The weight _W_ resting at _G_ is 20,000 pounds. A part of this weight is carried by _BC_ as a simple timber beam, while the remainder of the load will be carried on the triangular frame _BCD_ acting as a truss, the elastic deflection of the latter throwing a part of the load on _BC_ acting as a beam. According to the principle of least work the division of the load will be such as to make the work performed in straining the different members of the system a minimum. That part of _W_ which rests on _BC_ as a simple beam may be represented by _W₁_, while _W₂_ represents the remaining portion carried by the triangular frame. As _G_ is at the centre of the span, the beam reaction at either _B_ or _C_ is ½_W₁_. Hence the general value of the bending moment in either half of the beam at any distance _x_ from either _B_ or _C_ is _M_ = ½_W₁x._ Hence _M²dL_ = ¼_W₁²x²dx_. As there is but one member acting as a beam, whose moment of inertia _I_ is constant, the second term of the second member of equation (52) becomes, by the aid of the preceding equation, 1 1 ¹/₂ 1 _W₁²l³_ ----- ⌠ _M²dL_ = ---- ⌠ _¼W₁²x²dx_ = ---- --------. (54) 2_EI_ ⌡ _EI_ ⌡₀ _EI_ 96 [Illustration: FIG. 32.] The numerical elements of the expression for the work done in the members of the triangular frame are: Member. Stress. Length. Area of Section. _BC_ ½_W₂_ tan _α_ 360 inches = _l_ 140 square inches _DC_ ½_W₂_ sec _α_ 204.5 ” 4.14 ” ” _DG_ _W₂_ 96 ” 80 ” ” 10 × 14³ 27440 _I_ = -------- = ------- = 2286.7. 12 12 The substitution of those quantities in the first term of the second member of equation (52) will give 1 ⎲ _S²L_ 1 ( _W₂²_ tan² _α_.360 _W₂²_.96 ) -----⎳ ----- = ---------- (------------------ + ----------) 2_E A_ 2,000,000 ( 4 × 140 80 ) 2 _W₂²_ sec² _α_ .204.5 + ----------- --------------------- = .000,003,73 _W₂²_. 56,000,000 4 × 4.14 The substitution of numerical quantities in equation (54) gives 1 _W₁²l³_ ---- -------- = .000,213_W₁²_. _EI_ 96 Or, since _W - W₂ = W₁_, _e_ = .000,003,73_W₂²_ + .000,213(_W - W₂)²_. (55) Hence _de_ ---- = .000,007,46_W₂_ - .000,426(_W - W₂_) = 0. (56) _dW₂_ The solution of this equation gives _W₂_ = .893_W_ = 19,660 pounds. _W₁_ = 340 ”