Aci318m-11code Requirements for Beams Under Combined Shear and Torsion

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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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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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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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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> 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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> Check for cross section

adequacy due tocombined shear

and torsion.


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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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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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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 =

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Aci318m-11code Requirements for Beams Under Combined Shear and Torsion