Statics review • Example: a) Find reactions at supports A and C. Hinge connection 1300 N/m B b) Find internal forces and moment, N, V, and M at Y=1.0 m on column AB. C 680 N.m Roller 3.7 m Y X A 2.75 m 1 Statics review Global FBD Solution a) Reactions at A , C Hinge connection - Draw the FBD 1300 N/m B C 680 N.m 4 unknowns: MA , Ax , Ay , Cy Cy 3 equilibrium equations for the global FBD: M A =0, F x = 0, F y 3.7 m =0 Not enough equations for finding unknowns → draw FBD for substructures! Y X Ay A MA Ax 2.75 m 2 Statics review - Draw the FBD for members AB and BC Global FBD Hinge connection 1300 N/m B By Bx 1300 N/m B C Bx By Note: the internal forces must be in opposite directions! Ay 680 N.m Cy FBD for BC FBD for AB 3.7 m C Cy 3.7 m 680 N.m Y X Ay A MA Ax 2.75 m A MA Ax 2.75 m 3 Statics review - Equilibrium equations for BC: M B = 0 ➔ -680 + 2.75 Cy = 0 ➔ Cy = 247 N BC Fx = 0 ➔ Bx + Cx = 0 ➔ Bx = 0 Fy = 0 ➔ -By + Cy = 0 ➔ BY = 247 N For moment equation, CCW is assumed positive. FBD for BC By B C Bx 680 N.m Cy 2.75 m 4 Statics review - Equilibrium equations for AB: M A = 0 ➔ -2405(23 × 3.7 ) + MA= 0 ➔ MA= 5932 Nm AB Fx = 0 ➔ Ax + 2405 = 0 ➔ Ax = -2405 Fy = 0 ➔ By + Ay = 0 ➔ AY = -247 N For moment equation, CCW is assumed positive. FBD for AB 1300 N/m Bx=0 Bx=0 By=247 N 2405 N Note: statically equivalent concentrated forces for triangularly distributed force: By=247 N ≡ 3.7 m 2 × 3.7 𝑚 3 Ay A MA Ax Ay A MA Ax 5 Statics review Global FBD Solution b) Internal forces and moment, N, V, and M at Y=1.0 m on column AB. Hinge connection 1300 N/m B - Cut a section through the AB at Y=1 m, draw created FBD and write equilibrium : M AD D =0 ➔ 𝑀 + 𝑀A + Ax 1 + Fx = 0 ➔ V= -Ax - 1 F y 1 × 1300 2 3.7 = 0 ➔ N =-Ay= 247 1 × 1300 3.7 1300 1 1 3 M V D 680 N.m Cy × 1 = 0➔ M= -3585 Nm 1 = 0 ➔ V=2229 3.7 m Y X N 1m 1 1 × 2 3.7 C Ay A MA Ax 2.75 m Y X Ay A MA Ax 6 Shear and Normal Stress • Example: a) Find average shear stress in the plate b) Find the average compressive stress in the punch. punch Plate Solution a) Shear area in the plate: circumference of the hole times the thickness of the plate: As = 𝜋𝑑𝑡 = 𝜋 20 8 = 502.7 𝑚𝑚2 Average shear stress in the plate: 𝑃 110 × 1000 𝜏𝑎𝑣𝑒𝑟 = = = 219 MPa 𝐴𝑠 502.7 𝐴𝑝𝑢𝑛𝑐ℎ = 𝜋𝑑 2 /4 As = 𝜋𝑑𝑡 Solution b) Average compressive stress in the punch: 𝑃 110 × 1000 𝜎𝑐 = == = 350 MPa 𝐴𝑝𝑢𝑛𝑐ℎ 𝜋𝑑 2 /4 7