//////////////////////////////// QUESTION: WHY DO THEY USE THE WHOLE 3.25 KIPS F

Question

//////////////////////////////// QUESTION: WHY DO THEY USE THE WHOLE 3.25 KIPS FOR THE FORCE AND NOT HALF IT FOR EACH ITEM LIKE THE PIN AND LINK IF THEY ARE USING THE ULTIMATE TENSILE FORCE \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

I DONT WANT ANSWER ONLY EXPLANATION PLEASE

0 in. PROBLEM 1.53 l6 in Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section ¼ in, thick and 1 in. wide. and is made of a steel with an ultimate strength in tension of 60 ksi. The pins at C and F each have a -in. diameter and are made of a steel with an ultimate strength in shear of 25 ksi. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members. 2 kips SOLUTION Use member EFG as free body For 0 in 2kips FCF 3.25 kips Failure by tension in links CF. (2 parallel links) Net section area for 1 link. A = (b-d)t-a-)4)-0.125 in. F 2AoU()0.125)(60) - 15 kips Failure by double shear in pins =-d-=--|-| 0.196350 in Fu-2A = (2)(0.196350)(25)-9.8 175 kips Actual ultimate load is the smaller value. Fy - 9.8175 kips F.S. = -9.8175 F.S.-3.02 Factor of s FcF 3.25

Explanation / Answer

Force acting on each link is different. Maximum force acting on the link CF than the link BE. So to determine factor of safety maximum force is considered to prevent failure. So we use whole 3.25 kips for the force and not half of it.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.