Jurg Conzett – Traversina Bridge Add moment diagram Loading Moment Jurg Conzett – Traversina Bridge

Riccardo Morandi – Santa Barbara Power Station Add moment diagram Riccardo Morandi – Santa Barbara Power Station

We do all of these crazy shapes and forms to make sure that materials do not reach their capacity, which would cause a failure.

Materials Review

Stress-Strain curve fy This is review of last semester from materials and methods. Modulus of elasticity. = Modulus of Elasticity = E

Stress-Strain curve

Comparison of materials Yield Stress (fy) Material Modulus of Elasticity (E) bending compression tension Steel 29,000 ksi 36 ksi 36 ksi 36 ksi Concrete 3100 ksi 0.5 ksi 3 ksi 0.3 ksi Wood 1700 ksi 1.0 ksi 1.5 ksi 0.7 ksi Glass 10,000 ksi 24 ksi 145 ksi 24 ksi

Comparison of materials Yield Stress (fy) Material Modulus of Elasticity (E) bending compression tension Steel 17 36 24 50 Concrete 2 0.5 2 0.5 Wood 1 1 1 1 Concrete uses reinforcing steel for tension Glass 6 24 97 34

Allowable Stress Design Make sure that materials do not reach their yield stress by providing a factor of safety (FOS).

Factor of Safety Steel: 0.6

Allowable flexural stress = factor of safety x yield stress Steel: 0.6 Allowable flexural stress = factor of safety x yield stress Fb = 0.6 x fy

Allowable flexural stress (Fb)= factor of safety x yield stress Steel: 0.6 Allowable flexural stress (Fb)= factor of safety x yield stress Fb = 0.6 x fy Fb = 0.6 x 36 ksi Fb = 21.6 ksi

Moment = bending stress (fb) x SECTION MODULUS What is section modulus?

Moment = bending stress x SECTION MODULUS What is section modulus? Property of the cross sectional shape. It is what allows us to make the connection between the moment and stress.

Moment = bending stress x SECTION MODULUS What is section modulus? Property of the cross sectional shape. Where do you find it? Look it up in the tables OR calculate it

b h2 Section Modulus = S = 6 b b h h neutral axis

Deflection

the measured amount a member moves depends upon: Deflection the measured amount a member moves depends upon: Rigidity or stiffness of the material Property of the cross sectional shape Length of beam Load on beam

Deflection Rigidity or stiffness of the material Modulus of Elasticity (E) Property of the cross sectional shape Moment of Inertia (I)

Moment of Inertia Property of the cross sectional shape Where do you find it? Look it up in tables OR calculate it

b h3 Moment of inertia = I = 12 b b h h neutral axis

14” 14” 14” Area = 14 in2 I = 485 in4 Area = 14 in2 I = 229 in4 Area = 14 in2 I = 1.2 in4

P L Bigger S, bigger moment capacity

P L P M Rx Ry L Bigger S, bigger moment capacity

P P M Rx Ry P L3 Deflection = 3 E I L L Bigger S, bigger moment capacity P L3 Deflection = 3 E I

w w M Rx Ry w L4 Deflection = 8 E I L L Bigger S, bigger moment capacity w L4 Deflection = 8 E I

w w Rx Ry Ry 5 w L4 Deflection = 384 E I L L Bigger S, bigger moment capacity 5 w L4 Deflection = 384 E I

P P Rx Ry Ry P L3 Deflection = 48 E I L L Bigger S, bigger moment capacity P L3 Deflection = 48 E I

Moment of Inertia Property of the cross sectional shape Where do you find it? Look it up in tables OR calculate it Bigger Moment of Inertia, smaller deflection

STRUCTURAL ANALYSIS : Determining Structural Capacity

From Structural Analysis we have developed an understanding of all : Actions - Applied forces such as dead load, live load, wind load, seismic load. Reactions - Forces generated at the boundary conditions that maintain equilibrium. Internal forces - Axial, shear and moment (P V M) in each structural element.

Determination of Structural Capacity is based on each element’s ability to perform under the applied actions, consequent reactions and internal forces without : Yielding - material deforming plastically (tension and/or stocky compression). Buckling - phenomenon of compression when a slender element loses stability. Deflecting Excessively - elastic defection that may cause damage to attached materials/finishes – bouncy floors.

