Calculation of Sealant Tension Stress and Elongation Caused by Loads Q and F

Assuming a force Q acting on a point along the center axis with distance d from the bonding interface, the force then causes an inclination of the structural point. The resulting tension stress in the sealant shows a certain stress distribu­tion with a stress and elongation maximum located at one side of the structural point. The following derivation follows assumptions for both linear and nonlin­ear stress distributions (see Fig. 26).

Achieving a basic momentum balance around the bottom edge of the struc­tural point (y = 0) yields the following formula for small inclination angles fi

b(y) = 2 – л/D ■ y — y2

Assuming a linear function for the sealant extension over y provides

Copyright by ASTM Int’l (all rights reserved); Tue May 6 12:07:08 EDT 2014 Downloaded/printed by

Rochester Institute Of Technology pursuant to License Agreement. No further reproductions authorized.

k(y) — 1 + De(y)/e — (kmax – h)-y/D + k0 (A8)

with k0 and kmax as extension maxima at the lower and upper edge of the struc­tural point.

This also provides a function for the inclination angle fi

tan b — (kmax – ko)-e/D (A9)

For the following derivation a linear distribution for tension stress in the sealant may be used as shown in Eq A10

r(y) —E ■ — E ■(k – 1) (A10)

e

Alternatively, a nonlinear stress function (see lit [56].) may be used, as shown in Eq A11

r(y) — G ■ (k – kj) (A11)

Basic momentum and force equations (Eqs A5 and A6) combined with the rela­tionships defined in Eqs A7 and A8 can be solved by numerical integration and iteration and yield k0 and kmax.

The maximum tension stress values are obtained for k0 and kmax by using the linear or nonlinear stress functions (Eqs A10 or A11) above.

The maximum sealant strain in tension and inclination of the structural point are calculated as shown in Eqs A12 and A13

Demax — e ■ (k max 1) (A12)

tan b — (kmax – ko) e/D (A13)

The location of the rotational axis is given by Eq A14

1k

t/D — ——- ^ (for k max — kO) (A14)

kmax – k0

For the simple case of linear stress function and assuming no normal force (F — 0) acting on the fixing point, Eq A15 can be derived

Equation A15 allows a simple approximation of the maximum tension stress as shown in Eq A16

Copyright by ASTM Int’l (all rights reserved); Tue May 6 12:07:08 EDT 2014 Downloaded/printed by

Rochester Institute Of Technology pursuant to License Agreement. No further reproductions authorized.

rmax = 10.185 • Q • d/D3 (A16)