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Lecture 9 shear force and bending moment in beams
 

Lecture 9 shear force and bending moment in beams

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    Lecture 9 shear force and bending moment in beams Lecture 9 shear force and bending moment in beams Presentation Transcript

    • Unit 2- Stresses in BeamsTopics Covered  Lecture -1 – Review of shear force and bending moment diagram  Lecture -2 – Bending stresses in beams  Lecture -3 – Shear stresses in beams  Lecture -4- Deflection in beams  Lecture -5 – Torsion in solid and hollow shafts.
    • Why study stresses in beams
    • What are beams  A structural member which is long when compared with its lateral dimensions, subjected to transverse forces so applied as to induce bending of the member in an axial plane, is called a beam.
    • Objective  When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam.  To determine these stresses and strains, the internal forces and internal couples that act on the cross sections of the beam must be found.
    • Beam Types  Types of beams- depending on how they are supported.  
    • Load Types on Beams  Types of loads on beam   Concentrated or point load   Uniformly distributed load   Uniformly varying load   Concentrated Moment
    • Sign Convention for forces and moments P M M Q Q“Happy” Beam is +VE +VE (POSITIVE)
    • Sign Convention for forces and moments M M P Q Q“Sad” Beam is -VE -VE (POSITIVE)
    • Sign Convention for forces and moments  Positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts, and an internal moment that causes compression, or pushing on the upper arm of the member.  Loads that are opposite to these are considered negative.
    • SHEAR FORCES AND BENDING MOMENTS  The resultant of the stresses must be such as to maintain the equilibrium of the free body.  The resultant of the stresses acting on the cross section can be reduced to a shear force and a bending moment.  The stress resultants in statically determinate beams can be calculated from equations of equilibrium.
    • Shear Force and Bending Moment in a Beam
    • Shear Force and Bending Moment  Shear Force: is the algebraic sum of the vertical forces acting to the left or right of the cut section  Bending Moment: is the algebraic sum of the moment of the forces to the left or to the right of the section taken about the section
    • SF and BM formulasCantilever with point load W x Fx= Shear force at X A Mx= Bending Moment at X B LW SF Fx=+W BM Mx=-WxWxL at x=0=> Mx=0 at x=L=> Mx=-WL
    • SF and BM formulas Cantilever with uniform distributed load w Per unit length x Fx= Shear force at X Mx= Bending Moment at X A B L Fx=+wx BM at x=0 Fx=0 wL at x=L Fx=wL Mx=-(total load on right portion)* Distance of C.G of right portionwL2/2 Mx=-(wx).x/2=-wx2/2 at x=0=> Mx=0 at x=L=> Mx=-wl2/2
    • SF and BM formulas Cantilever with gradually varying load Fx= Shear force at X w wx/L Mx= Bending Moment at X A B x wx 2 L Fx = 2L at x=0 Fx=0 at x=L Fx=wL/2wL/2 € Parabola C Mx=-(total load for length x)* Distance of load from X wx 3 Cubic Mx = 6L at x=0=> Mx=0 at x=L=> Mx=-wl2/6 €
    • SF and BM formulas Simply supported with point load W x Fx= Shear force at X C Mx= Bending Moment at X A B W W RA = 2 L RB = 2 Fx=+W/2 (SF between A & C)€ € Resultant force on the left portion W/2 SF Baseline B ⎛ W ⎞ W Constant force A C ⎜ − W ⎟ = − ⎝ 2 ⎠ 2 between B to C SF W/2 € BM WL/4 B C B
    • SF and BM formulas Simply supported with point load W x Fx= Shear force at X C Mx= Bending Moment at X A B W W RA = L RB = for section 2 2 between A & C W€ € M x = RA x = x W/2 SF 2 Baseline B at A x=0=> MA=0 W L A C at C x=L/2=> MC = × SF W/2 2 2 € for section between C & B BM €W × ⎛ x − L ⎞ = W x − Wx + W L M x = RA x − ⎜ ⎟ WL/4 ⎝ 2 ⎠ 2 2 W L B C B = − x +W 2 2 WL W MB = − L =0 2 2 €
    • SF and BM formulas Simply supported with uniform distributed load w Per unit length x Fx= Shear force at X Mx= Bending Moment at X A B RA C RB L wL RA = RB = BM 2 wL/2 wL Fx = RA − w.x = − w.x A C B 2 wL w.0 wL x = 0 ⇒ FA = − = wL/2 2 2 2 L wL wL x = ⇒ FC = − =0 wL2 2 2 2wL2/2 8 wL wL x = L ⇒ FB = − wL = 2 2 € €
    • SF and BM formulas Simply supported with uniform distributed load w Per unit length x Fx= Shear force at X Mx= Bending Moment at X A B C x RA RB M x = RA x − w.x L 2 wL w.x 2 BM = x− wL/2 2 2 A C B wL w.0 x = 0 ⇒ MA = .0 − =0 2 2 wL/2 L 2 wL L w ⎛ L ⎞ wL2 wL2 wL2 x = ⇒ Mc = . − ⎜ ⎟ = − = 2 2 2 2 ⎝ 2 ⎠ 4 8 8wL2/2 wL2 wL w 8 x = L ⇒ MB = L − L2 = 0 2 2 € €
    • SF and BM diagram P Constant Linear Load Constant Linear ParabolicShear Linear Parabolic CubicMoment
    • SF and BM diagram 0 0 Constant Load M Constant Constant LinearShear Linear Linear ParabolicMoment
    • Relation between load, shear force and bending moment 1 2 x w/m run dFA B = −w dx 1 C 2 L The rate of change of shear force is equal to the rate of loading M M+dM € F F+dF dM dx =F dx The rate of change of bending moment is equal to the shear force at the section €