Aci318m-11code Requirements for Beams Under Combined Shear and Torsion
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Transcript of Aci318m-11code Requirements for Beams Under Combined Shear and Torsion
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8/12/2019 Aci318m-11code Requirements for Beams Under Combined Shear and Torsion
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Structural Design Calculation
Published by:
E.G. NavarroCE - Mindanao University of Science and Technology
April 3, 2014
ACI318M-11 CODE REQUIREMENTS
FOR BEAMS UNDER COMBINED SHEAR
AND TORSION
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Structural Design Calculation
TYPES OF TORSION
1. EQUILIBRIUM TORSION- The torsional moment cannot be reduced by redistribution of internal
samp e o eam su ec o equ r um ors on
forces . This is referred to as equilibrium torsion, since the torsional
moment is required in the structure to be in equilbrium.( ACI 11.5.2.1)
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Structural Design Calculation
2. COMPATIBLE TORSION- The torsional moment can be reduced by redistribution of internal
twisting to maintain compatibilty of deformation. This torsion
is referred to as compatible torsion. ( ACI 11.5.2.2)
forces after cracking If the torsion arises from the member
sample of beam subject to compatible torsion
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Structural Design Calculation
TORSION CAN BE NEGLECTED IF IT WILL COMPLY SECTION 11.5.1 OF THE CODE
If Tu is lesser than the
> threshold torsion thenit can be neglected
because it will not
have significant effect
on the structure.
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Structural Design Calculation
> Tu reduction is notallowed for member
subject to equilibrium
torsion.
> Tu reduction isallowed for member
subject to compatible
torsion because of
redistribution.
Tu is permitted to be
reduced equal to
for nonpressed
members.
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Structural Design Calculation
> Check for cross section
adequacy due tocombined shear
and torsion.
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Structural Design Calculation
TORSIONAL REINFORCEMENT SHOULD BE ACCOUNTED IF IT DOES NOT COMPLY SECTION 11.5.1
(THE TORSIONAL MOMENT IS GREATER THAN THE THRESHOLD TORQUE)
TORSIONAL REINFORCEMENT SHOULD BE COMPUTED AS STATED BELOW.
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Structural Design Calculation
1. A 250mmX600mm BEAM IS SUBJECTED TO A TORSIONAL MOMENT (Tua) = 20 KN.m
SHEAR (Vu)=168 KN AND BENDING MOMENT(Mu) = 120 KN.m. CHECK IF THE SECTION
IS ENOUGH CARRY SUCH LOADAND HOW MUCH REINFORCEMENT IS NEEDED.
Tua =
Vu =
Mu =
H d
PARAMETERS:
fc' = 27.5 MPa
fy = 414.0 MPa
= 0.75
H = 600.0 mm (total height of the beam)
bw = 250.0 mm (total width of the beam)
d = 550.0 mm (effective depth of the beam)
Cc = 35.0 mm
xo = bw-(2xCc) = 180.0 mmyo = H-(2xCc) = 530.0 mm
Acp = bw x H = 150000.0 mmPcp = (2xbw)+( 2xH) = 1700.0 mm
Aoh = xoXyo = 95400.0 mm
Pn = 2 x (xo+yo) = 1420.0 mm
(center to center length of stirrups in y-direction)
(cylinder compressive strength of concrete)
(yield strength of reinforcement)
(strength reduction factor for shear in concrete)
(center to center length of stirrups in x-direction)
120.0 kN.m
SAMPLE CALCULATION:
B
xo
yO
168.0 kN
(concrete cover to center of stirrups)
(area enclosed by outside perimeter )(outside perimeter of concrete section)
(area enclosed by transverse reinforcement )
(perimeter of transverse reinforcement )
20.0 kN.m
Cc
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Structural Design Calculation
1.Check for type of torsion :
Type of torsion = Equilibrium torsion
Therefore: Tu = Tua = 20.0 kN.m (ACI 11.5.2.1)
2.Check for design torsion(Tu) versus threshold torsion(Ts) :
Ts = Tcr/4 Ts is one quarter of cracking torque
(ACI 11.5.1)
Tcr = (0.75)(0.33)(27.5)((150,000.0)^2/(1,700.0))Tcr =
Ts =