Pure Bending Normal Stress Formula. Normal stress on cross-sections: o = M, y/I, : Maximum normal stress on cross-sections:Omax = M, Ymax /I, = M. /W,W, = I. /ymax = 2I, /h: bending section modulus: Remarks:. The neutral axis passes through the centroid of the cross-sectionalarea when the material follows Hooke's law and there is no axialforce acting on the cross section.: Our discussion is limited to beams for which the y axis is an axis ofsymmetry. Consequently, the origin of coordinates is the centroid.: Because the y-axis is an axis of symmetry, it follows that the y-axisis a principal axis. So is the z-axis11
• Normal stress on cross-sections: : bending section modulus • Maximum normal stress on cross-sections: . M y I z z max 2 W I y I h z z z max max M y I M W z z z z Pure Bending Normal Stress Formula 11 • The neutral axis passes through the centroid of the cross-sectional area when the material follows Hooke’s law and there is no axial force acting on the cross section. • Our discussion is limited to beams for which the y axis is an axis of symmetry. Consequently, the origin of coordinates is the centroid. • Because the y-axis is an axis of symmetry, it follows that the y-axis is a principal axis. So is the z-axis. • Remarks:
Sample Problem: A strain gauge is placed under cross-section C of a simply supportedbeam shown. Under the concentrated load P, the strain gauge reads = 6× 10-4. Find the magnitude of P for E = 200 GPaPB40 mmC D20 mm-0.4:m-0.5m.m112
• A strain gauge is placed under cross-section C of a simply supported beam shown. Under the concentrated load P, the strain gauge reads ε = 6× 10-4 . Find the magnitude of P for E = 200 GPa. Sample Problem 12 P C D A B 1 m 0.5 m 0.4 m 20 mm40 mm
: Solution:Oc = E= 200×103 ×6×10-4=120 MPaMc =cW, =640 N·mMc =0.5R = 0.5×0.4P =0.2P =640 N·mP=3.2kN13
• Solution: 3 4 200 10 6 10 120 MPa 640 N m 0.5 0.5 0.4 0.2 640 N m 3.2 kN C C C z C A E M W M R P P P 13
Sample Problem. Find the support position (a) at the condition of minimummaximum normal stress" for the overhanging I-beam shown belowunder uniformly distributed load q. Solution:FAFBq1.reaction force at thesupports. Due to symmetryF = Fβ = ql/27x2. Equation of bending1moments:xe[0,a]qx21Mxe[a,l-a2214
• Solution: • Find the support position (a) at the condition of minimum “maximum normal stress” for the overhanging I-beam shown below, under uniformly distributed load q. 2 F F ql A B 2 2 1 0, 2 1 1 , 2 2 M qx x a M qx ql x a x a l a a l a FA q FB x Sample Problem 1. reaction force at the supports. Due to symmetry: 2. Equation of bending moments: 14
3. Diagram of bending moments:. Equating the absolutevalue of the negative andqagqpositive moment2extremities results inminimum bending+moments and henceminimum maximumqlaqlnormal stress.""82qaala21229q?q12qla0=ao22824= a ~ 0.207.115
2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 1 1 0 2 8 2 4 2 0.207 qa l l q ql a qa ql qla a la l a l l l a l ⊕ 2 8 2 ql qla 2 2 qa 2 2 qa • Equating the absolute value of the negative and positive moment extremities results in minimum bending moments and hence minimum “maximum normal stress.” 3. Diagram of bending moments: 15