1. Jurg Conzett – Traversina Bridge
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Jurg Conzett – Traversina Bridge

3. Riccardo Morandi – Santa Barbara Power Station
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Riccardo Morandi – Santa Barbara Power Station

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

5. Materials Review

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

7. Stress-Strain curve

8. 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

9. 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

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

11. Factor of Safety
Steel: 0.6

12. 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

13. 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

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

15. 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.

16. 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

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

18. Deflection

19. 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

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

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

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

23. 14”
14”
14”
Area = 14 in2
I = 485 in4
Area = 14 in2
I = 229 in4
Area = 14 in2
I = 1.2 in4

24. P L
Bigger S, bigger moment capacity

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

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

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

28. 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

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

30. 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

31. STRUCTURAL ANALYSIS :
Determining Structural Capacity

32. 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.

33. 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.

34. TENSILE YIELDING and ALLOWABLE STRESS :

35. stress
Plastic Range
FY = yield stress
Elastic Range
deformation

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

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

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

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

40. 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

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

42. 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

43. 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

44. 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

45. 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

46. FLEXURAL YIELDING and ALLOWABLE BENDING STRESS :

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

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

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

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

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

53. 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

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

55. 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

56. 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)

57. 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

58. 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

59. 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

64. 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

65. BUCKLING and ALLOWABLE COMPRESSION STRESS :

66. 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)

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

68. 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)

69. 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 **

71. 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

72. W12x65 A = 19.1 in2
ryy = 3.02 in

75. 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 = in2
Use W12x65 Area = 19.1 in2
check actual stress: fC = P/A
fC = 240 kips / 19.1 in2 = 12.6 ksi OK!

76. BUCKLING and ALLOWABLE COMPRESSION STRESS :

77. Allowable Compression Stress depends on slenderness ratio = kl/r

78. 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

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

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

81. 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

82. 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 = in2

84. 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 = in2
Use W12x65 Area = 19.1 in2
fC = Pmax/Area = 238 k / 19.1 in2 = 12.5 ksi

85. 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 = 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

86. 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 = 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

87. 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 = 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

88. 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 = 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

89. 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

90. 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

91. 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)

92. ALLOWABLE BENDING + COMPRESSION:

94. 80 kips
40 kips
40 kips
200 kips
200 kips

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

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

102. 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) +

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

104. 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

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

110. 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

111. 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

112. 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= l = 180 in
kl/r = (2.0)(180 in)/3.78 in = 95

115. 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= 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

116. 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= 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

117. Assume 70% capacity of bending (fb)

118. 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

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

121. 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

122. 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

123. 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= l = 180 in
kl/r = (2.0)(180 in)/3.56 in = 101

125. 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= l = 180 in
kl/r = (2.0)(180 in)/3.56 in = 101, using Table C-36
Fc = ksi
fc/Fc = 3.4 ksi/12.85 ksi = 0.26

126. 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= l = 180 in
kl/r = (2.0)(180 in)/3.56 in = 101, using Table C-36
Fc = ksi
fc/Fc = 3.4 ksi/12.85 ksi = 0.26
fb/Fb + fc/Fc = 0.99 < 1.0, therefore ok