Mechanics of Engineering Structures |
From inside the book
Results 1-3 of 35
Page 69
... normal strain original length . Thus , - ΔΙ L of any fiber is the change in ... normal stress and normal strain given by Hooke ; * namely , Sn = Een ( 5 ) where E is Young's modulus of ... STRESS RESULTANTS AND STRESS INTENSITIES 69.
... normal strain original length . Thus , - ΔΙ L of any fiber is the change in ... normal stress and normal strain given by Hooke ; * namely , Sn = Een ( 5 ) where E is Young's modulus of ... STRESS RESULTANTS AND STRESS INTENSITIES 69.
Page 71
... stress resultants N , M ,, and M , are given , the normal stress sn at any point ( y , z ) of the cross section may be found directly from equation 2.2.13 . In a given situation , we begin by finding the purely geometric properties of ...
... stress resultants N , M ,, and M , are given , the normal stress sn at any point ( y , z ) of the cross section may be found directly from equation 2.2.13 . In a given situation , we begin by finding the purely geometric properties of ...
Page 73
... normal resultant force . We have z and yk = ( 3.00 - 1.33 ) = -167 in . - kips , My = normal stress is given by or Sn = + = = = ( 3.00 -0.67 ) = 2.33 in . 1.67 in . , and N -100 kips , Mz - ( 100 ) ( 1.67 ) ( 100 ) ( 2.33 ) = - 233 in ...
... normal resultant force . We have z and yk = ( 3.00 - 1.33 ) = -167 in . - kips , My = normal stress is given by or Sn = + = = = ( 3.00 -0.67 ) = 2.33 in . 1.67 in . , and N -100 kips , Mz - ( 100 ) ( 1.67 ) ( 100 ) ( 2.33 ) = - 233 in ...
Contents
1 Definitions 1 1 2 Equations of equilibrium 5 1 3 Stability | 45 |
STRESS RESULTANTS AND STRESS INTENSITIES | 65 |
plastic membersaxial force 80 2 5 Normal stresses in elastoplastic beams | 98 |
Copyright | |
17 other sections not shown
Other editions - View all
Common terms and phrases
40 kips assumed axes axial force bar forces beam of Figure bending moment bending moment diagram Betti's law centroid coefficients collapse load column conjugate beam considered constant cross section curve damping displacement distribution elastic elastoplastic element equal to zero equilibrium Example F₁ fibers Find the bending Find the reactions fixed end moments forces acting free body diagram ft-kips ft-kips girder given beam hinge Hooke's law in.-kips indeterminate influence diagram joint kips/ft left support length M₁ M₂ mass matrix method method of moment midspan moment of inertia normal force normal stress plane positive Problem R₁ R₂ redundant right support rigid rotation shear force shear stress shown in Figure slope-deflection equations solution span statically determinate station strain energy strain energy stored stress resultants structure shown symmetry truss twisting vector vertical yield ΕΙ ΕΙΟ