Question

Please explain how to get angle for neutral axis and find highest tensile stress not compression

Problem 4- For the following cross section (all dimensions in mm) determine: a) The location and magnitude of the greatest tensile stress on the cross-section; b) Stress at point A; c) The location of the neutral axis, and sketch it on the cross section. Given: M = 100 kN.m 150 mm 150 mm 300 mm 150 mm 150 mm -150 mm

Answer 1

(c) Draw the given cross-sectional area as follows: 300 mm 300 mm NA ....NA 150 mm 150 mm | 150 mm

Calculate the location of the neutral axis as follows: 1 - 4y; – 4,92 4 – A, - (300x600x300) – (150x150x225) (300x600)-(150x150) y=310.71 mm = = 0 mm (due to symmetry) (a) Take the component of the moment about the y and z axis, then find the region of compression and tension as follows:

Compresssion NA...... 'M Compresssion Tension Tension D NA Compression and tension are shown in the figure above. Then maximum tension is at the point C.

Calculate the moment of inertia about the neutral axis as follows: , _ 600x3009150x1503 12 12 1, = 1.3078125x109 mm (300x600 10° +300 x 600x (310.71–300)? 20 1. = { Į 12 150x1509_150x150x(310.71–225)*| 12 1. = 5.2131696x10° mm Components of the moment are as follows: M, = M sin 30° M, = 100xsin 30° M, = 50x10° N·mm

M. = M cos 30° M. = 100x cos 30° M. =86.6x10“ N.mm +- max = Calculate the maximum tensile stress as follows: M,x150 M.310.71 1, 1. 50x106x150 86.6x100 x310.71 °max 1.3078125 x 10° 5.2131696x10° Omax = 10.896 N/mm² Omax = 10.896 MPa

(b) 0A- 1, Calculate the stress at point A as follows: M,,x150 M x310.71 41, 1. 50x10° x150 86.6x10º 310.71 04 – 1.3078125x10° 5.2131696 x 10° 0=-10.896 N/mm² CA=-10.896 MPa Negative sign indicates compressive stress.