Chapter 1: SIMPLE TRUSSES (Revision) 1.1 Definition A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. A truss is a structure composed of slender members joined together at their end points. Fig. 1 1.2. The Method of Sections The method of sections is a process used to solve for the unknown forces acting on members of a truss. The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane. To remain each section in equilibrium, the cut members will be replaced by forces equivalent to the internal load transmitted to the members. Each section may constitute of non-concurrent force system from which three equilibrium equations can be written. NB: Because we can only solve up to three unknowns, it is important not to cut more than three members of the truss. Depending on the type of truss and which members to solve, one may have to repeat Method of Sections more than once to determine all the desired forces. Example 1 Using the method of sections, determine the force in members BD, CD, and CE of the roof truss shown in Fig. 3 Fig. 3 Solutions Step 1: Sketch a free-body diagram of the structure. Step 2: Determine the reactions at the supports, using the two conditions for static equilibrium if they are required or necessary for further calculations. Step 3: Draw a section line through the structure, cutting those members in which the forces are required. A maximum of three members should be cut by a section line. Step 4: - The external forces on only one side of the section line must be considered. - Indicate this direction on the section line. - Indicate the assumed force directions in the members that have been cut by the section line. - Take moments about any joint on the structure, considering only the external forces and forces in the members that have been cut as indicated in steps 3. The selected joint must have only one unknown force causing a moment about that joint. This force may then be calculated using the second condition for static equilibrium. NB: Tie = Tension and Strut = Compression 1.3. The Method of Tensions Coefficients The tension coefficient for a member of a frame is defined as the pull or tension in that member is divided by its length. Consider the following truss member having coordinates of (x1, y1, z1) and (x2, y2, z2) and force F acting on it: Fig.4 Unit of tension coefficient (t) = kN/m Similarly, FAB sinΘ = tAB (y2-y1) - Force in any member in any direction is equal to the product of tension coefficient in that member and difference in the co-ordinates in that direction. - Using the above formula and the conditions ΣFx = 0, ΣFy = 0, ΣFz = 0 at any joints of the truss three equations can be formed and hence three unknown values of tension coefficient can be found. - After finding each of the tension coefficients in the truss it can be multiplied with its length forces in that particular member can be found. Example 2 Determine the forces in the members of the pin-jointed truss shown in Fig. 2. Fig. 5 The equations of equilibrium at joint A are Substituting the values of RA,H, RA,V and the joint coordinates into the above equations tAC = 2 kN/m and from Eq. (y) tAB(1.5−0) + tAC(0−0)+1=0 tAB= −0.67kN/m We can now proceed to joint B at which, since tBA (= tAB) has been calculated, there are two unknowns x - direction: tBA(xA−xB) + tBC(xC−xB) + tBD(xD–xB) +3 = 0 y - direction: tBA(yA−yB) + tBC(yC−yB) + tBD(yD−yB) = 0 Substituting the values of the joint coordinates and tBA in Eqs (x) and (y) we have, from Eq. (x) −0.67(0−0) + tBC(1.5−0) + tBD(1.5−0) + 3=0 which simplifies to (x): 1.5tBC + 1.5tBD + 3 = 0 and from Eq. (y): −0.67(0−1.5) + tBC(0−1.5) + tBD(1.5−1.5)=0 tBC= 0.67 kN/m Hence, from Eq. (y) tBD= −2.67 kN/m There are now just two unknown member forces at joint D. Hence, at D x - direction: tDB(xB−xD) + tDF(xF−xD) + tDC(xC−xD) = 0 y - direction: tDB(yB−yD) + tDF(yF−yD) + tDC(yC−yD) – 5 = 0 Substituting values of joint coordinates and the previously calculated value of tDB (= tBD) in Eqs (x) and (y) we obtain, from Eq. (x) −2.67(0−1.5) + tDF(3.0−1.5) + tDC(1.5−1.5)−5 = 0 tDF= –2.67kN/m and from Eq. (y) −2.67(1.5−1.5) + tDF(1.5−1.5) + tDC(0−1.5) = 0 tDC= –3.33kN/m The solution then proceeds to joint C to obtain tCF and tCE or to joint F to determine tFC and tFE; joint F would be preferable since fewer members meet at F than at C. Finally, the remaining unknown tension coefficient (tEC or tEF) is found by considering the equilibrium of joint E. Then, tFC=2.67kN/m, tFE=−2.67kN/m, tEC=0 The forces in the truss members are now calculated by multiplying the tension coefficients by the member lengths, i.e. TAB = tABLAB = −0.67×1.5 = −1 kN (compression) TAC = tACLAC = 2×1.5 = 3 kN (tension) TBC =tBCLBC = 2.12 m TBC = 0.67×2.12= 1.42kN (tension) Note that in the calculation of member lengths it is immaterial in which order the joint coordinates occur in the brackets since the brackets are squared. Also TBD = tBDLBD = −2.67×1.5= −4 kN (compression) TDF =−4 kN (compression) TDC =−5.0kN (compression) TFC = 5.67kN (tension) TFE = −4 kN (compression) TEC = 0