Ship Strength Stress & Strain Bending & Shear Moment of Inertia & Section Modulus

STRESS: STRESS: Force/unit area (lbs/sq.in.) Load F Reaction = F Compression (e.g., pier pylon) Cross- section area A Compressive Stress  c = F/A Load F Reaction = F Tensile Stress  t = F/A Cross- section area A Tension (e.g., bridge suspension cable)

STRAIN: STRAIN: deformation in length Load F Reaction = F Compression Strain  negative) Load F Reaction = F Tensile Strain (Strain)  positive LL   STRAIN:  /L (dimensionless)

Hooke’s Law: Hooke’s Law: Stress ∝ Strain STRAIN:  /L STRESS: F/A y = mx  = E  /L Slope, E = Modulus of Elasticity Structural steel Yield point Proportional Limit Ultimate strength Fracture point Maximum working stress Factor of Safety Re: to lower stress:  Reduce LOAD, or  Increase AREA

BENDING HOGGING: SAGGING: NA CT CT Top side in TENSION Bottom side in COMPESSION Bottom side in TENSION Top side in COMPESSION

Since there is both tension & compression in bending … we need to look at Bending Moments: Distributed load: w lbs/ft L wL/2 LOAD diagram xx SHEAR diagram Cumulative Load: Vx = ∫ w dx x 0 x BENDING diagram Cumulative Area: Mx = ∫ V x dx x 0 Note: Bending is maximum at the location where Shear is zero

STRESS in Bending: Remember: Stress = Force/Area (psi) And Stress (in bending) increases as we move away from the neutral axis NA C T Still need to know: Distance from NA (c) Something about how the cross-sectional area is distributed N & S of the neutral axis … Moment of Inertia, I c1c1 c2c2

Moment of Inertia: For rectangular beams, I = bh 3 /12 NA b h Thus for a 2x4: I = bh 3 /12 = 2 x 4 3 /12 = 32/3 = And for a 2 x 12: I = 2 x 12 3 /12 = 288! Note that a 2x12 lying “flat” (b=12; h=2) I = bh 3 /12 = 12 x 2 3 /12 = 8 (less than a 2x4 standing up)

Moment of Inertia: (cont’) In general, the more area farther from the NA, the greater the moment of inertia Distributing the same area at a greater distance from the NA produces a stronger beam of the same weight

STRESS in Bending: (reprise) Maximum Stress (in bending),  max = Mc/ I Where:M = maximim bending moment unit: lb-in (lb-ft x 12) c = maximum distance from NA unit: inches I = moment of inertia of beam cross-section unit: inches 4 Shorthand notation for I /c is Z (section modulus) in units of in 4 /in = in 3 Thus,  max = M / Z (in psi—lb-in/in 3 = lb/in 2 )

STRESS in Bending: (reprise) Consider the sections examined under a bending moment of 1200 ft-lb (14400 in-lb) Upright 2x4 I = in 4 c = 2 in Z = 5.33 in 3  max = 2700 psi Upright 2x12 I = 288 in 4 c = 6 in Z = 48 in 3  max = 300 psi Flat 2x12 I = 8 in 4 c = 1 in Z = 8 in 3  max = 1800 psi If the maximum allowable working stress for this material were 3000 psi, which would you choose?

Moment of Inertia: (cont’) There are tables of I (and Z) for standard steel “sections” I-beam TeeChannel L And I for any shape can (with some effort) be calculated

How about this shape? Ship sections can be reduced to an equivalent (if irregular) I-beam The section modulus, Z, can then be calculated … and the I-beam examined under different bending load conditions Yes, it is complicated

Remember: Remember: Load  Shear  Bending Moment? Distributed load: w lbs/ft L wL/2 LOAD diagram x Cumulative Load: Vx = ∫ w dx x 0 Cumulative Area: x SHEAR diagram x BENDING diagram Mx = ∫ V x dx x 0

How Complicated? How Complicated? Consider … Irregular loading: downward forces of cargo & structure Buoyant (upward) forces proportional to underwater volume of hull … and that’s in still water Shear forces & bending moments vary as wave crests amidships or fore & aft

If fact, stresses are dynamic … rollingpitching yawing racking hogging & sagging

And sometimes, stuff happens …