CHAPTER 5 Statically Determinate Plane Trusses

TYPES OF ROOF TRUSS

ROOF TRUSS SETUP

OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse and calculate the FORCES in truss members To calculate the DEFORMATION at any joints

INTRODUCTION A truss is a structure of slender members joined at their end points by bolting or welding to gusset plate. Members commonly used: wooden struts, metal bars, angles or channels

ROOF TRUSSES Transmitting loading from roof to columns by means of series of purlins. Trusses used to support roofs are selected on the basis of the span, slope and roof materials. Metal Gusset Top Chord Truss Web Bottom Chord

ROOF TRUSSES

BRIDGE TRUSSES

ASSUMPTIONS IN DESIGN The members are joined together by smooth pins (NO friction) All loadings and reactions are applied at the joints The centroid for each members are straight and concurrent at a joint Each truss member acts as an axial force member. If the force tends to elongate tensile (T) If the force tends to shorten compressive (C)

SIGN CONVENTION Tensile (positive) Compressive (negative)

STABILITY & DETERMINACY In terms of stability, the most simple truss can be constructed in triangle using three members. This shape will provide stability in both x and y direction. Each additional element of two members will increase one number of joint

STABILITY & DETERMINACY There are 3 types of stable trusses: 1. Simple Truss 2. Compound Truss combination of two or more simple trusses together 3. Complex Truss one that cannot be classified as being either simple or compound

STABILITY & DETERMINACY Simple Truss

STABILITY & DETERMINACY Compound Truss

STABILITY & DETERMINACY Complex Truss

DETERMINACY Let consider a simple truss. There will be 6 unknown values: 3 internal member forces and 3 reactions m + R m + 3 And for every joint, 2 equilibriums can be written ( F x = 0 and F y =0) no rotation or moment at joint 2 j By comparing the total unknowns with total number of available equation, we can check the determinacy.

DETERMINACY The determinacy of truss should be checked internally and externally The external determinacy is given by: R = 3 (provided that the support reactions have no lines of action that are either concurrent or parallel) If R > 3 Statically indeterminate (external) R = 3 Statically determinate (external) R < 3 Unstable truss system

DETERMINACY The internal determinacy is given by m = 2j 3 (provided that the components of the truss do not form a collapsible mechanism) If m > 2j 3 Statically indeterminate (internal) m = 2j 3 Statically determinate (internal) m < 2j 3 Unstable truss system

EXAMPLE 1 Determine the stability and determinacy of the truss shown in the figure below.

Externally: EXAMPLE 1 Solution R = 3 ; R 3 = 3 3 = 0 OK Internally: m = 9, j = 6, 9 = 2(6) 3 = 9 OK Therefore, the truss is determinate (externally and internally)

EXAMPLE 2 Determine the stability and determinacy of the truss shown in the figure below.

Externally: EXAMPLE 2 Solution R = 3 ; R 3 = 3 3 = 0 OK Internally: m = 9, j = 6, 9 = 2(6) 3 = 9 OK Therefore, the truss is determinate (externally and internally)

EXAMPLE 3 Determine the stability and determinacy of the truss shown in the figure below.

EXAMPLE 3 Solution Externally: R = 4 ; R 3 = 4 3 = 1 1 degree redundant Internally: m = 10, j = 6, 10 > 2(6) 3 : 10 > 9 1 degree redundant Therefore, the truss is internally and externally indeterminate (1 degree redundant)

MEMBER FORCES There are several methods of calculating the member forces for the truss i. Method of Joints ii. Method of Sections iii. Method of Force Resolution

METHOD OF JOINTS Suitable to be used to determine all the member forces in the truss In this method, every joint will be analysed by drawing the Free Body Diagram, limiting the unknown values to TWO only. The selected joints must only consisted concurrent and coplanar forces Using the equilibrium of F x = 0 and F y =0, we can start and solve the problems.

