EGR 106 – Project Description (cont.) Analysis of truss structures A simple example Factor of Safety calculations This week’s assignment

A function, analyze_truss.m, has been developed and is available for download from the course web page: Input variables: joint_def (joint definitions) and member_def (member definitions) Output variables: disp (joint displacements) and stress (member stresses) Truss Analysis [displacements,stress]=analyze_truss(joint_def,member_def)

Truss Analysis – A simple example C x = 100 N C y = -50 N 300 mm 150 mm

A simple example – joint definitions C x = 100 N C y = -50 N 300 mm 150 mm joint_def=[0, 0, 1, 0, 0; 50, 50, 0, 0, 0; 150, 150, 0, 100, -50; 250, 50, 0, 0, 0; 300, 0, 3, 0, 0; 200, 0, 0, 0, 0; 100, 0, 0, 0, 0]; Column 1 – x coordinates of joints Column 2 – y coordinates of joints Column 3 – bcdof, where bcdof=0 for unconstrained bcdof=1 for X&Y–displacement=0 bcdof=2 for X-displacement=0 bcdof=3 for Y-displacement=0 Column 4 – force in X-direction Column 5 – force in Y-direction

A simple example – member definitions C x = 100 C y = mm 150 mm member_def=[1, 2, 10, 10; 2, 3, 10, 10; 3, 4, 10, 10; 4, 5, 10, 10; 5, 6, 10, 10; 6, 7, 10, 10; 7, 1, 10, 10; 2, 7, 10, 10; 7, 3, 10, 10; 3, 6, 10, 10; 6, 4, 10, 10]; Column 1 – first joint in member Column 2 – second joint in member Column 3 – width of member Column 4 – thickness of member

Joint Displacments displacments = C x = 100 N C y = -50 N 300 mm 150 mm Simple example – displacement results } X & Y displacements – joint 1 (A) } X & Y displacements – joint 2 (B) } X & Y displacements – joint 3 (C) } X & Y displacements – joint 4 (D) } X & Y displacements – joint 5 (E) } X & Y displacements – joint 6 (F) } X & Y displacements – joint 7 (G)

C x = 100 C y = mm 150 mm Member Stresses stress = } stress in member AB } stress in member BC } stress in member CD } stress in member DE } stress in member EF } stress in member FG } stress in member AG } stress in member BG } stress in member CG } stress in member CF } stress in member DF Simple example – stress results

The stress results give the force per unit area in each member (N/mm 2 ) If the stress is positive, the member is being stretched in tension If the stress is negative, the member is being compressed These stresses need to be compared to material strengths To quantify the relation between stress and strength, we introduce the Factor of Safety (FOS) Strength Analysis

For members in tension (stress > 0), the factor of safety is the ratio of the material’s tensile strength, S T, to the stress in that member: FOS = S T / stress For example, if stress = 100 N/mm 2 and S T =200 N/mm 2, then FOS =2 (i.e. the material is twice as strong as needed) Factor of Safety – members in tension

For members in compression (stress < 0), two failure modes must be considered: – Failure may occur when the stress reaches the compressive strength – Or, failure may occur due to buckling of long slender members Hence, we must compute two Factor of Safety values The smaller of these values is the critical Factor of Safety to be used in designing the truss Factor of Safety - members in compression

For members in compression (stress 0): FOS 1 = - S C / stress Factor of Safety – members in compression (cont.)

The second compression factor of safety is the ratio of the material’s buckling strength to the stress in the member. The buckling strength is inversely proportional to the length of the member and is defined as a buckling coefficient divided by the member length, K B /L 2. Hence, the second factor of safety for compression members is given by: FOS 2 = - ( K B /L 2 ) / stress Factor of Safety – members in compression (cont.)

The critical failure mode is determined by comparing FOS 1 and FOS 2. The smaller value is the one to be used in designing the truss FOS = minimum (FOS 1, FOS 2 ) Factor of Safety – members in compression (cont.)

If a member has a stress of zero, then its Factor of Safety in infinite: FOS =  Factor of Safety – zero stress members

Overall Factor of Safety After determining the Factors of Safety for each member, the overall Factor of Safety is the smallest of the individual member Factors of Safety. If the Overall Factor of Safety is less than one, than the truss is expected to fail when the forces are applied. The larger the Overall Factor of Safety, the safer the design.

This Week’s Assignment* 100 mm For the truss shown, write a Matlab program that: Determines the joint displacements Determines the member stresses Computes a factor of safety for each member Computes an overall factor of safety Displays the joints and elements in a figure window Displays the deformed truss in a 2 nd figure window * See assignment handout for more details