Lecture 6 - Flexure: September 13, 2002 Cven 444

Lecture 6 - Flexure September 13, 2002 CVEN 444

Lecture Goals • • • •

Basic Concepts Rectangular Beams Non-uniform beams Balanced Beams

Flexural Stress Example Consider a simple rectangular beam( b x h ) reinforced with steel reinforcement of As. (1) Determine the centroid ( neutral axis, NA ) and moment of inertia Izz of the beam for an ideal beam (no cracks). (2) Determine the NA and moment of inertia, Izz, of beam if the beam is cracked and tensile forces are in the steel only.

Example Ec – Modulus of Elasticity - concrete Es – Modulus of Elasticity - steel

As – Area of steel d – distance to steel b – width h – height

Es n Ec

Example yA  Centroid (NA) y A I   I    y  y  A Moment of Inertia i

i

i

2

i

i

i

Example (uncracked) Concrete Steel

Area

yi

y iA

I

yi - y

(y i -y)2 A

bh (n-1)As

h/2 d

bh2/2 d(n-1)As

bh3/12 ---

(h/2-y) (d-y)

(h/2-y)bd (d-y)2(n-1)As

bh 2   n  1 As d y A  i i 2 y  Ai bh   n  1 As I   I i    yi  y 

2

2

bh  h 2  Ai     y  bh   d  y   n  1 As 12  2  3

Example - (cracked) For a cracked section the concrete is in compression and steel is in tension. The strain in the beam is linear.

1 C  yb f c 2 T  As fs

Example - (cracked) Using Equilibrium

T C 1 As fs  yb f c 2  yb  fs    fc  2 As 

Example - (cracked) Using Hooke’s law

f  E

 yb   c  Es   2 As  2nAs Es s       Ec c    s  Ec   yb  yb  2 As  However, this is an indeterminate problem to find y . We will need to use a compatibility condition.

Example - (cracked) Using a compatibility condition.

s

c

c y    d y y s d  y Substitute into the first equation.

Example - (cracked) Substitute in for the strain relationship.

2nAs y  dy yb Rearrange the equation into a quadratic equation.

 2nAs   2nAs  y   y  d  0  b   b  2

Example - (cracked) Use a ratio of areas of concrete and steel.

As  bd

y  2n d y  2n d  0 2

2

Modify the equation to create a non-dimensional ratio. 2

y y    2n     2n   0 d d

Example - (cracked) Use the quadratic formula

 y  2n    2n    8n    2 d 2 y     n    2n   n  d 2

Solve for the centroid by multiplying the result by d.

Example - (cracked) The moment of inertia using the parallel axis

I   I i    yi  y  Ai 2

2

by  y  2     by   d  y  nAs 12  2  3 by 2    d  y  nAs 3 3

Example For the following example find centroid and moment of inertia for an uncracked and cracked section and compare the results. Es = 29000 ksi Ec = 3625 ksi d = 15.5 in b = 12 in. h = 18 in.

Use 4 # 7 bars for the steel.

Example E s 29000 ksi n  8 Ec 3625 ksi A #7 bar has an As = 0.6 in2

As  4  0.6 in

2

  2.4 in

2

Example The uncracked centroid is bh 2   n  1 As d y 2 bh   n  1 As

12 in 18 in  

2

  8  1  2.4 in 2  15.5 in 

2 12 in 18 in   8  1  2.4 in 2 

2204.4 in 3   9.47 in 2 232.8 in

Example The uncracked moment of inertia 2

bh  h 2  I    y  bh   d  y   n  1 As 12  2  3

12 in 18 in   18 in      9.47 in  12 in 18 in  12  2  3

2

 15.5 in  9.47 in   8  1  2.4 in 2  2

 6491 in 4

Example The cracked centroid is defined by: As 2.4 in 2    0.0129 bd 12 in 15.5 in  y   d 

 n 

2

 2n   n 

 8 0.0129  

2

 2  8  0.0129    8  0.0129 

 0.3627 y  0.3627 15.5 in   5.62 in

Example The cracked moment of inertia is 3

by 2 I   d  y  nAs 3 1 3  12 in  5.62 in  3

 15.5 in  5.62 in   8   2.4 in 2

 2584.2 in

4

2



Example Notice that the centroid changes from 9.47 in. to 5.62 in. and the moment of inertia decreases from 6491 in4 to 2584 in4 . The cracked section loses more than half of its’ strength.

