CAPÍTULO V: DISCUSIÓN
Anexo 15. Plan de segregación de residuos sólidos de construcción
Dead load combination.
Refer to table 2 to obtain the concentrated load acting on the haunch portal frame at the purlins point shown in figure 12 and calculate free bending moment diagram.
Take moments, MB5 = 2.93 k x 1.5
= 4.4 k N.m
MB4 = (2.93 k x 3) + (4.84k x 1.5)
= 16.05 k N.m
MB3 = [2.93 k x 4.5] + [4.84 k x (3 + 1.5)]
= 34.97 k N.m
MB2 = [2.93 k x 6] + [4.84 k x (4.5 + 3 + 1.5)]
A B1
B B1 B2 B3 B4 B5 C
‘cut’
Figure 12
= 61.14 k N.m
MB1 = [2.93 k x 7.5] + [4.84 k x (6 + 4.5 + 3 + 1.5)]
= 94.58 k N.m
MB = [2.93 k x 9] + [4.84 k x (7.5 + 6 + 4.5 + 3 + 1.5)]
= 135.27 k N.m MB = MA
Since the concentrated load value for the other side of the haunch portal frame is the same. The free bending moment is the same. Figure 13 show the free bending moment diagram for haunch portal frame under dead load with partial fraction, γf 1.4.
Figure 13
When the haunch portal frame is ‘cut’ at the apex point, there are 3 equal and opposite reaction;
moment ‘M’, horizontal shear force ‘H’ and vertical shear force ‘V’ as showing in figure 14. By using the equilibrium equation, maximum plastic moment occur (plastic hinge) on the rafter of the structure will be calculate.
Refer [8:p161], figure 15 and figure 16 shows two possible collapse mechanisms for a haunch portal frame. To determine the plastic moment, reactant moment must determine 1st. Thus, reactant moment equation are required at A, E, B1, B1 and B5.
hr
hc
hh
A E
W
B1 B5
H H
V
V
M M D5 D1
Figure 14
B1 D1
Figure 15
Collapse mechanism ‘A’
A E
B5 D5
B1 D1
Where, hr = 3 m hh = 0.5 m hc = 2.5 m W = 18 m
Taking moment about possible plastic hinge may occur on haunch portal frame (refer to figure 14, figure 15 and figure 16):
(Assume anti-clockwise moment positive for left hand side and clockwise positive for right hand side of ‘cut’ at apex)
At, A; reactant moment = M + 6H + 9V
At, A1; reactant moment = M + 5H + 9V At, A2; reactant moment = M + 4H + 9V At, B1; reactant moment = M + 3.5H + 9V
At, B; reactant moment = M + 3H + 9V
At, B1; reactant moment = M + 2.5H + 7.5V At, B2; reactant moment = M + 2H + 6V At, B3; reactant moment = M + 1.5H + 4.5V At, B4; reactant moment = M + H + 3V At, B5; reactant moment = M + 0.5H + 1.5V At, D5; reactant moment = M + 0.5H - 1.5V At, D4; reactant moment = M + H - 3V At, D3; reactant moment = M + 1.5H - 4.5V At, D2; reactant moment = M + 2H - 6V At, D1; reactant moment = M + 2.5H – 7.5V
At, D; reactant moment = M + 3H - 9V
At, D1; reactant moment = M + 3.5H - 9V At, E2; reactant moment = M + 4H - 9V At, E1; reactant moment = M + 5H - 9V
At, E; reactant moment = M + 6H - 9V
Figure 16
Collapse mechanism ‘B’
A E
B5 D5
B1 D1
Table 11
Reactant Moment at purlin point
Investigation of the collapse mode ‘A’ under dead load by using equilibrium equation.
Refer to figure 17, figure 15 and table 11.set up the equilibrium equation.
At A;135.27 = reactant moment = M + 6H + 9V
135.37 - M - 6H - 9V = 0 equation 1
At B1; 135.27 = reactant moment + Mp = M + 3.5H + 9V + Mp
135.27 - M - 3.5H - 9V = +Mp equation 2
At B5; 4.4 + Mp = reactant moment = M + 0.5H + 1.5V
4.4 - M - 0.5H - 1.5V = -Mp equation 3
At E; 135.27 = reactant moment = M + 6H - 9V
135.27 - M - 6H + 9V = 0 equation 4
Solve the simultaneous equation;
Consider equation 4 and equation 1.
Equation 1;
M = 135.37 – 6H – 9V equation 5
Figure 17
Free bending moment and reactant moment under collapse mechanism ‘A’
Equation 4;
M = 135.27 - 6H + 9V equation 6
Solve equation 6 and equation 5, 135.37 – 6H – 9V = 135.27 - 6H + 9V
V = 0
Substitute equation 5 into equation 2,
135.27 – (135.37 – 6H – 9V) - 3.5H - 9V = +Mp
135.27 – 135.27 + 6H + 9V -3.5H -9V = +Mp
2.5H = +Mp equation 7
Substitute equation 5 and equation 7 into equation 3, 4.4 – (135.37 – 6H – 9V) - 0.5H - 1.5V = -2.5H 8H = 130.97 H = 16.37 k N
Substitute H into equation 5 and equation 7, to obtain M and Mp. M = 135.37 – 6(16.37)
= 37.15 k N.m Mp = 2.5 (16.37)
= 40.93 k N.m
Mp > M (obey fundamental yield condition of plastic theory).
Location for the plastic hinge to occur on the portal frame is at B1, B5, D1 and D5. To check whether the plastic hinge is located at the correct position on the rafter, calculation of the reactant moment on all the purlin point is needed to check the maximum plastic moment can occur along the rafter. Table 12 show the amount of plastic moment along the rafter on each point of purlin under influence of dead load.