TENSILE YIELDING and ALLOWABLE STRESS :

stress Plastic Range FY = yield stress Elastic Range deformation

(fA = P/Area of Section) stress FY fA P1 Force on the spring generates an axial stress and elastic deformation deformation

stress FY When Force is removed, the spring elastically returns to its original shape deformation

stress fA FY A Larger Force may generate an axial stress sufficient to cause plastic deformation deformation P2

stress fA FY When the larger force is removed, the plastic deformation remains (permanent offset) deformation

To be certain that the tension stress never reaches the yield stress, Set an ALLOWABLE TENSILE STRESS : FTension = 0.60 FY stress FY fT deformation

If using A36 Steel : FY = 36 ksi Allowable Tensile Stress (FT ): FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

fA stress If using A36 Steel : FY = 36 ksi Allowable Tensile Stress FT: FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area (actual axial stress fA = P/A) Aarea P force

FT stress If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired Areq Pmax

FT stress If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired AreaRequired = Pmax/FT Areq Pmax

21.6 ksi If using A36 Steel : FY = 36 ksi Allowable Tensile Stress : FT = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi P = 5,000 lb or 5 kips fA = P/Area FT = Pmax /AreaRequired AreaRequired = Pmax/FT = 5k / 21.6 ksi = .25 in2 Areq 5k

FLEXURAL YIELDING and ALLOWABLE BENDING STRESS :

stress Plastic Range FY = yield stress Elastic Range deformation fb = M/S S = Section Modulus

P1 stress FY Force on the BEAM generates an bending stress (tension and compression) and elastic deformation (fb = Mmax/S) fb deformation

stress FY When Force is removed, the BEAM elastically returns to its original shape deformation

P2 stress fb FY A Larger Force may generate an bending stress sufficient to cause plastic deformation deformation

stress fb FY When the larger force is removed, the plastic deformation remains. deformation

To be certain that the bending stress never reaches the yield stress, Set an ALLOWABLE BENDING STRESS : Fbending = 0.60 FY stress FY Fb deformation

If using A36 Steel : FY = 36 ksi Allowable Bending Stress (Fb) : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi

If using A36 Steel : FY = 36 ksi Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in

If using A36 Steel : FY = 36 ksi Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S (actual bending stress fb = M/S)

If using A36 Steel : FY = 36 ksi Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired

If using A36 Steel : FY = 36 ksi Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired SRequired = Mmax / Fb

If using A36 Steel : FY = 36 ksi Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in fb = M/S Fb = Mmax / SRequired SRequired = Mmax / Fb = 3792 k-in / 21.6 ksi = 176 in3

If using A36 Steel : FY = 36 ksi Mmax = 316 k-ft Mmax = 316 k-ft (12 in / ft) = 3792 k-in Allowable Bending Stress : Fb = 0.60 FY = 0.60 (36 ksi) = 21.6 ksi fb = M/S Fb = Mmax / SRequired SRequired = Mmax/Fb = 3792 k-in / 21.6 ksi = 176 in3 Use W24x76 : SX-X = 176in3

BUCKLING and ALLOWABLE COMPRESSION STRESS :

PC Buckling is a compressive phenomenon that depends on : ‘unbraced length’ of the compression element: (k x l) shape of the section: (radius of gyration ryy) Allowable Material compressive stress: (Fc)

l ‘unbraced length’ (kxl) depends upon the boundary conditions of an element

The radius of gyration (ryy) is a property of a members cross section. It measures the distance from the neutral axis a member’s area may be considered to be acting I = Ar2 r = (I/A)0.5 (I = moment of inertia)

Allowable Compression Stress Fc depends on ‘kl/r’ l = 15 ft = 180 in assume ryy = 3.0 in.** kl/r = 60 Fc = 17.4 ksi ** we must always come back and verify this assumption **

If using A36 Steel : FY = 36 ksi Pmax = 240 kips (typ. read this from your P diagram] Allowable Compression Stress (Fc) : FC = 17.4 ksi fC = P/Area FC = Pmax/AreaRequired AreaRequired = Pmax/FC = 240k / 17.4 ksi = 13.8 in2

W12x65 A = 19.1 in2 ryy = 3.02 in

If using A36 Steel : FY = 36 ksi Pmax = 240 kips Allowable Compression Stress : FC = 17.4 ksi fC = P/Area FC = Pmax/AreaRequired AreaRequired = Pmax/FC = 240k / 17.4 ksi = 13.8 in2 Use W12x65 Area = 19.1 in2 check actual stress: fC = P/A fC = 240 kips / 19.1 in2 = 12.6 ksi OK!

BUCKLING and ALLOWABLE COMPRESSION STRESS :

Allowable Compression Stress depends on slenderness ratio = kl/r

Slenderness Ratio = kl/r k = coefficient which accounts for buckling shape for our project gravity columns, k=1.0 for moment frames see deformed shape

Slenderness Ratio = kl/r l = unbraced length (inches)

Slenderness Ratio = kl/r r = radius of gyration (inches) typical use ry (weak direction) rx > ry

Allowable Compression Stress (Fc) slenderness ratio = kl/r assume r = 2 in., k = 1.0 lcolumn = 180 in kl/r = 90 use Table C-36 to determine Fc = 14.2 ksi

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Use W12x65 Area = 19.1 in2 fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Use W12x65 Area = 19.1 in2 fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi check ry for W12x65 and verify FC

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Use W12x65 Area = 19.1 in2 fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi check ry for W12x65 and verify FC ry (W12x65) = 3.02 kl/r = (1.0)(180 in)/3.02in = 60

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Use W12x65 Area = 19.1 in2 fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi check ry for W12x65 and verify FC ry (W12x65) = 3.02 kl/r = (1.0)(180 in)/3.02in = 60, using Table C-36 Fc = 17.4 ksi

AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Pmax Column 2 = 238 kips (assume columns are continuous from foundation to roof, total length 30 feet) FC = 14.2 ksi AreaRequired = Pmax/FC = 238k / 14.2 ksi = 16.8 in2 Use W12x65 Area = 19.1 in2 fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi check ry for W12x65 and verify FC ry (W12x65) = 3.02 kl/r = (1.0)(180 in)/3.02in = 60, using Table C-36 Fc = 17.4 ksi > fc , therefore ok

Column 2, Efficiency Check: W12x65 fC = 12.5 ksi (actual stress fc = P/A) FC = 17.4 ksi [allowable stress from chart C-36] fC/FC < 1.0

Column 2, Efficiency Check: W12x65 fC = 12.5 ksi FC = 17.4 ksi fC/FC = 12.5 ksi/17.4 ksi = 0.72 < 1.0

Column 2, Efficiency Check: W12x65 fC = 12.5 ksi FC = 17.4 ksi fC/FC = 12.5 ksi/17.4 ksi = 0.72 < 1.0 (72% of capacity is used)

ALLOWABLE BENDING + COMPRESSION:

80 kips 40 kips 40 kips 200 kips 200 kips

Axial Diagram Moment Diagram 900 k-ft 80 kips 80 kips - compression + tension - compression fb=M/S fa=P/Area

Combined Stress (fa+fb) = Axial Stress (fa) Bending Stress (fb) Combined Stress (fa+fb) + =

To be certain that the combined stress (bending + axial) never reaches the yield stress, use the INTERACTION EQUATION fb/Fb + fa/Fa < 1.0 + Bending Stress (fb) Axial Stress (fa) +

Mmax = 900 k-ft Pmax = 200 kips Assume 50% capacity of bending (fb)

Mmax = 900 k-ft Pmax = 200 kips Assume 50% capacity of bending (fb) 50% Fb = (0.5)(21.6 ksi) = 10.8 ksi SREQ = Mmax/50%Fb = 900k-ft (12in/1ft) / 10.8ksi SREQ = 1000in3

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi Fb = 21.6 ksi fb/Fb = 11.3ksi/21.6ksi = 0.52

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi Fb = 21.6 ksi fb/Fb = 11.3ksi/21.6ksi = 0.52 fc = Pmax/Area = 200 kips/76.5 in2 = 2.6 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.78 in = 95

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi Fb = 21.6 ksi fb/Fb = 11.3ksi/21.6ksi = 0.52 fc = Pmax/Area = 200 kips/76.5 in2 = 2.6 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.78 in = 95, using Table C-36 Fc = 13.6 ksi fc/Fc = 2.6ksi/13.6ksi = 0.19

TRY W36x260, Sx-x = 953 in3 A = 76.5 in2 ry-y = 3.78 in fb = Mmax/S = 900 k-ft (12in/1ft) / 953in3 = 11.3 ksi Fb = 21.6 ksi fb/Fb = 11.3ksi/21.6ksi = 0.52 fc = Pmax/Area = 200 kips/76.5 in2 = 2.6 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.78 in = 95, using Table C-36 Fc = 13.6 ksi fc/Fc = 2.6ksi/13.6ksi = 0.19 fb/Fb + fc/Fc = 0.71 < 1.0, therefore ok

Assume 70% capacity of bending (fb)

Assume 70% capacity of bending (fb) 70% Fb = (0.7)(21.6 ksi) = 15.1 ksi SREQ = Mmax/70%Fb = 900 k-ft (12in/1ft) / 15.1 ksi SREQ = 720 in3

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi Fb = 21.6 ksi fb/Fb = 14.2 ksi/21.6 ksi = 0.73

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi Fb = 21.6 ksi fb/Fb = 14.2 ksi/21.6 ksi = 0.73 fc = Pmax/Area = 200 kips/59.1 in2 = 3.4 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.56 in = 101

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi Fb = 21.6 ksi fb/Fb = 14.2 ksi/21.6 ksi = 0.73 fc = Pmax/Area = 200 kips/59.1 in2 = 3.4 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.56 in = 101, using Table C-36 Fc = 12.85 ksi fc/Fc = 3.4 ksi/12.85 ksi = 0.26

TRY W33x201, Sx-x = 684 in3 A = 59.1 in2 ry-y = 3.56 in fb = Mmax/S = 900 k-ft (12 in/1 ft) / 684 in3 = 14.2 ksi Fb = 21.6 ksi fb/Fb = 14.2 ksi/21.6 ksi = 0.73 fc = Pmax/Area = 200 kips/59.1 in2 = 3.4 ksi Slenderness ratio: k=2.0 l = 180 in kl/r = (2.0)(180 in)/3.56 in = 101, using Table C-36 Fc = 12.85 ksi fc/Fc = 3.4 ksi/12.85 ksi = 0.26 fb/Fb + fc/Fc = 0.99 < 1.0, therefore ok