M-(1zy.z– ly.y) + M,(1;.z – lzy-y)Iz.ly - Izyloa - Ol dyPrMyFormüller; o =-• 01• Ox2PrOx + dyo - Ga0, + o3 < of , Jort• 022tTmaks222wOz + OyTc d?y M(x)- a.2Omaks.min+ Ty , 8 = aAT, 1 =.EI(**-%)* + r3y · tejp -MalanıcpXc, @pcEIMalanipcEIA(0xTầy • tc/pTmaks2ВƏMMap dx, PervQItEICalculate the reaction at support A for the beam and loading shown in the figure.w = 5 N/mm and L = 1.2m2250 N3520 N1930 N2680 N3210 N

Answered: Image /qna-images/answer/a19aa788-bc2b-41bb-8899-23077b868a95.jpg

Formüller; a Iz.ly – Izy² Pr ož - G,0, + oš < o , Jort = Oz + Oy 2 Tmaks 2t 2 w Tc d²y M(x) Omaks.min I dr? El A Tmaks MalanıcpXc , @pjc EI Malanipe El 2 В n²El ƏM M ap dx. Per vQ It alculate the reaction at support A for the beam and loading shown in the figure. = 5 N/mm and L = 1.2m 2250 N 3520 N 1930 N 2680 N 3210 N

Formüller; a Iz.ly – Izy² Pr ož - G,0, + oš < o , Jort = Oz + Oy 2 Tmaks 2t 2 w Tc d²y M(x) Omaks.min I dr? El A Tmaks MalanıcpXc , @pjc EI Malanipe El 2 В n²El ƏM M ap dx. Per vQ It alculate the reaction at support A for the beam and loading shown in the figure. = 5 N/mm and L = 1.2m 2250 N 3520 N 1930 N 2680 N 3210 N

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

M2(Izy.Z– ly-y) + M,(1z.z – lzy-y) 1z.ly - Izy? Pr Мy Formüller; o =- • Oy = Oz + Oy o - ạ0p + o < oỷ , Jort = , Tmaks 2 Pr •02 = 2 2t 2 Çelik Og + Oy Te d²y_M(x) Omaks.min + τ , ε αΔΤ, τ,. EI MalanıcoXc , @pjc EI 2 Malanıpc 2000 mm ) + ry tejp = Tmaks = EI Bakır ƏM dx. Pgr VQ M It If a copper ring with an outer diameter of 2000 mm and a thickness of 20 mm and a steel ring with an inner diameter of 2000 mm and a thickness of 10 mm are intertwined and heated up to 100 °C; a) Calculate the maximum stress in the steel ring. b) Calculate the pressure exerted by the copper ring on the steel ring. Copper = Ebakır= 100 GPa, abakıF 16.10-6/°C Steel = Eçelik= 200 GPa açelik= 12.10-6/ºC
M-(1zy.z– ly -y) + M,(1,.z – Izy -y) I4.ly – Izy Pr Мy Formüller; a -- • Ox = Pr o - 0g0, + o
M-(zy.z-ly.y) + My,(1;.z – lzy-y) .ly - Izy Pr My Formüller; a =- Pr o - đạ0p + of < of , Ox + Oy Oort = 2 •Imaks = 2t dz + Øy + + Ty , 8 = aAT, 1=. Tc d'y M(x) Omaks.min 2 I dx? EI C MalanicoXe, epc EI Malanıpc El Tmaks = 1 y = ƏM -dx. Per aP VQ It L/2- L/2- AB beam is a W section drawn profile, P= 50 kN, L= 1,25 m ve E= 200 GPa since it is ( Iw= 26,9.106 mm4) b) Find the slope at A. a) Find the deflection at C. 0,0015 rad 0,886 mm 0,0012 rad 1,859 mm 0,0007 rad 0,554 mm 0,0009 rad 1,246 mm 0,0005 rad 0,378 mm
M-(Izy z-ly.y) + My(1z.z – lzy -y) Iz.ly - Izy Pr My Formüller; a-- Pr Og + dy oå - da0p + o < oỷ , Jort = Tmaks 2t 0z + Oy + Ty , E = aAT, T-. Tc d'y M(x) Omaks.min 2 I dx? El + Ty tc/D MalanicoXc, @pỊc El Malanipc EI Tmaks = aM Mp dx, Par = VQ %3D It Find the critical buckling load if the length of the 10 cm * 12 cm sectioned bar with both ends is 4 m? E = 1.106 N/cm2 92.45 kN 75.64 kN 88,82 kN 61,68 kN 53,52 kN
Guén く。 aiven Jeetor A = 508L t 202 4.3 k - 6 - 2.5 i - o4j • 6 C 5.1 t 3.7 k d a Ax B. c b) A.(BxC)
%3D Galcukle Ihe voiide re by an Icked gog going rom ARDiake A hoc for each b) ABG o AC direchlys ABE ->
Select one: O a. X=1,498 m O b. X 0,498 m O c. X=0,754 m O d. X 0,0498 m k = m x(t) 3EI 3EI 13 13 y(t)
Formül; TL 60 MPa T Tmaks = Gab = vQ Sab?c: 25 için g = g =(1- 0,630b/a) Cyab'G 45 MPa M;(Izy.z-ly.y) + M,1..z- Izy.y) My T It 2tA 100 MPa 100MPA +dy ¯dy cos20 + Tzysin20 , Jort = cos20 + Trysin20, dort - *Oy, tan28, - 2 2 O-dy sin20 + I cos20, Gmaksmin -2 2 Ty -- .min- 45 MPa 60 MPa Tmaks Find the maximum principal stress for the stress case given in the figure. O 80.00 MPa O 138,75 MPa O 129,24 MPa O 114,25 MPa O 49.24 MPa
What is… Vertical reaction at A: Horizontal reaction at A: Vertical reaction at B: Horizontal reaction at B:
Detanneined the magnitude. end direction ad Petanninel tha magnitude.and direction at farce iP if the Resullant ofip andc120NLis to be all6ON) vertio ce. Eメュー 120 N 120 120 SIMz5 S Px=20 SI25 = 50.71N -TEFY=Ry=R= lbo. 0-ニRy120 ces 25 > Pye51.25N
I need solutions to this question
R:37)