300_Mechanics Homework Mechanics of Materials Solution

# 300_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: r’ﬁﬁr‘gi—jm m mm ijr—Jrﬁ 1—1 ! 4.14 Knowing that a beam of the cross section shown is ﬂat glam-g vertical axis and PROBLEM 4.14 that the bending moment is 4 kN ° m, determine the total force acting on the shaded 9 portion ofthe lower ﬂange. ~ \ “ 15mm SOLUTION 15 mm 45mm :i: The. shess dis-I-h‘lnaJHOIn OVEN“ ﬁe «+qu 15mm cross SaaHon 1'5 given by Hue. bending ﬁves: {law anqu - Mun—- when y is o. awarding/‘2 Wf'l‘ln {+3 origin an Hue new-I‘M! sun's and I is He momewf o‘F inewh‘a. 0'? H38 en'Hre ems sea-Hana! area. The "Force on ‘Hne shaded 1‘: caﬁcuﬁmhd_¥row +lu‘s shes: JIS'I-riBUHon. (New on area. ‘epemrm-I' AA We'J-‘ov‘ée is air: :- 6‘," JA =— A?! AA Time {vial} ‘Pof‘ce on He Shac’ec‘ area. is 'H'Iela F: SAP = h§%¥JA 3 -%SY¢I = —%3\$A" where 3* II: 44.9 cen'f'mrdra! coordu'naaLe 0'?“ +ke shades» Par'h'ém and A’ 125 H: area... If I = 71555.3 -.- 7505“”? = 0.52734wo‘ m. .1. V -I ‘31“: "’ {£05wa 3 0.012%»«10‘ an». I= I. + I; + 1:3 = {.0672‘wo‘ m“ = 1.0672 ”0“». N" = (any-7.5)(Is) = 450 m“: 450on m g"? -..- Ji(37.5"+ 7‘5) 1 22-5'1vnm 7' 0-0227 M F': M", (Hts/(oakdozzs) 45mm") I wanna“ 237.9%!03N = 37.91:» «I: ...
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