CE 317: Design of Concrete Structures II 3.00 credit, 3hrs/week

Dr. Tahsin Reza Hossain Professor, Room No-536

Email: tahsin@ce.buet.ac.bd

Web: trhossain.info

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Syllabus • Design of column supported slabs • Introduction to floor systems • Design of columns under uniaxial and biaxial

loading, Introduction to slender column • Structural design of footings, pile caps • Seismic detailing • Shear wall • Structural forms • Introduction of prestressed concrete, Analysis

and preliminary design of prestressed beam section

New

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Class routine

• A section- Tuesday (3)

• B section- Tuesday (1)

• C section- Sunday (2)

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Books

• Design of Concrete Structures

– Nilson, Darwin, Dolan 15th Ed

• Structural Concrete- Theory and Design

– Hassoun, Al-Manaseer 4th Ed

• Reinforced Concrete- Mechanics & Design

– Wight & McGregor 5th Ed

Many more……..

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COLUMNS

Short columns Chapter 8

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• Columns are members that carry loads chiefly in compression

• Usually also carry bending moment. Tensile stresses may be produced over a part of the cross section

• Columns are generally referred to as compression member. Compression dominates behavior

Introduction: Axial Compression

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• Generally vertical members

• Also arches, truss members,

• Column= compression member

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Gravity load Lateral load

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Three types of column

1. Members reinforced with longitudinal bars and lateral ties

2. Members reinforced with longitudinal bars and continuous spirals

3. Composite compression members

• 1 and 2 are common

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Types- 1. Tied Column

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2. Spirally reinforced column

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3. Composite columns

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Main reinforcement • Main reinforcement is longitudinal, parallel to

load, square, rectangular or circular arrangement

• Reinf ratio- 0.01 to 0.08 (1 to 8 %)

• Lower limit to ensure resistance to bending not accounted for and reduce effects of creep and shrinkage

• Higher limit – not economical and difficult due to congestion

• Normally 2-3 %, preferably not exceed 4%.

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Main reinforcement

• Higher size (No 5 and more) used

• Even No 14 and No 18, even bundled

• Minimum 4 longitudinal bars when rectangular or circular ties

• Minimum 6 bars when continuous spiral is used

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Short column and Slender column

• Secondary effects - buckling

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Reinforcements –usual sizes

• Slab- No 3, 4, 5 (10mm, 12mm, 16mm)

• Beam- No 5,6, 7, 8 (16 20 22 25mm)

• Stirrup/tie- No, 3 4 (10 12mm)

• Column –No 5, 6 7 8 9 10 11 14 18 (16 20 22 25 28 32 ….)

• Mat- No 4,5,6,8 (12 16 20 25 mm)

• Smaller sizes preferred as long as there is no congestion

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NominalStrength

Elastic range

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Design strength

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8.2 Lateral ties and spiral

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• Large P, small M – longitudinal bars uniform (a to d)

• Large M – bars at maximum distance from axis of bending

• Bundled bar- 2,3,4

• Bundled bars act as a unit

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Purpose of Lateral ties and spiral

• Hold longitudinal bar in position while concrete is placed

• Prevent longitudinal bars from buckling

• Shear steel

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ACI provisions for ties

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ACI provisions for spirals

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• 𝑃𝑢 ≤ 𝛼∅𝑃𝑛 Demand< Capacity

• 𝑃𝑛 = 0.85𝑓𝑐′ 𝐴𝑔 − 𝐴𝑠𝑡 +𝐴𝑠𝑡𝑓𝑦

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Axially loaded column

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• 𝛼 = 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡𝑎𝑙 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦

=0.80 for tied column

=0.85 for spiral column

• ∅ = 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

=0.65 for tied column

=0.75 for spiral column

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Purpose of Lateral ties and spiral

• Hold longitudinal bar in position while concrete is placed

• Prevent longitudinal bars from buckling

• Shear steel

• Spiral gives confinement

• All bars shall be enclosed by lateral tie

• #3 tie bar for up to longitudinal bar #10

• #4 tie bar for longitudinal bar #11 and above

• #4 tie bar for bundled longitudinal bar

• Spacing of tie shall be less than

– 16dlong, 48dtie, Least dim

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ACI provisions for ties

• Every corner and alternate bar shall have lateral support provided by corner of tie (<135)

• Clear spacing between supported bar and unsupported bar shall be less than 6in.

