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Integrated Project: Design of main truss and columns

  1. Description of Facility

The building is composed of a single main floor on top of two staircases with support serving as its main frame. The building is simplified to a rectangular shape with a gable type roof.

The building has a total floor area of 280 m2, which is meant for gallery show. No division will be constructed within the building’s walls to accommodate more people coming into the gallery or exhibit. The main floor is supported by beams on both ends where a staircase is also situated. It is 8 meters above the floor level, to provide an outside parking area at the bottom of the building while saving land space.

Objectives:

The main objective of this project is to design a gallery building that is similar to the previous term project and demonstrate understanding of the concepts learned throughout the course. Among the specific objectives of this project are:

  • Determine the loads produced on the structure based on materials chosen.
  • Calculate the load and design the critical members of the structure such as the slab, beam, column and truss.
  1. Structural Idealization

Figure 1 shows the common load acting on a building. Snow load (S) Wind load (W) and dead load (D) are expected to act on the roofing, while dead load and live load (L) acts on the bottom part of the truss. Wind action is directed towards the roof, while there is also wind load acting internally or from the inside of the building. Dead load and live load act on the first floor of the building.

Figure 1. Load Analysis

 Estimation of Loads

Loads are determined based of the structural idealization presented previously. Load analysis is performed considering the main truss and columns only assuming that other components are affected insignificantly. Among the loads considered are live load, dead load, wind load and snow load. Upon performing some calculations, the combination of each loads was determined and the design of critical members are prepared.

 Roof Truss Load Analysis

Given the design specifications of the roof truss:

Total Span of the Truss = 4@1.75 m = 7 m

Truss Spacing = 10 m

Pitch Angle = 15⁰

The amount of load acting on the roof truss is determined. Among the loads considered are the wind load, dead loads due to weight of materials used and live loads.

  • Wind Load Analysis

The roof is designed to be inaccessible to people except for usual maintenance operations such as painting and minor repairs (Category H). For a sloping roof, and category H (EBCS-1 EN 1995).

qk = 0.25 kN/m­2

Qk = 1.0 kN

External Wind Pressure

The wind pressure acting on the external surface of the structure is computed using the formula:

We = qref  x Ce(Ze) x Cpe

Where, qref = reference wind pressure

qref = ρ/2 Vref2

where, ρ = air density (from site with elevation from sea level > 2000m)

ρ = 0.94 Kg/m3

Vref = reference wind velocity

Vref = Cdir x Ctem x Calt x Vref0 = 1x1x1x22 m/s

Thus,

qref = ρ/2 Vref2

qref = 227.48 N/m2­­

Ce = pressure coefficient that accounts the effect of terrain roughness, topography, and height above the ground given the mean wind speed. It is computed as:

Ce(z) = Cr2(Z)Ct (Z){1+[7KT/Cr(Z)xCt(Z)]}

Where KT is the terrain factor.

For the site location with at least 20% of the surface is covered with buildings with a mean height of 15 m,

KT = 0.24

Cr(Z) is the roughness coefficient, computed as:

Cr(Ze) = KT ln (Ze/Z0) for Zmin < Z

Cr(Ze) = Cr(Zmin) for Z < Zmin

Where:             Z0 is the roughness length (1 m)

Zmin­ is the minimum height (16 m)

Z = height of the building, roof level (17 m)

Computing for Cr(Ze),

Cr(Ze) = KT ln (Ze/Z0) for Zmin < Z

Cr(Ze) = 0.24 ln (17/1)

Cr(Ze) = 0.68

For zones, unaffected by topography, Ct(Z­e) = 1.

Ce(Ze) = 0.682 x 12 x {1+[(7 x 0.24)/(0.68 x 1)]}

Ce(Ze) = 1.602

Cpe is the external pressure coefficient from the Appendix A of EDBC 1 EN 1995.

For 0 degrees, for the wind angle on the roof, the value for Cpe is equal to 0.20.

Solving for the wind pressure on the external surface:

We = qref  x Ce(Ze) x Cpe  = 227.48 x 1.602 x (0.20)

We = 0.073 kPa

For 90 degrees, for the wind angle on the roof, the value for Cpe is equal to -1.96.