Flexural Stress Basic Assumptions in Flexure Theory • Plane sections remain plane ( not true for deep beams h > 4b) • The strain in the reinforcement is equal to the strain in the concrete at the same level, i.e. s = c at same level. • Stress in concrete & reinforcement may be calculated from the strains using s curves for concrete & steel.

Flexural Stress Additional Assumptions for design (for simplification) • Tensile strength of concrete is neglected for calculation of flexural strength. • Concrete is assumed to fail in compression, when c (concrete strain) = cu (limit state) = 0.003 • Compressive s relationship for concrete may be assumed to be any shape that results in an acceptable prediction of strength.

Flexural Stress The concrete may exceed the c at the outside edge of the compressive zone.

Flexural Stress The compressive force is modeled as Cc = k1k3f’c b*c at the location x = k2*c

Flexural Stress The compressive coefficients of the stress block at given for the following shapes. k3 is ratio of maximum stress at fc in the compressive zone of a beam to the cylinder strength, fc’ (0.85 is a typical value for common concrete)

Flexural Stress The compressive zone is modeled with a equivalent stress block.

Flexural Stress The equivalent rectangular concrete stress distribution has what is known as a b1 coefficient is proportion of average stress distribution covers.

b1  0.85 for f c  4000 psi  f c  4000  b1  0.85  0.05 *    0.65  1000 

Flexural Stress Requirements for analysis of reinforced concrete beams [1] Stress-Strain Compatibility – Stress at a point in member must correspond to strain at a point.

[2] Equilibrium – Internal forces balances with external forces

Flexural Stress Example of rectangular reinforced concrete beam. (1) Setup equilibrium.

F

x

0 

TC

As f s  0.85 f c ab a   M  0  T d  2   M n

Flexural Stress Example of rectangular reinforced concrete beam. (2) Find flexural capacity.

T  As f s C  0.85 f c ab a

As f y 0.85 f cb

Flexural Stress Example of rectangular reinforced concrete beam. (2) Find flexural capacity.

M n  T  moment arm  a   As f y  d   2 

Flexural Stress Example of rectangular reinforced concrete beam. (3) Need to confirm s > y

y  c

s

sy Es a

b1

 d  c   c

c

 y

Flexural Stress Example Example of rectangular reinforced concrete beam. Given a rectangular beam fc = 4000 psi

fy = 60 ksi (4 #7 bars) b = 12 in. d = 15.5 in. h= 18 in. Find the neutral axis. Find the moment capacity of the beam.

Example Determine the area of steel, #7 bar has 0.6 in2.

As  4  0.6 in 2   2.4 in 2 The b value is b1 = 0.85 because the concrete has a fc =4000 psi.

Example From equilibrium

C T 0.85 f cba  f y As a

f y As 0.85 f cb



2 60 ksi 2.4 in   

0.85  4 ksi 12 in 

 3.53 in.

The neutral axis is a 3.53 in. c   4.152 in. b1 0.85

Example Check to see whether or not the steel has yielded.

fy

60 ksi y    0.00207 Es 29000 ksi Check the strain in the steel

 d c s     0.003 Steel yielded!  c   15.5 in.  4.152 in.     0.003  0.0082  0.000207 4.152 in.  

Example Compute moment capacity of the beam.

a  M n  As f y  d   2  3.53 in.     2.4 in   60 ksi  15.5 in.   2    1979 k-in.  164.8 k-ft. 2