Tarek Hegazy, Univ. of Waterloo 2 t / m 2 t 1m 2m Xa Ya Yb Xb Yc Step 1: Stability Check No. of Equilibrium Equations: 3 No. of Extra Conditions: 2 (two intermediate pins) Since the unknowns 5 = Equations (3 + 2) Then, Structure is Stable & Statically Determinate Assumed reactions No. of Unknown Reactions? 5 Draw the B.M.D. & the S.F.D.

Tarek Hegazy, Univ. of Waterloo 2m Yc 2 t / m 2 t 1m Xa Ya Step 2: Reactions Sign Convention =0  M +  X +  Y + Any point For the whole structure considering all forces including reactions 2m Yb Xb e d Start writing an equation with least number of unknowns: Start writing an equation with least number of unknowns:  M + Since (e) is an intermediate pin, then at (e) right side only = 0 Since (e) is an intermediate pin, then at (e) right side only = 0

Tarek Hegazy, Univ. of Waterloo  M + at (e), right side only = 0 2 t 2m Yb Xb e Xb. 0 = 0 +Xb. 0 = 0 +Yb. 4 +Yb. 4 Solve, Yb = +1 i.e., correct direction Solve, Yb = +1 i.e., correct direction Sign Force Distance

Tarek Hegazy, Univ. of Waterloo 2m Yc 2 t / m 2 t 1m Xa Ya 2m 1 Xb  M + Since (d) is an intermediate pin, then at (d) right side only = 0 Since (d) is an intermediate pin, then at (d) right side only = 0 Let’s write another equation: Let’s write another equation: d

Tarek Hegazy, Univ. of Waterloo  M + at (d), right side only = 0 2 t 2m 1 Xb d Xb. 1 = 0 +Xb. 1 = Solve, Xb = 0 Solve, Xb = 0 1m Sign Force Distance

Tarek Hegazy, Univ. of Waterloo 2m Yc 2 t / m 2 t 1m Xa Ya 2m 1 Now, = 0, +Xa - 0 = 0, or Xa = 0  X + 0  M + Since (d) is an intermediate pin, then at (d) left side only = 0 Since (d) is an intermediate pin, then at (d) left side only = 0 0 Let’s now write two equations to get the last two unknowns: Let’s now write two equations to get the last two unknowns: d

Tarek Hegazy, Univ. of Waterloo - Ya.4 - Ya.4 2m Yc 2 t / m Ya d  M + at (d), left side only = 0 Equivalent = 2 x 4 = 8 Equivalent = 2 x 4 = 8 - Yc.2 - Yc = = 0 Or, 4 Ya + 2 Yc = (1) Or, 4 Ya + 2 Yc = (1) Sign Force Distance

Tarek Hegazy, Univ. of Waterloo 2m Yc 2 t / m 2 t 1m Ya 2m 1 Now, then,  Y=0 + Solve (1) & (2), Ya = -1, Yc = 10 Opposite direction Same direction Solve (1) & (2), Ya = -1, Yc = 10 Opposite direction Same direction +Ya +Yc +1 -2x4 -2 = 0 +Ya +Yc +1 -2x4 -2 = 0 4 Ya + 2 Yc = (1) Ya + Yc = 9... (2) Ya + Yc = 9... (2) All Up All Down

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Let’s first define the beam sections with changes in load or beam’s Shape. Let’s first define the beam sections with changes in load or beam’s Shape. We are now ready to draw the B.M.D. and the S.F.D Now, we analyze each section separately, considering only one side of the structure. Now, we analyze each section separately, considering only one side of the structure. Final Reactions: Shear is Paralle to the section & Perpendicular to the beam Axial (i.e., Normal) force is parallel to the beam Section 2 Section 3

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Section Analysis: Analyze the right side or the left side, whichever has less calculation. Analyze the right side or the left side, whichever has less calculation. +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Section 1 Left Section 1 Left Section 1 B.M. = = 0 0 S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. 1 2 Section 2 B.M. = = - 6 Section 2 B.M. = = Section 2 Left Section 2 Left S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D Section 3 B.M. = = -6 Section 3 B.M. = = -6 Section 3 Left Section 3 Left S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D Section 4 B.M. = = 0 Section 4 B.M. = = 0 Section 4 Left Section 4 Left S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D Section 5 Left Section 5 Left Section 5 B.M. = 0 (same as section 4) Section 5 B.M. = 0 (same as section 4) S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Section 10 Right Section 10 Right Section 10 B.M. = = 0 Section 10 B.M. = = S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Section 9 Right Section 9 Right Section 9 B.M. = = +2 Section 9 B.M. = = S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Section 8 Right Section 8 Right Section 8 B.M. = = +2 Section 8 B.M. = = +2 S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D. Sections 7 & 6 Right Sections 7 & 6 Right Sections 7 & 6 B.M. = 0 Sections 7 & 6 B.M. = S S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m B.M.D w.L 2 /8 = 1 Now, we connect the moment values Now, we connect the moment values

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m S.F.D. Section 1 Left Section 1 Left Section 1 S.F. = -1 + Shear is a force perpendicular to the beam S

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 2 S.F. = = - 5 Section 2 S.F. = = - 5 Section 2 Left Section 2 Left S.F.D. -5 S +

Tarek Hegazy, Univ. of Waterloo m 2 t/m 2 t 1m 2m Section 3 S.F. = = +5 Section 3 S.F. = = +5 Section 3 Left Section 3 Left S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 4 S.F. = = +1 Section 4 S.F. = = +1 Section 4 Left Section 4 Left S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 5 Left Section 5 Left Section 5 S.F. = 0 Section 5 S.F. = 0 S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 6 Left Section 6 Left Section 6 S.F. = 0 Section 6 S.F. = 0 S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 10 Right Section 10 Right Section 10 S.F. = -1 Section 10 S.F. = -1 S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 9 Right Section 9 Right Section 9 S.F. = -1 Section 9 S.F. = -1 S.F.D S +

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Section 8 Right Section 8 Right S.F.D S + Section 8 S.F. = = +1

Tarek Hegazy, Univ. of Waterloo 2m 2 t/m 2 t 1m 2m Sections 7 Right Sections 7 Right Sections 7 S.F. = = +1 Sections 7 S.F. = = +1 S.F.D S

Tarek Hegazy, Univ. of Waterloo S.F.D. 2m 2 t/m 2 t 1m 2m Now, we connect the shear values Now, we connect the shear values

Tarek Hegazy, Univ. of Waterloo S.F.D. 2m 2 t/m 2 t 1m 2m w.L 2 /8 = B.M.D Final Answer