Question

The cantilever T-beam shown below is 2 meters long and cemented to a wall. The origin of the Cartesian axis shown is at point A on the wall. The cross-section dimensions are shown at the front of the beam.

Find the following: 1) Location of the beam's cross section centroid w.r.t. the origin A 2) Area moment of inertia of the beam's cross section about the Z-axis shown above 3) All support reaction components of the fixed reaction at the wall holding the beam in static equilibrium

Answer 1

Part 1 Draw the cross section of the beam as follows: 25 mm 37.5 mm 37.5 mm 100 mm 25 mm =- Calculate the location of centroid with respect to A as follows: x = 1000 mm (As length of beam is 2 m) - (25x100)*(-12.5)+(10025)+(-75) (25x100)+(10025) =-43.75 mm - (25x100)x(50)+(100 x 25)x(50) (25x100)+(100 x 25) = = 50 mm Part 2 Calculate the area moment of inertia of upper rectangular section about z-axis as follows: Apply the Parallel axis theorem, 1 = 6* + Ax? 1 - 100x 25° +(100 x 25)x12.52 12 1, = 520833.33 mm Calculate the area moment of inertia of lower rectangular section about z-axis as follows: Apply the Parallel axis theorem, le = bal} 12= 12 - + Ar2 _ 25x100 1= _+(100x 25)x 752 1, = 16145833.33 mm

Calculate the area moment of inertia of rectangular cross-section about z-axis as follows: I=1, +1, I = 520833.33+16145833.33 I=1.67x10 mm Part 3 Calculate the reaction forces as follows: F =0 Ri+R, j+R_k+i-4j-8k+2j – 2k – 2j + 2k = 0 (R+1)i+(R. -4+2 – 2)j +(R. -8-2+2)k = 0) (R +1)i+(R, -4)j +(R. -8)k = 0 R =-1 kN R = 4 kN R = 8 kN Reaction Force, R=(-li+4j+8k) kN Write the position vector of the forces as follows: = (2i – 0.125j +0.0375k) m r2 = (21+0j +Ok) m r2 = (2i – 0.025j +0.1k) m Calculate the moment of forces about A as follows: M = {(rxF) MA = (r; x F )+(v *F)+(rz xF;) i j k | i j k li j k| M= |2 -0.125 0.0375 +2 0 0 +2 -0.025 0.1 1 -4 -8 0 2 -2 0 -2 2| M. = (1.15i+16.0375j – 7.875k)+(0i+4j+ 4k)+(0.150i – 4 ; – 4k) MA = (1.3i+16.0375 j – 7.875k) kN·m The reaction moments should balance the moment about A, then: Reaction Moment, M=(-1.31–16.0375j+7.875k) kN.m.