Strength of materials - Chapter 10: Combined loadings - Nguyễn Sỹ Lâm

Nội dung text: Strength of materials - Chapter 10: Combined loadings - Nguyễn Sỹ Lâm

1. CHAPTER 10: COMBINED LOADINGS 10.1 Unsymmetric Bending 10.2 Eccentric Axial Loadings 10.3 Torsion and Bending 10.4 General Combined Loadings
2. 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING y • 0 Fx  xdA  m dA c or 0 y dA neutral axis passes through centroid y • M M z y  m dA Wish to determine the conditions under c which the neutral axis of a cross section σ I or M m I I moment of inertia of arbitrary shape coincides with the c z axis of the couple as shown. defines stress distribution • The resultant force and moment from the distribution of y • 0 M y z xdA z  m dA elementary forces in the section c must satisfy or 0 yzdA I yz product of inertia Fx 0 M y M z M applied couple couple vector must be directed along a principal centroidal axis
3. 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING M z y M y z  x I z I y • Along the neutral axis, M z y M y z M cos y M sin z  x 0 I z I y I z I y y I tan z tan z I y • Equation of Neutral Axis M y I M y I y z z y z z M z I y M z I y
4. 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING – SUMMARY - -  min h x Compression + + stress y  max b Mx x  - + z h x M M y y - + u Tension stress M y I x y • Equation of Neutral Axis y x M I b (N.A.) x y M M x y (Diagram of Equation : y = ax)  z y x I x I y
5. 10.1 UNSYMMETRIC BENDING EXAMPLE 10.01 • Resolve the couple vector into components and calculate the corresponding maximum stresses. M z 1600lb in cos30 1386lb in M y 1600lb in sin 30 800lb in I 1 1.5in 3.5in 3 5.359in 4 z 12 I 1 3.5in 1.5in 3 0.9844in 4 y 12 The largest tensile stress due to M z occurs along AB M y 1386lb in 1.75in  z 452.6psi 1 4 Iz 5.359in The largest tensile stress due to M z occurs along AD M y z 800lb in 0.75in  609.5psi 2 4 I y 0.9844in • The largest tensile stress due to the combined loading occurs at A. max 1 2 452.6 609.5 max 1062psi
6. 10.2 ECCENTRIC AXIAL LOADINGS ECCENTRIC AXIAL LOADING IN A PLANE OF SYMMETRY • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment  x  x centric  x bending P M y A I • Eccentric loading • Validity requires stresses below proportional F P limit, deformations have negligible effect on geometry, and stresses not evaluated near points M Pd of load application.
7. 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.02 • Normal stress due to a centric load A c2 0.25in 2 0.1963in 2 P 160lb  0 A 0.1963in 2 815psi • Equivalent centric load • Normal stress due to and bending moment bending moment P 160lb I 1 c4 1 0.25 4 4 4 M Pd 160lb 0.6in 3.068 10 3 in 4 104lb in Mc 104lb in 0.25in  m I .068 10 3 in 4 8475psi
8. 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.03 The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the link. SOLUTION: • Determine an equivalent centric load and bending moment. • Superpose the stress due to a centric load and the stress due to bending. • Evaluate the critical loads for the allowable Given: tensile and compressive stresses. 3 2 A 3 10 m • The largest allowable load is the smallest Y 0.038m of the two critical loads. I 868 10 9 m4
9. 10.2 ECCENTRIC AXIAL LOADINGS GENERAL CASE OF ECCENTRIC AXIAL LOADING • Consider a straight member subject to equal and opposite eccentric forces. • The eccentric force is equivalent to the system of a centric force and two couples. P centric force M y Pa M z Pb • By the principle of superposition, the combined stress distribution is P M z y M y z  x A Iz I y • If the neutral axis lies on the section, it may be found from M M P z y y z Iz I y A
10. 10.2 ECCENTRIC AXIAL LOADINGS 2 2 SUMMARY M u M x M y M x M y N z  z y x M M y - -  x y x I x I y A z I I h x x y Mx + + y  min b x Compression stress - + h x Mu y - +  max x y My b  z N z M N + +  u y Tension stress z z A h x M y I N I + + Equation of Neutral Axis (N.A.): y x x z x M x I y A M x y (Diagram of Equation : y = ax +b)
11. 10.3 TORSION AND BENDING CIRCULAR SHAFTS Mz = M0  max  max 2 2  max 4 max   Mu z or 2 2  max 3 max   u  min v 2 2  min 4 max   or 2 2  min 3 max  
12. 10.3 TORSION AND BENDING RECTANGULAR SECTIONS 2 2   S y   4 TB3 1 1 2 2n OR 2 2   S y  TB4  1 3 1 3 3n
13. 10.4 GENERAL COMBINED LOADINGS RECTANGULAR SECTIONS 2 2   S y   4 TB3 1 1 2 2n OR 2 2   S y  TB4  1 3 1 3 3n