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Moment and shear diagram of a beam under dead and live loads are
shown below.

- Concrete is assumed to resist compression only, tension shall be resisted by reinforcements. Reinforcements shall be placed at the side of the beam that has tension. For a simply supported beam, tension is at the bottom of beam. For a cantilever end, tension is at the top of the beam.
- Shear is at its maximum at edge of supports. Diagonal shear cracks is normally developed close to the support. Stirrup for shear reinforcement is normally placed vertically to intercept the crack. They are normally closer spaced near the support and gradually spread out toward center of the beam.

Design assumption:

- Strain distribute linearly across the section.
- Concrete resists only compressive stresses.

Therefore, the stress distribution across the section of the beam is as shown below.

At an ultimate strain of 0.003, the stress at extreme fiber of the
beam reaches ultimate strength of concrete f_{c}’. The
distribution of the compressive stresses is a complex curve. For
calculation purpose, a stress block of 0.85fc’ spread over a depth,
a, is used. Therefore,
the total compressive stress in a rectangular beam is

C = 0.85f_{c}’ab

Where b is the width of the beam.

At ultimate stress situation, the concrete at top portion is subjected to compression. The compressive stresses distribute uniformly over a depth a. The resultant of compressive stress, C is located at a distance, a/2, from the top surface. Tensile force is taken by rebars at an effective distance, d, from the top surface. By equilibrium, the tensile force is equal to the compression resultant,

T = A_{s}f_{y} =
C = 0.85f’_{c} ab

where f_{y} is
the yield strength of reinforcing steel and A_{s} is
the area of steel. Therefore,

The depth of stress block,

a = A_{s}f_{y}/(0.85f’_{c} b),
or a = A_{s}f_{y}d/(0.85f’_{c} bd),

Let the reinforcement ratio, r = As/bd, then

a = rf_{y}d/0.85f’_{c}

Let m = f_{y}/0.85f’_{c} ,
then, a = rdm..The
nominal moment strength of the section,

M_{n} = C (d-a/2) =
0.85f’_{c} ab(d-a/2)

Then, The nominal moment strength of the section,

M_{n} = A_{s}f_{y} (d-a/2)
= A_{s}f_{y} (d-rdm/2)
= A_{s}f_{y} d-
A_{s}f_{y} drm/2

ACI code requires that the factored moment,

M_{u} £ f M_{n}

Where, f =
0.9, is the strength reduction factor for beam design. Let M_{u} = f M_{n} ,
We have M_{u} = f (A_{s}f_{y} d-
A_{s}f_{y} drm/2)

Divide both side by bd^{2}, we have M_{u}/fbd
= (A_{s}/bd)f_{y} -(A_{s}/bd)
f_{y} rm/2)
= rf_{y} -
f_{y} r^{2}m/2)

Let R_{n} = M_{u}/fbd^{2},
and we can rewrite the equation as

r^{2}(m/2) - r -
R_{n}/f_{y} =
0

Solving the equation, the reinforcement ratio,

r =
(1/m)[1-(1-2mR_{n}/f_{y})^{1/2}]

The area of reinforcement is A_{s} = rbd

There are two situations when a reinforced concrete beam fails due
to bending. One
is when the reinforcing steel reaches its yield stress, f_{y}. The
other is when the concrete reach it maximum compressive stress, f’_{c}. When
a reinforced concrete beam fails in yielding of steel, the failure
is ductile because the steel can stretch for a long period of time
before it actually breaks. When
it fails in concrete, the failure is brittle because concrete breaks
when it reach maximum strain.

When concrete reaches its maximum strain at the same time as the steel reach is yielding stress, it is called a balance condition. Using a maximum strain, 0.003 of concrete and assume a linear distribution of strain across beam section, one can determine the reinforcement ratio at balanced condition. The reinforcement ratio based on ACI code is

r_{b} =
(0.85f’_{c}/f_{y}) b_{1} [87000/(87000+f_{y})] [f’_{c} and
f_{y} are in psi
(lb/in^{2})]

r_{b} =
(0.85f’_{c}/f_{y}) b_{1} [600/(600+f_{y})] [f’_{c} and
f_{y} are in MPa
(MN/m^{2})]

Where b_{1} =
0.85 for 4000 psi (30 Mpa) concrete, and reduce 0.05 for each 1000
psi of f’_{c} in
excess of 4000 psi.

To ensure a ductile failure of beam, ACI code limits the maximum
reinforcement ratio to 0.75r_{b}. On
the other hand, when the amount of steel is too small, the beam will
fail when concrete reach its tensile strength. It needs to have a
minimum amount of steel to ensure a ductile failure mode. The
minimum reinforcement ratio in ACI code is r_{min} =
200/f_{y} (psi).

