Chapter A15 Sample Test


1. For the idealized airfoil and loading shown and assuming the webs are ineffective in carrying bending, determine and draw

(a) the resisting shear flow pattern,
(b) the resisting shear flow pattern, if a .03-in thick, flat vertical web is added between stringers 1 and 4.

The internal transverse shear load is 1000 lb down.

NOTE:The thickness of web 2-3 is .01" and that of web 3-4 is .02".


q1-2 = 1.103 lb/in from 2 to 1;
q2-3 = 48.897 lb/in from 2 to 3;
q3-4 = 1.103 lb/in from 4 to 3;
q4-1 = 51.103 lb/in from 1 to 4.

q1-2 = 13.62 lb/in from 2 to 1;
q2-3 = 36.38 lb/in from 2 to 3;
q3-4 = 13.62 lb/in from 4 to 3;
q4-1 (flat web) = 58.2 lb/in from 1 to 4
q4-1 (curved web) = 5.42 lb/in from 1 to 4.

2. For the two-cell, three-stringer cantilever beam shown, if the applied load passes through the shear center, determine

(a) the shear flow distribution at the root,
(b) the axial force in each stringer at the root if the beam has a length of 96",
(c) the magnitude and location of maximum shear stress,
(d) the x coordinate of the shear center, ex


(a) qAB (semi-cir.) = 108.61 lb/in upward, qAB = 127.96 lb/in upward, qAC = 13.43 lb/in to the left, and qBC = 13.43 lb/in to the right.
(b) FA = 24,000 lb (T), FB = 24,000 lb (C), FC = 0 lb.
(c) maximum shear stress = 4,265.33 psi in flat web AB.
(d) ex = 0.72 in to the right of stringer A.

3. For the 2-cell box beam of problem 2, determine the y coordinate of the shear center.
Hint: Keep in mind that although the stringers are distributed symmetrically about the horizontal centroidal axis, the webs are not.


ey = 1.334 in above the centroidal horizontal axis.
NOTE: if web AC and BC had equal thickness, then ey would have been zero.

4. Consider the tapered single-celled, skin-stringer wing box and loading shown. Using delta P method, over a distance of 20" measured from the root section along the span, determine at the root section

(a) the shear flow distribution in the webs,
(b) the axial force distribution in the stringers,
(c) the maximum shear stress and its location.


(a) qab = 7.44 lb/in (downward), qbc = 4.94 lb/in (to the left), qcd = 17.56 lb/in (downward), and qda = 4.94 lb/in (to the right)
(b) Fa = Fd = 6,056 lb (C), Fb = Fc = 6,056 lb (T)
(c) max. shear stress = 439 psi (in web cd)

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