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__The problem:__

Determine the cross-sectional area and the position of the centroid for the sections shown in Figures 1a to 1e. Please consider that the origin of the coordinate system is located at the lower left corner of every section (see each Figure below).

__Tip__: It is given that the total height and the area of the 457 x 152 x 52 UB section are: h = 449.8mm and A = 66.6cm^{2}, respectively.

__The solution:__

a) i) cross-sectional area

A = 240 x 15 + 360 x 10 = 7200mm^{2}

ii) position of centroid

x̄ = 240/2 =120mm

ȳ = [(240 x 15) x 367.5 + (360 x 10) x 180] / 7200 = 273.75mm

b) i) cross-sectional area

A = 240 x 15 + 400 x 8 + 180 x 15 = 9500mm^{2}

ii) position of centroid

x̄ = 240/2 =120mm

ȳ_{ }= [(240 x 15) x 422.5 + (400 x 8) x 215 +(180 x 15) x 7.5] / 9500 = 234.66mm

c) i) cross-sectional area

A = 350 x 10 + 200 x 12 + 300 x 8 + 200 x 12 = 10700mm^{2}

ii) position of centroid

x̄ = 350/2 =175mm

ȳ_{ }= [(350 x 10) x 329 + (200 x 12) x 318 + (300 x 8) x 162 + (200 x 12) x 6] / 10700 = 216.63mm

d) i) cross-sectional area

A = 220 x 8 + 82 x 15 + 82 x 15 + 6660 = 10880mm^{2}

ii) position of centroid

x̄_{ }= 220/2 =110mm

ȳ _{=} [(220 x 8) x (449.8 + 4) + (82 x 15) x (449.8-41) + (82 x 15) x (449.8-41) + 6660 x 449.8/2] / 10880 = 303.51mm

e) i) cross-sectional area

A = 1420 x 20 + 500 x 10 + 500 x 10 + 1220 x 12 = 53040mm^{2}

ii) position of centroid

x̄_{ }= 1420/2 =710mm

ȳ = [(1420 x 20) x 522+ (500 x 10) x 262 + (500 x 10) x 262 + (1220 x 12) x 6] / 53040 = 330.56mm

Structural AnalysisTruss deflection using the unit load methodMethod of sectionsBending stress in a beam elementOverhanging beam: shear force and bending moment calculationCalculation of the second moments of area

Buckling

1 pages

Confinement (structural)

Elasticity and Inelasticity

Fatigue

Moment distribution

Nonlinear analysis

Structural stability

1 pages

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