Solving for the wind pressure on the external surface:

We = qref  x Ce(Ze) x Cpe  = 227.48 x 1.602 x (-1.96)

We = -0.714 kPa

Internal Wind Pressure

Wi = qref  x Ce(Ze) x Cpi

For closed buildings, Cpi is equal to 0.8 and -0.5, as the extreme values. Thus, the wind load at the roof is given as:

Wi = 0.292 kPa

Wi = – 0.182 kPa

Critical Wind Load

The net negative wind pressures, (external + internal) is equal to -1.006 kPa, and the net positive wind pressure is equal to 0.255 kPa.

  • Load on Purlin

The material of construction selected for the roofing is EGA – 300, as shown in the figure below.

Figure 2. EGA – 300

For a thickness of 0.4 mm, the load per area brought by the roofing material (treated as dead load):

DL1 = 0.0382 kPa

For the purlins, the self-weight is equal to (DL2) 0.0286 kPa.

Consider the live load as: LL = 0.25 kPa

Wind Load acting on the purlin:

WL = 0.255 kPa x cos 15⁰

WL = 0.246 kPa

Total Load computation:

Total Load = 1.3DL + 1.6LL + WL

Total Load = 0.733 kPa

Using the data for the allowable load, and the thickness of the EGA roof cover, the purlin spacing based on table is:

Purlin spacing = 1.75 to 2.0 m

Considering the given design parameters for this project, with roof width span of 7.0 m and the pitch angle of 15⁰, the suggested purlin spacing is 1.81 m.

Figure 3. Roof Truss Design Layout (Drawn to Scale)

  • Load on Sag rods

The maximum axial force on the sag rod is equal to 5/8 Wind Load. Choosing a sag rod with a diameter of 10 mm, the weight of the sag rod will be about 0.05 kPa, which is within the maximum limit for a maximum stress of 138 MPa.

  • Roof Total Load Analysis

Dead Load

Consider the total weight of the roof = roofing + purlins + sag rods

= 0.0286 kPa + 0.0382 kPa + 0.05 kPa

= 0.1168 kPa

Purlin Spacing, Sp = 1.81 m

Truss Spacing, St = 10 m

Thus, the concentrated roof loads on truss joints is given by:

= 0.1168 kPa x Sp x St

= 2.11 kN

Assume that the weight of the truss itself is equal to 1 kN/m, Therefore the concentrated self-weight on truss joints is equal to:

= 1 x 7/4 = 1.8 kN

Dead load on top joints,

Ptop = 2.11 + 1.80 = 3.91 kN

Dead load on bottom joints,

Pbot = 1.80 kN

Total Dead load on end Joints

Pend = 2.11/2 + 1.80 = 2.86 kN

Figure 4. Dead Load Analysis

Wind Load

Based on the previous computation, positive wind acting on the purlin is equal to 0.255 kPa, while the negative wind is -1.006 kPa. Thus, the reaction in the purlin is computed as:

Positive wind, R = 0.255 kPa x Sp x S= 4.6155 kN

Rx = 1.1946 kN

Ry = 4.4582 kN

Figure 5. Wind Load Analysis

Combination Load Acting on the Roof Truss:

Member Dead Load, kN Wind Load, kN Length, m Total Load, kN
Axial Shear Axial Shear Axial Shear
Top Diagonal 1 12.794 -1.8267 -13.952 2.4599 1.82 -1.158 0.6332
2 14.841 0.75905 -18.596 -1.1981 1.82 -3.755 -0.43905
3 13.862 -0.92053 -18.596 1.1981 1.82 -4.734 0.27757
4 10.251 1.8301 -13.952 -2.4599 1.82 -3.701 -0.6298
Bottom Horizontal Cord 5 -7.4643 -1.8112 12.739 2.3668 1.75 5.2747 0.5556
6 -11.181 0.68747 16.491 -1.093 1.75 5.31 -0.40553
7 -11.794 -0.84106 16.491 1.093 1.75 4.697 0.25194
8 -9.3539 1.7164 12.739 -2.3668 1.75 3.3851 -0.6504
Vertical Cord 9 -1.8122 0.29702 4.0037 0 1 2.1915 0.297021
10 2.4987 3.7167 -1.6599 3.752 0.5 0.8388 7.4687
11 2.5575 -2.4397 -1.6599 3.752 0.5 0.8976 1.3123
Diagonal inside 12 -0.86375 0.28404 1.6499 -0.48061 1.82 0.78615 -0.19657
13 -0.50205 -0.40116 1.6499 0.48061 1.82 1.14785 0.07945

  • Other Structural Load Considerations

The following loads were considered for designing the slab and beams, based on the ACI 318-05.

