Question

Problem 2 Problem Statement A W310x33 wide-flange A36 steel beam is used for the application shown below. 10 kN 20 kN 20 KN 10 KN B --Im-eIm-e-Im -Im -Im-- Using beam theory, determine the maximum tensile and compressive bending stress you would expect in this problem and determine where those stresses would occur. Using the theory of elasticity finite element model available here, create a plot which verifies the stress distribution and the values predicted by beam theory at the cross-section at which the maximum tensile and compressive bending stresses occur.

Answer 1

Tensile and compression bending stresses are find as follows

In this I section wide flange is used of W310*33 wide-flange so dimensions of wide flange are taken from American wide flange Beam standards.
→ Given Beam Designation is w310x33 wide- flange, According to standard volue from American coide flange Beam (screenshot of standard value table is given below) -> W310 X33 Designation is close to standard designahon W310x32.7 properties are: 10.8mm width = 102 mm Depth = 312 mm thickness=10.8 mm web thickness = 6.60 mm. Ixx = 64.9x108 mm 4 10.8 mm Iyy = 1.94x100mm 4 6.6omm 302mm 102mm - Moterial for beam is A 36 steeel From google standard properties value are. Sut= 400 Mpa= Syt = 25ompa > Position of Neutral axis. Y=302 = 15mm -.-- symmetrical flange Moment of Inertia Ixx = 64.9x106mm 4 --- from table. Moment on given beom: - 20KN 10KN LOKN 20KN alm xx Im Im e im nk i'm l. & lm t a) Reaction EMA= REX 5000 - 10000 x 4000 - 20,000x3000 - 20000 x 2000-10000 x 1000 Rf = 30,000 N RE+FA = 60OOON, RA= 30,000N

W b) BM EMA = EME=o Emo - 30,000x1000 = 30x106 Hmm. 2 = 3 0 0 0 0 x 2 0 00 - 10000 < | 0 0 0 = 03 10 N O Mo = 30,000 x 2000 - 10,000 X 2000 - 20,000x1000 50x106 Nmm 2 Me - 30,000 X 4000 - 10.000X3000 - 20,000 x 2000 - 20,000x1000 = 30.0106 Nmm. From above maximum BM is at c&D max = 50x106 Nmm.. → Note, above part of Neutral axis is subjected to compression and below cross sectional part of beam from N.A is subjected to tensile, because beam is simply supported. - We know that at outermost surface shear stress is remain zero and at NA. maximum, ie where, is zero shear stress their is pure bending so that, Maximum tensile & compression bending stresses occur at outermost fibre, ie Top of the beam & bottom of beam - Flexure formula. 64.98 106 151 6. = 67 116.3328 N/mm² stress in compression & tension remain same because distance of outermost fibre from N.A. is same because of symmetrical I-beam I I - 66=116.33 23 Ni Ye=151 mm -NA ZG YE/151mm. 6==116 33 23 N/mm² stress Distribution diagram