Lecture #2Shear stresses and shear centerin multiple closed contour

SHEAR STRESSES RELATED QUESTIONS

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- shear flows due to the shear force, with no torsion;

- shear center;

- torsion of closed contour;

- torsion of opened contour, restrained torsion and deplanation;

- shear flows in the closed contour under combined action of bending and torsion;

- twisting angles;

- shear flows in multiple-closed contours.

SHEAR FLOW IN MULTIPLE-CLOSED CONTOUR.STATIC INDETERMINACY

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The total shear flow is represented as a sum of variable part qf for an opened contour and shear

flows in separate cells q0i taken with certain sign ̅qi :

Here ̅qi = ±1 and is determined according to positive or

negative tangential coordinate direction.

0f iii

q q q q

If the contour has i closed contours (cells), the problem has i unknown cell flows q0i but it is statically

indeterminate only i-1 times because one unknown could be evaluated from equilibrium equation.

EQUILIBRIUM EQUATION

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An equilibrium equation includes i unknown flows in cells q0i :

Here Mq0 , Mqf – moments from constant and variable

parts of shear flow, respectively;MQ – moment from resultant shear force;

i – double area of separate i-th contour.

0i i qf Qi

q M M

0q qf QM M M

DETERMINATION OF RELATIVE TWIST ANGLES

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We use the formula derived at last lecture:

1; .t

q qdt q

G

Substituting the sum for shear flow, we get

or

0f j i j

iij jt t

q q q qdt q

G G

0iF ij ii

q

THE SYSTEM OF EQUATIONS

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The system of equations includes one equilibrium equation and j = i relative twist angles equations:

0

01

;

, 1, .

i i qf Qi

n

ij i iFi

q M M

q j n

EXAMPLE – GIVEN DATA

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EQUIVALENT DISCRETE CROSS SECTION

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EXAMPLE – DISCRETE APPROACH

, kN mfq t

1q

2q

= 6.59·10-6 °/N;

= - 9.65·10-7 °/N;

= - 7.07·10-7 °/N;

= 5.84·10-6 °/N;

F= - 0.87 °/m;

2F= - 3.13 °/m.

EXAMPLE – DISTRIBUTED APPROACH

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For equilibrium equation, we have:- moment from shear flows qf : Mqf = -47.3 kN·m ;

- moment from resultant shear force Qy :

MQ = -15 kN·m .Solving the system of equations, we get- shear flows in contours q01 = 114.2 kN/m and

q02 = 455.0 kN/m ;

- relative twist angle = - 0.556 °/m (compare to = - 0.473 °/m which we calculated for similar single- closed contour)

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, kN mfq t

011 , kN mq q

022 , kN mq q

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, kN mq

EXAMPLE – FINAL DIAGRAMFOR DISCRETE APPROACH

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, kN mq

EXAMPLE – CONTINUOUS APPROACH , kN mfq t

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, kN mq

EXAMPLE – CONTINUOUS AND DISCRETE APPROACHES BEING COMPARED

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, kN mq

EXAMPLE – SINGLE-CLOSED AND DOUBLE-CLOSED CONTOURS BEING COMPARED

SHEAR CENTER CALCULATION

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Firstly, we solve the following system with arbitrary chosen torsional moment MT :

Next, we find torsion rigidity G·I = MT / T and

0

01

;

, 1, .

i i Ti

n

ij i Ti

q M

q j n

SC Qy

G IX X

Q

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EXAMPLE – SHEAR CENTER CALCULATION

shear center

shear center

TOPIC OF THE NEXT LECTURE

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Torsion of opened cross sections

All materials of our course are availableat department website k102.khai.edu

1. Go to the page “Библиотека”2. Press “Structural Mechanics (lecturer Vakulenko S.V.)”