EXAMPLE 4 Determine all the member forces for the given truss below. 100 kn B C E 6m A D F G 8m 150 kn 50 kn 3m 3m

EXAMPLE 4 Solution 1. Check the Stability and Determinacy Externally: R = 3 ; R 3 = 3 3 = 0 OK Internally: m = 11, j = 7, 11 = 2(7) 3 = 11 OK Therefore, truss is determinate (externally and internally)

EXAMPLE 4 Solution 2. Calculate the Reactions at the Support + M A = 0 100 (6) + 150 (8) + 50 (11) R G (14) = 0 R G = 167.9 kn ( ) + F y = 0 ; R A 150 50 167.9 kn ; R A = 32.1 kn ( ) + F x = 0 ; H A = 100 kn ( )

EXAMPLE 4 Solution 3. Analyse Every Joints At Joint A F AB 100 A F AD + F x = 0 ; 100 + FAD = 0 F AD = 100 kn (T) + F y = 0 ; F AB + 32.1 = 0 F AB = 32.1 kn (C) 32.1

EXAMPLE 4 Solution At Joint B 100 B F BC + F y = 0 ; 32.1 F BD (6/10) = 0 F BD = 53.5 kn (T) 32.1 F BD + F x = 0 ; F BC + 100 + 53.5 (8/10) = 0 F BC = 142.8 kn (C)

EXAMPLE 4 Solution At Joint C 142.8 C + F x = 0 ; 142.8 + F CE (3/ 18) = 0 F CE = 201.9 kn (C) F CD F CE + F y = 0 ; F CD ( 201.9) (3/ 18) = 0 F CD = 142.8 kn (T)

EXAMPLE 4 Solution At Joint D 142.8 53.5 F DE + F y = 0 ; 142.8 + 53.5 (6/10) 150 100 D F DF + F DE (3/ 18) = 0 F DE = 35.2 kn (C) 150 + F x = 0; 100 53.5 (8/10) + (-35.2)(3/ 18) + F DF = 0 F DF = + 166.7 kn (T)

EXAMPLE 4 Solution At Joint E 142.8 35.2 E F EF F EG + F x = 0 ; F EG (3/ 18) + 201.9 (3/ 18) + 35.2 (3/ 18) = 0 F EG = 237.1 kn (C) + F y = 0; F EF 201.9 (3/ 18) + 35.2 (3/ 18) ( 237.1)(3/ 18) = 0 F EF = 49.8 kn (T)

EXAMPLE 4 Solution At Joint F 49.8 + F x = 0 ; 167.7 + F FG = 0 F FG = 167.7 kn (T) 166.7 F F FG 50

EXAMPLE 4 Solution At Joint G (Checking) 237.1 166.7 G + F y = 0 ; 167.9 237.1 (3/ 18) = 0 OK! 167.9 + F x = 0; 167.7 + 237.1 (3/ 18) = 0 OK!

EXAMPLE 4 Solution Summary: Internal Member Forces Member Force AB 32.1 C AD +100 T BC 142.8 C BD +53.5 T CD +142.8 T CE 201.9 C Member Force DE 35.2 C DF +167.7 T EF +49.8 T EG 237.1 C FG +167.7 T

METHOD OF SECTIONS When only some of the member forces need to be calculated, it is suitable to use this method. However, it can also used to determine all the member forces in truss. The method of sections consists of cutting through the truss into two parts, provided that the unknown values are not more than three The unknown forces will be assumed to be either in tension or compression Three equilibriums ( F x = 0, F y = 0, M = 0) will be used to solve the problems.

EXAMPLE 5 Determine the member forces for BD, DE and CE. 100 kn A B D 2 2 m 1 80 kn C E F 40 kn 4m 4m 4m

EXAMPLE 5 Solution Stability and Determinacy Externally: R = 3 ; R 3 = 3 3 = 0 OK Internally: m = 9, j = 6, 9 = 2(6) 3 = 9 OK Therefore, the truss is determinate (externally and internally)

EXAMPLE 5 Solution Calculate the Reactions at the Support + M F = 0 (2/ 5)R A (12) + (2/ 5)R A (2) 100 (8) 40 (4) = 0 R A = 82.6 kn ( ) + F y = 0 ; R F + (2/ 5)(82.6) 100 40 = 0 ; R F = 66.2 kn ( ) + F x = 0 ; H F + (1/ 5)(82.6) 80 = 0 ; H F = 43.1 kn ( )

EXAMPLE 5 Solution Section 1: F DB D F DC F EC E F 43.1 M D = 0 40 kn 66.2 F EC (2) 43.1 (2) 66.2 (4) = 0 ; F EC = 175.5 kn (T) F y = 0 ; 40 + 66.2 F DC (2/ 20) = 0 ; F DC = 58.6 kn (T) F x = 0 ; F DB 58.6 (4/ 20) 175.5 + 43.1 = 0 ; F DB = 184.8 kn (C)

EXAMPLE 5 Solution F ED F FD Section 2: F EC E F 43.1 40 kn 66.2 M F = 0 ; F ED (4) 40 (4) = 0 ; F ED = 40 kn (T)

METHOD OF FORCE RESOLUTIONS This is an extended version from the method of joints Every single joint is carefully analysed by considering not more than two unknowns at each joints. In this method, we do not have to write all the equations and calculations. All member forces are solved directly on the diagram.