Slab

Dead Load = 0.5 kPa

Live Load = 3.0 kPa

Walls

Dead load = 1.5 kPa

Beam Loads

Live Load = 3 kN/m2 (due to load on Floor Slab)

Dead Loads = 0.5 kN/m2 (due to tiling finish on Floor Slab)

= 24 kN/m3 x 0.25m = 6 kN/m2 (due to self – weight of Floor Slab)

= 1.5 KN/m2 (due to Block work partitions)

Total Beam Load:

Wu = 1.2 DL + 1.6 LL + Anticipated Beam Weight

= 1.2 (1.5 KN/m2 x 4m + 6.5 kN/m2 x 5m) + 1.6 (3 kN/m2 x 5m) + 6 kN/m

= 46.2 KN/m + 24 KN/m + 6 KN/m

= 76.2 KN/m

  1. Design of Critical Members

 Truss Members

The design of truss members is carried out using the following equations.

Members under tension:

The acceptable design condition is .

Members under compression:

Slenderness ratio,

The acceptable design condition is .

 Design of Bottom Cord Members:

For the bottom cord members, the maximum tensile force is 5.2747 kN and the maximum compressive force is 0.6504 kN. Since all the bottom cord members are of equal length (i.e., if effective length factor k is taken = 1, then Le = 1.75 meters), the maximum forces are taken as the design forces.

Design of Top Cord Members:

For the top cord members, the maximum tensile force 4.734 kN and the maximum compressive force is 0.6504 kN. Since all the top cord members are of equal length (i.e., if k = 1, Le = 1.82 meters), the maximum forces are taken as the design forces.

Design of Vertical and Diagonal Members:

For the vertical and diagonal members, the maximum tensile force is 2.1915 kN and the maximum compressive force is 7.4687 kN. The maximum tension acts on each member while the maximum compression acts on one member, which is 0.5 m long. Although the first two described are slightly longer (i.e., 1.82 m long), the maximum compressive forces on them are much smaller (members 11, 12, and 13). Therefore, the effective length of the design members is taken as Le = 0.5 m. The sections are designed for both tension and compression although it is likely to be governed by compression.

  • Floor Slab

Figure 6. Slab Layout

Assumptions

Concrete f’c = 4000 psi

16mm Grade 60 main tension bars

12mm Grade 60 shrinkage bars

Concrete cover = ¾” (0.02m)

Concrete Density = 24 kN/m3

Live Load on Slab = 3 kN/m2

Dead Load on Slab = 0.5 KN/m2 (due to tiles)

Slab Thickness = L/20 = 5/20 = 0.25 m

Factored Uniform Load, Wu = 1.2 DL + 1.6 LL

= 1.2 (0.3m x 0.5 kPa + 0.3m x 0.25m x 24 kPa/m) + 1.6 (0.3m x 3 kPa)

= 2.34 kN/m + 1.44 kN/m

= 3.78 kN/m

Maximum Factored Moment, M­max

d = slab thickness – concrete cover – 0.5 diameter of reinforcement (ACI 318-05)

d = 0.25 – 0.02 – 0.5(0.016)

d = 0.22 m

Tension Bar Spacing

Tension bar spacing requirement for moment = As per one bar / As = 0.40 m apart.

Maximum tension bar spacing between tension bars

Thus, use 16 mm tension bars at 0.3 m c/c

  • Beam Design

Concrete f’c = 4000 PSI

3- 29mm Grade 60 main bars

10mm Grade 60 stirrups

Concrete cover = 1½” (38mm)

Concrete Density = 24 kN/m3

b = 1/2h

Minimum Suggested Thickness,

h = L/18.5 = 0.65 m

b = 0.5 x 0.65m = 0.33m or 0.35m

Mpos = 0.08wL2 = 0.08 x 76.2 kN/m x (6m)2 = 219.46 kNm

Mneg = -0.1wL2 = -0.1 x 76.2 kN/m x (6m)2 = 274.32 kNm

Based on (ACI 318-05 section 7.7.1)

d = h – concrete cover – stirrup bar diameter – ½ (main bar diameter)