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ACI provisions for ties

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Problem 1

• Determine nominal and design axial compression capacity of a column 12˝X12˝reinforced with 4 No 9 bars. Also check the ties No. 3 @ 12in c/c. Given: fc’= 4ksi and fy=60 ksi.

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Problem 2

• Design a tied column for

– PDL= 300 kip and PLL= 200 kip

– fc’= 3 ksi and fy=60 ksi

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Problem 3

• Design a tied column with section 10˝x10˝ for

– PDL= 60 kip and PLL= 30 kip

– fc’= 3 ksi and fy=60 ksi

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Problem 4. • Determine nominal and design axial

compression capacity of a circular pile 40inch diameter reinforced with 40 No. 10 (32mm) bars. Also design the ties. Given: fc’= 3ksi and fy=60 ksi. Note bundle bars. Alternate bars?

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Spirally reinforced column

• If filled with sand, load carrying capacity is due to hoop tension only

• If filled with concrete, it can carry without confinement

• Closely spaced spiral behaves like this drum, ie. it counteracts expansion of concrete

• Capacity of the core greatly increased

• Failure occurs when spiral yields and confinement greatly reduces

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Tied and spiral behavior • A tied column fails when load reaches Pn

• Concrete fails in crushing and shearing in inclined plane

• Longitudinal steel buckles between ties

• A spirally rein column, the outer shell spalls off at the same load Pn

• Depending on the amount of spiral, the failure load can be much higher than Pn

• Axial strain will be much higher- higher toughness

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Effect of confinement

• Increases strength

• Increases failure strain

• Ductility and toughness 49

Failure of tied and spiral columns

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ACI spiral (also BNBC 2020)

• Excess capacity is wasted

• ACI provides minimum amount of spiral that contributes to capacity slightly higher than that concrete shell

• This is ACI spiral-

• It hardly increases Pn

• It prevents instantaneous crushing, buckling of long steel, produces a gradual and ductile failure- a tougher column

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ACI spiral derivation

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Problem 5

• Design a circular spiral column to support

– PDL= 475 kip and PLL= 250 kip

– fc’= 4 ksi and fy=60 ksi

– Steel ratio of about 3%.

– Also design necessary spiral

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Compression plus Bending

Rectangular columns

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• Columns chiefly carries compression

• But bending is almost always present

– By continuity, part of monolithic frames

– By transverse loads, wind, earthquake

– By eccentric on bracket

– By inevitable construction imperfection

– Arch axis not coincides with pressure line

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Statically equivalent • Two loads are statically equivalent

• Columns can be classified by e

• Small e – Comp over entire section

– If overloaded, fails by crushing of concrete and yielding of steel in comp in overloaded side

• Large e – Some part in tension

– If overloaded, may fail in yielding of steel in tension at the farthest side from load

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8.4 Strain Compatibility Analysis and Interaction diagram

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• For large reinforcement ratio

• For large e

– Failure is initiated by yielding of tension steel fs=fy

– When εu is reached, compression steel may or may not have yielded, can be found from compatibility of strain

• For small e

– εu is reached before tension steel yields

– Stress in the other side of load may also be in compression, not in tension.

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Interaction diagram

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Balanced Failure

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Problem

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Read article- important 87

Why phi is low for column?

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Why alpha?

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DESIGN CHARTS

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Review problem

• Use the charts to determine the column strength ϕPn, of the short column shown in Fig (14x24, reinforced with 8 No 10 bars), Use fc’=4ksi and fy=60 ksi, e=12in

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BIAXIAL BENDING

• There are situations when axial compression is associated with simultaneous bending present about both principal axes of the section

• Corner column is such a case

• Interior column may also experience biaxial bending-irregular grid, lateral load

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Load contour method

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Reciprocal Load method

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Example 8.5

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For both axially loaded and biaxial problems, alpha phi Pn should be greater than max Pu .

Ignore Mu first, check two problems of last year

do not use graph paper for interaction, make hand sketch

don’t attach charts

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