__Situation__:

A simply supported reinforced concrete beam is supporting uniform dead and live loads

Design data:

Dead load: 1500 lb/ft

Live load: 800 lb/ft

Length of beam: 20 ft

Width of beam: 16 in

Depth of beam: 24 in

Minimum concrete cover: 1.5 in

Diameter of stirrup, 0.5 in

Compressive strength of concrete: 4000 psi

Yeild strength of steel: 60000 psi

__Requirement__: Design
flexural reinforcement for bending

__Solution__:

1. Calculate factored moment:

Weight of beam: W_{B} =
150 lb/ft x 1.33 ft x 2 ft = 400 lb/ft

Factored load: Wu = 1.4(400+1500)+1.7(800) = 4020 lb/ft

Factored moment: Mu = (4020)(20^{2})/8 = 201000 ft-lb

Assume the main reinforcement bar is 1" in diameter (#8 bar)

Effective depth: d:24-1.5-0.5-0.5 = 21.5 in

Factor: R_{n} =
(201000)(12)/[(0.9)(16)(21.5^{2})]=362.4 psi,

m = 60000/[(0.85)(4000)]=17.65

Reinforcement ratio

r =
(1/m)(1-2mR_{n}/f_{y})^{1/2})=0.0064

Minimum reinforcemnet ratio: r_{min} =
200/f_{y}=0.0033

Maximum reinforcement ratio; r_{min} =
(0.75)(0.85f’_{c}/f_{y}) b_{1} [87000/(87000+f_{y})]=0.021

Required reinforcement, As = rbd
= 2.2 in^{2}.

Use 4#8 bar area of reinforcement is 0.79 in^{2}x4 = 2.37 in^{2}.

The direct shear shrength according to ACI is

fv_{c} =0.85[1.9Öf_{c}’+2500r_{w}(V_{u}d/M_{u})] £ 0.85(3.5Öf_{c}’)

where r_{w} (» 0.002)
is reinforcement ratio, V_{u} is
factored shear stress, M_{u} is
factored moment at the critical section. Or

fv_{c} =0.85(2Öf_{c}’)

- When shear stress, v
_{u}£ ½ fv_{c},no shear reinforcement is required. - When ½ fv
_{c}< v_{u}£ fv_{c}, use minimum reinforcement

A_{v} = 50 b_{w }s
/f_{y}

Where s is spacing of web reinforcement, f_{y} is
yield strength of steel, A_{v} is
cross section area of web reinforcement, b_{w} is
width of beam web.

- When fv
_{c}< v_{u}, use v_{u}£ f(v_{c}+ v_{s}), where v_{s}is shear strength provided by shear reinforcement.

The shear force that is resisted by shear reinforcements is Vs = (Vu
- fV_{c}). Normally,
stirrup is spaced vertically at a spacing, s, for shear
reinforcement. Within
an effective depth d, the shear strength provided by A_{v}f_{y}d/s,
where A_{v} is
area of stirrup, f_{y} is
yield strength of reinforcing steel. The
shear strength multiply by a reduction factor, f,
needs to be larger than Vs. Therefore, V_{s} = f(A_{v}f_{y}d/s). The
spacing of stirrup is calculated as

s = (fA_{v}f_{y}d)/V_{s}

ACI code requirements for placing stirrup are as follows.

- When ½ fv
_{c}< v_{u}£ fv_{c}, max s = d/2 £ 24 in. - When fv
_{c}< v_{s}£ 4Öf_{c}, max s = d/2 £ 24 in. - When fv
_{c}< v_{s}£ 8Öf_{c}, max s = d/4 £ 12 in.

__Situation__:

Design shear reinforcement for the beam in the previous example

Support column size: 12”x12”

__Solution__:

1. Calculate factored shear:

Clear distance between support, L_{n} =
19 ft

Factor shear V_{u} =
W_{u}L_{n}/2 = 38.2 kips

Shear strength of concrete:

fV_{c} =
0.85(2Ö4000) d b = 37 kips

1/2fV_{c} =
18.5 kips

The length that required no shear reinforcement is

L_{1} = (L_{n} /2)(18.5/38.2)
= 4.6 ft

Distance from center of beam that required minimum
reinforcment is

L_{2} = (L_{n}/2)( fV_{c} /Vu)
= 9.2 ft close
to L_{n}/2 = 9.5 ft

Use #3 stirrup the area of stirrup, area of steel: A_{v} =
2(0.11 in^{2}) = 0.22 in^{2}.

Maximum spacing, s = (0.22 in^{2})(60000 psi) /[(50
psi)(16 in)] = 16.5 in

Maximum spacing d/2 = 10.75 in (Govern)

Use 6 stirrups at 10.75 inch spacing, with first stirrup at
5". Total length cover
by stirrups is L_{s} =
(5)(10.75 in)+5 in = 4.9 ft O.K.