ID NUMBER TO USE = 12-7234-499

NUMBER OF ABSENCES TO BE USE = 1

Following is a simple truss.

1. Analyze the structure using loads computed as instructed.

2. Choose 2 tension members and propose 3 possible design sections considering all 3 possible failures under ASD and LRFD. This totals to 6 proposed sections with each section checked against 3 possible failure modes under 2 design philosophies.

3. Each member must be have a proposal for a Channel section, an S section and a Plate section. Further, in using your Channel and S sections, use the webs for your first tension member and use the flanges for your second tension member.

4. Material strengths are to be assumed and specified.

5. You are to assume your own bolt diameter. For the plate sections and when using the web of your C and S sections, you are to use 2 columns of 3 bolts. Pitch and gage distances must at least be 1.5 times that of your bolt diameter. When using the flange of the C section, 1 column of 3 bolts shall be located at each of the flanges, pitch distance is to be maintained at 1.5 times the bolt diameter. When using the flange of the S sections, 1 column of 3 bolts each must be located in each of the 4 flange elements. Connections must be oriented in such a way that the columns are parallel to the direction of the load and no bolt must be located at a distance of less than 1.5 times the diameter from all free edges.

6. Prepare a drawing for all your final sections and indicate each section's capacity against the three modes of failure.

6 ft 6 ft -4 at 9 ft (m)P P P = y(1 + z), y = first 2 digits of your ID number and z = last 3 digits of your ID number divided by 1000; x = last 2 digits of your ID number divided by 100. If x is higher than z, use x instead of z. i.e. ID No: 15-321 - 046; 46100 = 0.46; x = 0.46 i.e. ID no: 15 - 321 - 046; y = 15, z = 46500 = 0.092, x is higher, use x. w = 15(1 + 0.46) = 21.9 kN/m n = the first two digits of your ID number divided by 10 with maximum value of 2. i.e. n = 15/10 = 1.5 m = your number of absences 6 ft 6 ft -4 at 9 ft (m)P P P = y(1 + z), y = first 2 digits of your ID number and z = last 3 digits of your ID number divided by 1000; x = last 2 digits of your ID number divided by 100. If x is higher than z, use x instead of z. i.e. ID No: 15-321 - 046; 46100 = 0.46; x = 0.46 i.e. ID no: 15 - 321 - 046; y = 15, z = 46500 = 0.092, x is higher, use x. w = 15(1 + 0.46) = 21.9 kN/m n = the first two digits of your ID number divided by 10 with maximum value of 2. i.e. n = 15/10 = 1.5 m = your number of absences