H 16.37
V 0
M 37.15
Point Reactant Moment F.B.M Reactant Moment Plastic Moment
A = M + 6H + 9V 135.27 135.37 -0.1
B^1 = M + 3.5H + 9V 135.27 94.445 40.825
B = M + 3H + 9V 135.27 86.26 49.01
B1 = M + 2.5H + 7.5V 94.58 78.075 16.505
B2 = M + 2H + 6V 61.14 69.89 -8.75
B3 = M + 1.5H + 4.5V 34.97 61.705 -26.735
B4 = M + H + 3V 16.05 53.52 -37.47
B5 = M + 0.5H + 1.5V 4.4 45.335 -40.935
C = M 0 37.15 -37.15
D5 = M + 0.5H - 1.5V 4.4 45.335 -40.935
D4 = M + H - 3V 16.05 53.52 -37.47
D3 = M + 1.5H - 4.5V 34.97 61.705 -26.735
D2 = M + 2H - 6V 61.14 69.89 -8.75
D1 = M + 2.5H - 7.5V 94.58 78.075 16.505
D = M + 3H - 9V 135.27 86.26 49.01
D^1 = M + 3.5H - 9V 135.27 94.445 40.825
E = M + 6H - 9V 135.27 135.37 -0.1
Plastic Moment under Dead Load All unit in (k N.m)
From table 12, the maximum plastic moment is at point B and D, but the plastic moment occurs at point B and point D is considered as moment in the haunch. It is assumed the plastic hinge that occurs at the eaves of the haunch is shifted to the column top immediately below the lower end of the haunch, point B1 and D1. Thus, the plastic hinge occurs at point D1, D5, B5 and B1. Figure 18 show the free bending moment line, reactant moment line and plastic moment base on the values that obtain from table 12.
Table 12
Investigation of the collapse mode ‘B’ under dead load will be carry on by using equilibrium equation.
Figure 19
Free bending moment and reactant moment under collapse mechanism ‘B’
Figure 18
Refer to figure 19, figure 16 and table 11.set up the equilibrium equation.
At A;135.27 = reactant moment = M + 6H + 9V
135.37 - M - 6H - 9V = 0 equation 8
At B1; 94.58 = reactant moment + Mp = M + 2.5H + 7.5V + Mp
94.58 - M - 2.5H – 7.5V = +Mp equation 9
At B5; 4.4 + Mp = reactant moment = M + 0.5H + 1.5V
4.4 - M - 0.5H - 1.5V = -Mp equation 10
At E; 135.27 = reactant moment = M + 6H - 9V
135.27 - M - 6H + 9V = 0 equation 11
Solve the simultaneous equation;
Consider equation 8 and equation 11.
Equation 1;
M = 135.37 – 6H – 9V equation 12
Equation 4;
M = 135.27 - 6H + 9V equation 13
Solve equation 12 and equation 13, 135.37 – 6H – 9V = 135.27 - 6H + 9V
V = 0
Substitute equation 12 into equation 9,
94.58 – (135.37 – 6H – 9V) - 2.5H – 7.5V = +Mp
94.58 – 135.27 + 6H + 9V -2.5H -9V = +Mp
3.5H - 40.69 = +Mp equation 14
Substitute equation 12 and equation 14 into equation 10, 4.4 – (135.37 – 6H – 9V) - 0.5H - 1.5V = -3.5H + 40.69
9H = 171.66
H = 19.07 k N
Substitute H into equation 12 and equation 14, to obtain M and Mp. M = 135.37 – 6(19.07)
= 20.95 k N.m Mp = 3.5 (19.07) – 40.69
= 26.06 k N.m
Mp > M (obey fundamental yield condition of plastic theory).
Location for the plastic hinge to occur on the portal frame is at B1, B5, D1 and D5. To check whether the location of the plastic hinge is correct or wrong, calculation of the moment at critical point can be done,
Let take B1,
At B1; Moment in envelope = free bending moment - reactant moment
= 135.27 - M - 3.5H - 9V
= 135.27 – 20.95 – 3.5(19.07)
= 47.57 k N.m (> Mp)
Yield violated the fundamental condition of plastic theory; mechanism ‘B’ is invalid.
Mechanism ‘A’ critical = 40.93 k N.m
Suitable cross-section of properties for the rafter to sustain the plastic moment occurs on collapse mechanism ‘A’.
Refer [1:p32: table 11], [1:p29: figure 5] and [1:p30: 3.5.2];
Refer to appendix, given section properties for the rolled I-Section,
The classification of this rolled I-section is determined by referring [1:p32: table11] and [1:p29:
figure 5]
b = 180/2 = 90 T = 16
d = 413 – (2 x 16) = 381 t = 9.65
Designation Area Depth Web
A thickness Width Thickness I S r I S r
x 10^ -3 d tw bf T x 10^ -6 x 10^ -3 x 10^ -6 x 10^ -3
mm x kN/m m^2 mm mm mm mm m^4 m^3 mm m^4 m^3 mm
W410 x 0.73 9.48 413 9.65 180 16 274 1.33 170 15.5 0.172 40.4
Flange Axis X-X Axis y-y
py = design strength = 275 N.mm-2 refer [1:table 9]
Outstand element of compression flange for rolled section;
625
Web of an I-Section, neutral axis at mid depth;
48
Class 1 Plastic – Cross-sections with plastic hinge rotation capacity.
Since the section is class 1 plastic, it may use for plastic design.
Given formula where,
Where Mp is maximum plastic moment and py is design strength.
From previous calculation,
The rafter section can sustain the plastic moment.