alculating Truss Forces alculating Truss Forces Principles of Engineering 22 Forces ompression body being squeezed Tension body being stretched Truss truss is composed of slender members joined together at their end points. They are usually joined by welds or gusset plates. opyright 2
alculating Truss Forces Simple Truss simple truss is composed of triangles, which will retain their shape even when removed from supports. Pinned and Roller Supports pinned support can support a structure in two dimensions. roller support can support a structure in only one dimension. Solving Truss Forces ssumptions: ll members are perfectly straight. ll loads are applied at the joints. ll joints are pinned and frictionless. Each member has no weight. Members can only experience tension or compression forces. What risks might these assumptions pose if we were designing an actual bridge? opyright 2 2
alculating Truss Forces Static eterminacy statically determinate structure is one that can be mathematically solved. 2J = M + R J = Number of Joints M = Number of Members R = Number of Reactions Statically Indeterminate Each pin connection contributes TWO reaction forces F = 5 lb truss is considered statically indeterminate when the static equilibrium equations are not sufficient to find the reactions on that structure. There are simply too many unknowns. Try It 2J = M + R 2(4) 5 + 4 Statically eterminate Is the truss statically determinate now? F = 5 lb truss is considered statically determinate when the static equilibrium equations can be used to find the reactions on that structure. Try It 2J = M + R 2(4) = 5 + 3 opyright 2 3
alculating Truss Forces Static eterminacy Example Each side of the main street bridge in rockport, NY has 9 joints, 35 members, and three reaction forces (pin and roller), making it a statically determinate truss. 2J = M + R ( ) 2 9 = 35 + 3 38 = 38 What if these numbers were different? Equilibrium Equations Σ M = The sum of the moments about a given point is zero. Equilibrium Equations Σ F x = The sum of the forces in the x-direction is zero. o you remember the artesian coordinate system? vector that acts to the right is positive, and a vector that acts to the left is negative. opyright 2 4
alculating Truss Forces Equilibrium Equations Σ F y = The sum of the forces in the y-direction is zero. vector that acts up is positive, and a vector that acts down is negative. Using Moments to Find R Y force that causes a clockwise moment is a negative moment. force that causes a counterclockwise moment is positive moment. F contributes a negative moment because it causes a clockwise moment about. R x R y 3. ft 7. ft 5 lb M = F (3. ft)+ R y (. ft)= 5 lb (3. ft)+ R y (. ft)= R y R y contributes a positive moment because it causes a counterclockwise moment around. 5 lb ft+ R y (. ft)= R y (. ft)=5 lb ft R y =5 lb Sum the y Forces to Find R y We know two out of the three forces acting in the y-direction. y simply summing those forces together, we can find the unknown reaction at point. 5. lb 5. lb Please note that a 5. lb+5. lb+ R y = negative sign is in front of 35. lb+ R y = F because the drawing shows the force as down. R y =35. lb R x R y F y = F + R y + R y = opyright 2 5
alculating Truss Forces Sum the x Forces to Find x ecause joint is pinned, it is capable of reacting to a force applied in the x-direction. However, since the only load applied to this truss (F ) has no x-component, R x must be zero. R x 35. lb 5. lb F x = 5. lb R x = Use cosine and sine to determine x and y vector components. θ ssume all members to be in tension. positive answer will mean the member is in tension, and a negative number will mean the member is in compression. s forces are solved, update free body diagrams. Use correct magnitude and sense for subsequent joint free body diagrams. Truss imensions 4. ft R x θ θ 2 3. ft 7. ft R y 5lb R y opyright 2 6
alculating Truss Forces Using Truss imensions to Find ngles 4. ft θ θ 2 3. ft 7. ft tanθ = opp adj tanθ = 4. ft 3. ft θ = 4. tan 3. θ = 53.3 4. ft Using Truss imensions to Find ngles tanθ = opp adj tanθ = 4. ft 7. ft θ = 4. ft θ θ 2 3. ft 7. ft tan 4. 7. 4. ft θ = 29.745 raw a free body diagram of each pin. R x 53.3 29.745 R y 5lb R y Every member is assumed to be in tension. positive answer indicates the member is in tension, and a negative answer indicates the member is in compression. opyright 2 7
alculating Truss Forces Where to egin hoose the joint that has the least number of unknowns. Reaction forces at joints and are both good choices to begin our calculations. R x 35lb R y 5lb 5lb R y 35 lb 437.5 lb 53.3 R y Σ = F Y + = 35lb + sin53.3 = sin53.3 = 35lb 35lb = sin53.3 = 438 lb y Update the all force diagrams based on being under compression. R x = R 35lb 5lb y = 5lb R y = opyright 2 8
alculating Truss Forces Σ FX = = 437.5 lb 53.3 262.5 lb 35 lb + = 437.5 lb cos53.3 + = = 437.5 lb cos53.3 = 262.5 lb x Σ FY = 32.33 lb 5 lb + sin 29.745 = sin 29.745 = 5lb 29.745 R y 5lb = 5 lb sin 29.745 = 32 lb + = y Update the all force diagrams based on being under compression. R x = R 35lb 5lb y = 5lb R y = opyright 2 9
alculating Truss Forces = 32.33 lb 29.745 262.5 lb 5 lb x Σ FX = 32.33 lbcos29.745 = = 32.33 lbcos29.745 = 262.5 lb = 5lb 5lb Σ = F Y F = 5lb = = 5lb opyright 2