EXAMPLE 6 Determine the member forces of the truss 50 kn 100 kn 50 kn 20 kn B D F 1.5 m A C E G H 2m 2m 2m 2m

ZERO FORCE MEMBERS Truss analysis using method of joints can greatly be simplified if one can first determine those member that support no loading (zero force member) The zero-force members can be determine by inspection of the joints. Normally, there are two cases where zero-force member can be identified

ZERO FORCE MEMBERS Case 1: If only two members form a truss joint, and no external load or support is applied, the members must be zeroforce members Case 2: If three members form a truss joint for which two of the members are collinear, the third member will be a zeroforce member (provided no external load or support reaction acting at the joint).

DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS The deformation of statically determinate plane truss can be determined using Virtual Work Method Consider to determine vertical deformation at joint C Due to external loads, point C will deform producing

DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS D B F A C E G H Actual Structure

DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS Now, eliminate all external loads and assign 1 unit load (vertical) at joint C. This structure is known as Virtual Structure C Virtual Structure

DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS Both structures will then combined, thus producing the concept of work. Therefore, the external work: W = 1. Say P is the member force due to external loads and u is the member force due to unit load If we consider one of the truss members, having force of P, this member will produce certain deformation which can be calculated, given as: = PL/AE

DEFORMATION OF STATICALLY DETERMINATE PLANE TRUSS Through combination, the amount of internal work is given by = u PL AE According to Energy Work Method: External load = Internal Load 1 = u PL AE

SOLUTION PROCEDURE 1. Calculate the member forces for Actual Structure. 2. Eliminate all external loads and assign ONE (1) unit load in the same direction of deformation. Then, calculate the member forces of Virtual Structures. 3. The deformation at point C can be then calculated using: u PL AE

VIRTUAL WORK METHOD When a structure is loaded, its stressed elements deform. As these deformations occur, the structure changes shape and points on the structure displace. Work the product of a force times a displacement in the direction of the force

VIRTUAL WORK METHOD External Work when a force F undergoes a displacement dx in the same direction as the force. Internal Work when internal displacements δ occur at each point of internal load u. P = Work of External Load u δ Work of Internal Load

VIRTUAL WORK METHOD When a bar is loaded axially, it will deform and store strain energy u. A bar (as shown in the figure) subjected to the externally applied load P induces an axial force F of equal magnitude (F = P). If the bar behaves elastically (Hooke s Law), the magnitude of the strain energy u stored in a bar by a force that increases linearly from zero to a final value F as the bar undergoes a change in length dl.

VIRTUAL WORK METHOD From Hooke s Law: dl = PL AE L F P F F P x dl x P Δ

DISPLACEMENT OF TRUSSES We can use the method of virtual work to determine the displacement of a truss joint when the truss is subjected to an external loading, temperature change, or fabrication errors. When a unit force acting on a truss joint, and resulted a displacement of Δ, the external work = 1 Δ.

DISPLACEMENT OF TRUSSES Due to the unit force, each truss member will carry an internal forces of u, which cause the deformation of the member in length dl. Therefore, the displacement of a truss joint can be calculated by using the equation of: 1. u. dl Virtual loadings Real displacements

50 kn B STEPS FOR ANALYSIS D 20 kn A C 1 1. Place the unit load on the truss at the joint where the desired displacement is to be determined. The load should be in the same direction as the specified displacement; e.g. horizontal or vertical.

STEPS FOR ANALYSIS 2. With the unit load so placed, and all the real loads removed from the truss, use the method of joints or the method of sections and calculate the internal force in each truss member. Assume that tensile forces are positive and compressive forces are negative. 3. Use the method of joints or the method of sections to determine the internal forces in each member. These forces are caused only by the real loads acting on the truss. Again, assume tensile forces are positive and compressive forces are negative.

Member STEPS FOR ANALYSIS Virtual Force, u Real Force, P (kn) L (m) u.pl (kn.m) AB 0.601 50 1.8 2 CB -0.507-30 1.8-3 DB 0.515 20 1.8 5 Total 4 Apply the equation of virtual work, to determine the desired displacement. It is important to retain the algebraic sign for ach of the corresponding internal forces when substituting these terms into the equation.

