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Sahithyan's S1
Sahithyan's S1 — Mechanics

Beams

Beam

  • long (L>>B,DL >> B,D)
  • axis of the beam is straight
  • constant cross-section throughout its length

Classified by supporting conditions

First 3 are the mandatory ones in s1.

u.d.l means uniformly distributed load.

TypeImageNotes
Simply supported beamSimply supported beamAt ends: Max SF and BM = 0
Cantilevered beamCantilevered beamAt fixed end: Max SF and BM
Overhanging beamOverhanging beam
Propped cantilevered beamPropped cantilevered beam
Continuous beamContinuous beam
Fixed beamFixed beamAt ends: Max SF and BM

At a section

At a section

  • PP - Normal force / Axial force
  • Sy,SyS_y, S_y - Shear forces along yy and zz axis
  • MxM_x - Twisting moment / Torque
  • My,MzM_y, M_z - Bending moments about yy and zz axis

Degress of freedom

A plane member have 3 degrees of freedom. Any of the 3 can be restrained.

  • Displacement in xx-direction
  • Displacement in yy-direction
  • Rotation about zz-direction

SFD & BMD

Sign convention

  • Bending moment
    • Hogging (curves upwards in the middle) is (+) ve
    • Sagging (curves downwards in the middle) is (-) ve
  • Shear force
    • Clockwise shear is (+) ve.
    • Counterclockwise shear is (-) ve.

Pure bending

A member is in pure bending when shear force is 00 and bending moment is a constant.

Point of Contraflexure

The point about which bending moment is 00, and changes its sign through the point.

Distributed load

Suppose a beam is under a distributed load of w=f(x)w=f(x) per unit length.

dSdx=w\frac{\text{d}S}{\text{d}x}=-w dMdx=S        d2Mdx2=w\frac{\text{d}M}{\text{d}x}=-S \;\; \land \;\; \frac{\text{d}^2M}{\text{d}x^2}=w

Deflection of a beam

Suppose a simply supported beam is applied a load of WW at mid-span.

Smax=WL4I        Dmax=WL348EIS_{\text{max}} = \frac{WL}{4I} \;\; \land \;\; D_{\text{max}} = \frac{WL^3}{48EI}

Here:

  • SmaxS_\text{max} - Maximum stress
  • DmaxD_\text{max} - Deflection
  • WW - Load
  • LL - Span length
  • EE - Young’s modulus
  • II - Second moment of cross-sectional area

Principle of Superposition

A beam with multiple loads can be split into multiple systems each with a single load. Reason for doing so is the ease of calculations.

Values will be the sum of each system’s corresponding value.