Calculating Truss Forces Method of Joints
Forces Compression 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.
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?
Static Determinacy 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 B D C Did you notice the two pinned connections? F D = 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
Statically Determinate B D F D = 5 lb Is the truss statically determinate now? truss is considered statically determinate when the static equilibrium equations can be used to find the reactions on that structure. Try It C 2J = M + R
Static Determinacy Example Each side of the main street bridge in Brockport, NY has 19 joints, 35 members, and three reaction forces (pin and roller), making it a statically determinate truss. 2 ( ) J = M + R 2 19 = 35 + 3 What if these numbers were different? 38 = 38
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. Do you remember the Cartesian coordinate system? vector that acts to the right is positive, and a vector that acts to the left is negative.
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 CY B force that causes a clockwise moment is a negative moment. force that causes a counterclockwise moment is positive moment. F D contributes a negative moment because it causes a clockwise moment about. R Cy contributes a positive moment because it causes a counterclockwise moment around. R x R y D 3. ft 7. ft 5 lb Σ M = F (3. ft ) + R (1. ft ) = D Cy 5 lb (3. ft) + R (1. ft) = Cy 15 lb ft+ R (1. ft) = Cy C R Cy R (1. ft) = 15lb ft Cy R Cy = 15lb
Sum the y Forces to Find R y We know two out of the three forces acting in the y-direction. By simply summing those forces together, we can find the unknown reaction at point. Please note that a negative sign is in front of F D because the drawing shows the force as down. R x R y B D 5. lb Σ F y = FD + RCy + Ry = 5. lb + 15. lb + R y = 35. lb + R y = Ry C 15. lb = 35. lb
Sum the x Forces to Find x Because joint is pinned, it is capable of reacting to a force applied in the x-direction. R x B D C However, since the only load applied to this truss (F D ) has no x-component, R x must be zero. 35. lb 5. lb Σ = F x 15. lb
Method of Joints 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. B s forces are solved, update free body diagrams. Use correct magnitude and sense for subsequent joint free body diagrams.
Method of Joints Truss Dimensions B 4. ft R x θ 1 θ 2 D 3. ft 7. ft C R y R Cy 5lb
Method of Joints Using Truss Dimensions to Find ngles B 4. ft θ 1 θ 2 D 3. ft 7. ft tanθ = opp 1 adj tanθ = 4. 1 3. θ 1 ft ft = tan 14. 3. C 4. ft θ 1 = 53.13
Method of Joints Using Truss Dimensions to Find ngles tanθ = opp 1 adj tanθ = 4. 1 7. θ 1 ft ft = tan 14. 7. B 4. ft θ 1 θ 2 D 3. ft 7. ft C 4. ft θ = 29.745 1
Method of Joints Draw a free body diagram of each pin. B R x 53.13 29.745 D C R y 5lb R Cy 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.
Method of Joints Where to Begin Choose the joint that has the least number of unknowns. Reaction forces at joints and C are both good choices to begin our calculations. B BD R x D D CD C 35lb R y 5lb 15lb R Cy
Method of Joints 35 lb 437.5 lb B 53.13 D D Σ F Y = 35lb + B sin 53.13 = B sin 53.13 = 35lb B = R y + B = y 35lb sin 53.13 B = 438 lb
Method of Joints Update the all force diagrams based on B being under compression. B BD R x = D D CD C R y = 35lb 5lb R Cy = 15lb
Method of Joints B 35 lb = 437.5 lb 53.13 D 262.5 lb Σ FX = 437.5 lb cos53.13 + D = D D = 437.5 lb cos53.13 = 262.5 B + D = lb x
Method of Joints R Cy Σ FY = + BC = y 32.33 lb 15 lb + BC sin 29.745 = BC C BC sin 29.745 = 15 lb CD 29.745 15lb BC = 15 lb sin 29.745 BC = 32 lb
Method of Joints Update the all force diagrams based on BC being under compression. B BD R x = D D CD C R y = 35lb 5lb R Cy = 15lb
Method of Joints Σ FX = BCx CD = BC = 32.33 lb CD 29.745 262.5 lb C 15 lb 32.33 lbcos29.745 CD = CD = 32.33 lbcos29.745 CD = 262.5 lb
Method of Joints 5lb BD Σ F Y = D BD F D = BD 5lb = 5lb BD = 5lb