95) / 3 = 49.97 KN Vhg = 2 (44.56) / 3 = 29.71 KN
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