Question

Mo 2Mo A B 7 1/4 [in] pin support It roller support a a a 1 1/2 [in]

a. Draw a free-body diagram for the beam shown above and derive expressions for the support reactions at A and B

b. Draw internal force (shear and bending moment) diagrams.

c. If a = 10 ft and M0 = 200 ft-lb, use the dimensions of the beam cross-section, provided on the previous page, to compute the maximum flexural and shear stresses on the beam cross-section.

d. If the allowable bending stress is 925 psi and the allowable shear stress is 175 psi, find the maximum value of the moment Mo that can be supported by the beam when a = 10 ft

Answer 1

Mo ” go 1 a V Free body diagram. Support reactions can be calculated by using equilibrium conditions. VATVB BM SMA ve + x3 a ta mot No- ch ve X 3 a 3Mo - Mola -ve tve a t. I true - ve VA Mola SF E Fx HA

- VB bo Shear force ak point B; SFB (-9% Mola load bekares As there is no romain A point and B; value of SF antil point A same SFA- Mo мо a MB Bending moment at point + VB x a BM at right Моха Mo BM at * O left Mo + 2M + 2M BM at с + VB X aa night Moxa a tamo

в м. Чех да + Qчо + Mo left Mo BNA Ve xЗа + моt Mo мох3 с * Змә 2 Mə 5) Но В Hola (42 Mo/a SFD Mo f ко (-) ко Вмр

M. = 200 ft-lb Griven a = 10 F E formula ; According to pure bending plino I 9 2 = Fly f Fleure Stress II f = M/z Z=) section modalus bd² sln ; for rectangular Ž= 6 2 1,5 X 7, 25 3 13.14 in Maac. Flescura stessa Q00X1 *12*13.14 182.64 effline alar secton; Shear stress is For rectangalar maoci mum at neutral axis. 3 Cmax = 5 bd s=Shear force &l

200 20 lb .: Maximum shear stress ; Inox 3 x 1.5x7.25 2.75 e blin? # For allo coable bonding shoess = 925 psi M= fxz. =925x13 14 - 12 154.5 lbf-in M. 1012.875 Ft For allowable shear stress ; T = 175psi value of moment is 2 maocimum T marc 3 Tang て masc S bd

(Mola) 3 x . Emax 3 6d G= Mola from fig Given 10 ft Imax 125 psi ft = 12 incha 125 3 x 2 (N70 /10x12) 1.5x 7.25 125 3 Ma 2 Glo M 108 750 lb-in Max Bonding moment; M = 9062,566ft