= 0.65m + 0.038m + 0.01m + ½ (0.029m) = 0.588m

Based on (ACI 318-05 section 7.6.1)

ρact = As/bd = 0.00962

ρact = 0.00962 > ρmin = 0.0033, and ρact = 0.00962 < ρmax = 0.0214

Thus:

Mu = 0.9 Asfyd{1-[0.59(ρact fy /fc )]}

Mu = 387.65 kNm

Mpos = 219.46 kNm < 387.65 kNm

Mneg = 274.32 kNm < 387.65 kNm

Using the original assumption of the beam:

Beam weight = 6 kN/m (Assumed)

Actual Beam dimensions = 0.35 x 0.65 m

Beam weight = 5.46 kN/m < 6 kN/m

  • Column Design

Assumptions

Concrete f’c = 4000 PSI

16mm Grade 60 main tension bars

10mm Grade 60 shrinkage bars

Concrete cover = 1.5” (0.04m)

Concrete Density = 24 kN/m3

3% Steel Reinforcement

 

Pu = Force from roof on column + Force from beams on column

= 342 kN + 2(1.1 x 76.2 kN/m x 6m)

= 1347.84 kN = 303006.49 lbf

Pn = Pu / ϕ = 466163.83 lbf

ϕ Pn = 0.80 x ϕ [ 0.85 f’c (Ag – Ast) + fy Ast]

0.65 x 466163.83 lbf = 0.80 x 0.65 [0.85 x 4000 (Ag– 0.03Ag) + 60000 (0.03Ag)

303006.49 lbf = 0.52 (3400Ag – 102Ag + 1800Ag)

303006.49 lbf = 2650.96 Ag

Ag = 114.3 in2

Thus, use a 12 x 12 column (144 ft2) = 0.3 m x 0.3 m column

Area of Steel Required = Ag x ρ= 0.3m x 0.3m x 0.01 = 0.0009 m2

Thus, use 6 – 16 mm diameter steel bars (0.0012 m2)

  1. Foundation Design

Square Pad Footing

Assumptions

Concrete f’c = 4000 PSI

25mm (#8) Grade 60 main bars

Concrete cover = 3”

qu =120 kN/m2 = 2506.25 psf

ρCONC. = 150 pcf

γs  = 100 pcf

Roof Slab = 0.3m thick

Roof dead load = 25 kN/m3 X 0.3m = 7.5 kN/m2

Roof live load = 1.5 kN/m2

Total Combined Load = 1.6 LL + 1.2 DL = 1.6 x 1.5 kN/m2 + 1.2 x 7.5 kN/m2 = 11.4 kN/m2

Force on Interior Column = 11.4 kN/m2 x 6m x 5m = 342 kN

ƩFactored Loads acting on footing = Load from roof + Load from beam + Self weight of column

= 342 kN + 2(1.1 x 76.2 kN/m x 6m) + (24 kN/m x 0.3m x 0.3m x 8m)

= 1365.12 KN = 306891.19 lbf

Unfactored Loads on Footing

Roof Dead Load = 7.5 kN/m2 x 5m x 6m = 225kN

Roof Live Load = 1.5 kN/m2 x 5m x 6m = 45kN

Loads on Beam = Floor Live load = 3 KN/m2 x 5m = 15 kN/m

Floor Dead Load = 6.5 kN/m2 x 5m = 32.5 KN/m

Dead Load from Partition = 1.5 kN/m2 x 4m = 6 kN/m

Load from Self – weight of Beam = 6 kN/m

ƩLoads from beams transferred to column = 2[1.1wl]

= 2 [ 1.1 x (15 + 32.5 + 6 + 6) x 6m]

= 785.4 kN

Self – weight of Column = 24 kN/m3 x 0.3m x 0.3m x 8m = 17.28kN

ƩUnfactored Loads = 225kN + 45kN + 785.4kN + 17.28kN = 1072.68kN = 241148.06lbf

Effective Depth, d = overall depth – concrete cover – 0.5(diameter main bar)

= 20” – 3” – 0.5(1”) = 16.5”

Footing weight calculation:

= 150 pcf x 20/12 = 250 psf

Weight of soil:

= 100 pcf x 24/12 = 200 psf

Area requirements:

Appendix 1. Drawings

 

 

Truss Layout

 


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