114 APPLIED MECHANICS 
the cantilever due to the curvature of the part HK. Let HK =dz, and 
‘ CD=du. Then, 0= dat = du therefore du= waar — Madx But M= Wa, 
BR. 2 EL ay: a 
thercfore du= a . The total deflection 
= SOW ada WS a? Wi? 
AB=u,={ EI m2 ae 
where / is the length of the cantilever. 
121. Cantilever Loaded Uniformly.—Let the load be w per unit of 
length. Using the same notation, and proceeding in the same way as in 
the preceding Article,du = = 2 “, but M=4uwe2, therefore du= i and 
wf’ 3,,_ wit WE nae tk ; 
sum,” da= SEI > SEP where W=wl=total load. The cantilever 
is supposed to be of uniform cross section. 
122. Beam Supported at the Ends and Loaded at the Centre.— 
U,= 
Referring to Fig. 155, and proceeding as in Art. 120, dw=CD= bea 
U 3 3 
but M = 4Wa2, therefore du = bitten 4, =AB= sel eda= pi = pil : 
w 
2 
i] 
1 
! 
! 
sf paar > 
=e] = ee eee ee ee al 
Fig. 155. 
123. Beam Supported at the Ends and Loaded Uniformly.— 
Referring to Fig. 155, but remembering that the load W is uniformly 
distributed over the length, and that its intensity is w per unit of 
length. 
Madx te $ P _whards _ watdx 
du=CD = ET’ but M = 4wLe - }w2x*, therefore du= IRL aRT- 
yy ant t ‘ i wh - w ey 
Hence My =a,” dn — sz, adaz Om "3 suica ; 
wh* whit _ wl (; :) _ Sol! _5Wie 
“48EI 128EL I6EI\3 8) 384EI 384EI° 
124. Slope of Bent Beam at any Point.—Referring to Figs. 154 
and 155, @ is the change in the slope of the beam between the points — 
