Could anyone please help me out with this problem? Problem 3. A beam with profil
ID: 1713261 • Letter: C
Question
Could anyone please help me out with this problem?
Problem 3. A beam with profile L 203x1 52x25.4 is fixed at the far end of L-shaped section in figure l·The beam is 3.0 meters in length and supports a load of 15 kN at its free end. The load is applied on the section centroid with a direction -0.6j+0.8k E- 200GPa. Determine: the neutral axis direction, . ANS: = degrees (w/ respect to positive z-axis). b) maximum tensile stress . and compressive stress max on the beam. ANS: .ax- MPa max A-8390 mm, 1,-33.7x106 m ; 1,-16. 1x 106 min. ; Izv = 13.4083 × 106 mm: t = 25.4 mm (thickness); c 419 mm ; Section properties: d 67.3 mm Figure 1. L-profileExplanation / Answer
Fz = 0.8 x 15 = 12 kN
Fy = -0.6 x 15 = -9 kN
My = 12 x 3 = 36 kNm
Mz = 9 x 3 = 27 kNm
Neutral axis will pass through the centroid of the L section.
Given section properties,
A = 8390 mm2
Iz = 33.7 x 106 mm4
Iy = 16.1 x 106 mm4
Izy = 13.4083 x 106 mm4
Thickness = 25.4 mm
Distance of centroid from left end = 41.9 mm
Distance of centroid from bottom = 67.3 mm
a) Angle of neutral axis with respect to positive z-axis = tan-1(- (My Izz / Mz Iyy)
= tan-1( -(36 x 33.7 x 106) / (27 x 13.4083 x 106)
= -73.3850
b) The region above the neutral axis will be in tension and region below the neutral axis will be in compression.
Maximum tensile stress will be at (z,y) : (41.9mm, 137.5mm)
Maximum value of tensile stress = Mz x 0.1375 / Izz - My x 0.0419 / Iyy
= (27 x 1000 x 0.1375 / 33.7 x 106 x 10-12 ) - (36 x 1000 x 0.0419 / 16.1 x 106 x 10-12 )
= 16.474 MPa
Maximum compressive stress will be at (z,y) : (41.9mm, -67.3mm)
Maximum value of tensile stress = Mz x 0.0673 / Izz + My x 0.0419 / Iyy
= (27 x 1000 x 0.0673 / 33.7 x 106 x 10-12 ) + (36 x 1000 x 0.0419 / 16.1 x 106 x 10-12 )
= 147.61 MPa