STEPS FOR ANALYSIS If the resultant sum of displacement is positive, the direction is same as the unit load or vice-versa. When applying any formula, attention should be paid to the units of each numerical quantity. In particular, the virtual unit load can be assigned any arbitrary unit (N, kn, etc.).

EXAMPLE 6 Determine the vertical displacement of joint C of the steel truss shown in Figure. The cross-sectional area of each member is A = 300 mm 2 and E = 200 GPa. F E 3 m A B C D 3 m 3 m 3 m 20 kn 20 kn

EXAMPLE 6 Solution Calculate the Member Forces due to Virtual Force -0.333-0.471-0.471 0.333 1-0.943 0.333 0.667 0.667 0.333 1.0 0.667 Virtual Structure: Virtual force, u

EXAMPLE 6 Solution Calculate the Member Forces due to Actual Forces -20 kn -28.3 kn 0 20 kn 20 kn -28.3 kn 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn Actual Structure: Real force, N

EXAMPLE 6 Solution Calculate the total deformation Member Virtual force, u Real force, P (kn) L (m) u.pl (kn.m) AB 0.333 20 3 20 BC 0.667 20 3 40 CD 0.667 20 3 40 DE -0.943-28.3 4.24 113 FE -0.333-20 3 20 EB -0.471 0 4.24 0 BF 0.333 20 3 20 AF -0.471-28.3 4.24 56.6 CE 1 20 3 60 Σ 369.6

EXAMPLE 6 Solution Calculate the final deformation 1 = u dl 1 kn cv = u PL AE = 369.6 AE cv = 369.6 300 10 6 200 10 3 cv = 6.16 mm

DISPLACEMENT OF TRUSSES (Due to Temperature Changes & Fabrication Error) In some cases, truss members may change their length due to temperature. If α is the coefficient of thermal expansion for a member and ΔT is the change in its temperature, the change in length of a member is: 1 = u α T L 1 = External virtual unit load acting on the truss joint in the stated direction of Δ u = Internal virtual normal force in a truss member caused by the external virtual unit load Δ = External joint displacement caused by the temperature change α = Coefficient of thermal expansion of member ΔT = Change in temperature of member L = Length of member

DISPLACEMENT OF TRUSSES (Due to Temperature Changes & Fabrication Error) Occasionally, errors in fabricating the lengths of the members of a truss may occur. Also, in some cases truss member must be made slightly longer or shorter in order to give the truss a camber. If a truss member is shorter or longer than intended, the displacement of a truss joint from its expected position can be determined from direct application: 1 = u L 1 = External virtual unit load acting on the truss joint in the stated direction of Δ u = Internal virtual normal force in a truss member caused by the external virtual unit load Δ = External joint displacement caused by the fabrication errors ΔL = Difference in length of the member from its intended size as caused by a fabrication error

EXAMPLE 7 Determine the vertical displacement of joint C of the steel truss as shown in the Figure. Due to radiant heating from the wall, member AD is subjected to an increase in temperature of ΔT = +60 C. Take α = 1.08 10-5 / C and E = 200 GPa. The cross-sectional area of each member is indicated in the figure. 2.4 m Wall 1.8 m D 1200 mm 2 1200 mm 2 900 mm 2 A 1200 mm 2 C 300 kn 1200 mm 2 B 400 kn

EXAMPLE 7 Solution Calculate the Member Forces due to Virtual Force 1 1.0 0.75 0.75 1-1.25 0 0.75 0 Virtual force, u

EXAMPLE 7 Solution Calculate the Member Forces due to Actual Force 400 kn 600 kn 600 kn 300 kn 400 kn -500 kn 400 kn 300 kn 0 Real force, N 400 kn

EXAMPLE 7 Solution Both loads and temperature affect the deformation, therefore: 1 = u dl + u α T L 0.75 600 1.8 1 kn cv = 1200 10 6 200 10 6 + 1 400 2.4 1200 10 6 200 10 6 1.25 500 3 + 900 10 6 200 10 6 + 1 1.08 10 5 60 2.4 cv = 0.0193 m = 19.3 mm

SUMMARY Forces in Member 1. Method of Joints (F x, F y ) 2. Method of Sections (F x, F y, M) 3. Force Resolution (F x, F y ) Stability and Determinacy (External & Internal) If OK, calculate reactions (3 equilibriums) A+ Virtual Work Method Actual structure (P) Virtual structure (u) To determine vertical/horizontal displacement