Question

4 kN/m BA D. 5 m (26 points) Calculate the reactions at A and B. Determine the internal loadings that act on the section at point E. Be sure to clearly indicate the direction of your assigned internal loadings in your FBD. For 2 points extra credit, determine the diameter of the double-sheared pin B to the nearest mm if the failure shear stress is 126 MPa. Assume a factor of safety of 1.75. -1.5 m - -1.5 m -3 m 3 m

Answer 1

ny (+) 4 KN m →X (+) 5m E i 1.5 m 1.5mi 3m 3m © = cos? cost () 53.13' BC is two force member. Let for be the force in two force member. Consider FBO of AB, (2x4x3) kN (2x4x3) KN UVL to point load 4 kn m ® Ax 3mil 3m FBC Ау K im im am 2m sign convention G (+) (1) EMF I. Fec cosex 6-(4x4x3]x2 - {{x4x3]xy = c 6 For COS 53.13' – 36 = 0

10 kn FBC Reaction at B, RB = FBC Kg lo KN (2) ЕҒх is Ax-FBc sine =0 Ax losin 53.13 =0 is Ax = 8 kN (A) (3) Efy :: Ay -[x4x3]-[ $x4x5] + foc coso =0 .: Ay – 6-6 + 10 C0553.13' =0 6 kN (1) Ay & Rescultant Reaction force. at A, RAJAR + Ay = 5 8² + 6² .. To kn RA

E Internal force at Section and taking consider section XX through point E FBD of left part of section xx (4X1.5XW) 4 EN WE X WE = 2 EN m UVL to 7 А. point load D NE Ax- WE AE 4 AD THE 1.5m "Ау 0.5m : WE 1.5 WE X 4 3 WE = 2 KN FBD Let VE, NE and Me be the internal shear force, Internal normal force and internal moment at E. Abbly equilibrium equationy, (1) Efy Ax-NE = 0 NE = AX 8 KN NE 8 kn (e) со .. (2) Efy Ay-( 5x1.58 WE) - VE 6-( 5x1.5x 2]-VE VE 4.5 kN (4)

0:50 (3) EME .: Me - Ay X1.5+ (4x1.57 WEIX 0:5 ME - 6x1.5 +1 1x1,5x2]x0.5 ME 8.25 knom (Anti-clockwise) Diameter of double sheared pin B (d) failure shear stress, Tp=126 MPa factor of safety, F.0.5 = 1-75 :. Allowable shear stress, Tallow If. F.o.s. Talion = 126 72 MPa 1.75 Shear force at bin B, = RB RB Tailow Area of shear pin :- Callow RB 2x1T x d? 72 lox 103 2X Ix d² d 9.403 mm