Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] U= 0.85 fracture path  ABC 2.5 [in] 3 [in] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 1.50 [in] A P 1.25 [in] Find the design strength, in pounds. [pause] In this problem, --- B D 1.25 [in] C L 6 x 6 x 1/2

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] U= 0.85 fracture path  ABC 2.5 [in] 3 [in] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 1.50 [in] A P 1.25 [in] a 6 by 6 by 1/2 L beam is subjected to a load P. The yield stress and ultimate stress ---- B D 1.25 [in] C L 6 x 6 x 1/2

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] U= 0.85 fracture path  ABC 2.5 [in] 3 [in] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 1.50 [in] A P 1.25 [in] of the steel are provided, as well as various dimensions --- B D 1.25 [in] C L 6 x 6 x 1/2

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] U= 0.85 fracture path  ABC 2.5 [in] 3 [in] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 1.50 [in] A P 1.25 [in] of the bolt hole size and their locations in the beam. [pause] The design strengh of the L beam, equals, --- B D 1.25 [in] C L 6 x 6 x 1/2

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] U= 0.85 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.50 [in] A P 1.25 [in] the smaller of the design strength of the beam yielding, and the design strength --- B D 1.25 [in] C L 6 x 6 x 1/2

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) resistance resistance of the beam fracturing. [pause] The resistance to yielding equals, --- to yielding to fracture

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 the resistance factor for yield type failures, which equals, 0.90, times, ----

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross the nominal resistance for a yield failure, which equals, the yield stress, ---

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross f y, times gross area of steel, A gross. The problem statement provides the yield stress as --- yield stress gross area

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross 36,000 pounds per inches squared, and we can look up --- yield stress gross area

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] L 6 x 6 x 1/2 fracture path  ABC φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross the gross area for a 6 by 6 by 1/2 L beam which equals, ---- yield stress gross area

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] Agross = 5.75 [in2] L 6 x 6 x 1/2 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross 5.75 inches squared. After plugging in these 3 variables, --- yield stress gross area

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] Agross = 5.75 [in2] L 6 x 6 x 1/2 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross the design strength resisting a yield failure, equals, --- yield stress gross area

Find: φ * Rn [lb] dhole=19/16 [in] fy= 36,000 [lb/in2] tdamaged=1/32 [in] fu= 58,000 [lb/in2] Agross = 5.75 [in2] L 6 x 6 x 1/2 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) φyield = 0.90 Rn,yield = fy * Agross 1.863*105 pounds. [pause] Next we’ll compute the design strength --- yield stress gross area φyield * Rn,yield=1.863 * 105 [lb]

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] resistance for a fracture failure. [pause] The resistance factor, for a fracture failure, --- to fracture

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] resistance equals, 0.75. [pause] The nominal resistance for a fracture failure, equals, ---- to fracture

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] the ultimate stress of the steel, times the net area, times U. From the problem statement, we know ---

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] the ultimate stress equals, 58,000 pounds per square inch. The variable U is the ---

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] reduction coefficient which accounts for inefficiencies of transferring loads, and equals, --- reduction coefficient

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] 0.85. [pause] The net area of the L beam, equals, --- reduction coefficient

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] the gross area minus the area from the holes, plus, --- s2 Anet=Agross-Aholes + t * reduction 4 * g coefficient

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] an area term accounting for diagonals in the failure path. From before, we already know the gross area, --- s2 Anet=Agross-Aholes + t reduction * 4 * g coefficient gross area of diagonal area holes term

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] equals 5.75 inches squared. The area of the bolt holes equals, the number of bolt holes, --- s2 Anet=Agross-Aholes + t reduction * 4 * g coefficient gross area of diagonal area holes term

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) fu * Anet * U 1.863 * 105 [lb] times the beam thickness, t, times, the quantity, the diameter of a single bolt hole plus 2 times the damaged thickness around the hole. s2 Anet=Agross-Aholes + t * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged)

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC Looking back at our L beam, the problem stated, that should a fracture failure occur, --- s2 Anet=Agross-Aholes + t * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged)

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC it would occur along Path A B C. Where Path A B C includes 3 holes, --- s2 Anet=Agross-Aholes + t * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged)

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC A, B and C. [pause] The thickness of a 6 by 6 by 1/2 L beam, --- s2 Anet=Agross-Aholes + t * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged) 3

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC equals, 1/2 of an inch. [pause] And the hole diamter and damaged thickness --- s2 Anet=Agross-Aholes + t t=1/2 [in] * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged) 3

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC are provided in the problem statement. [pause] The area from holes computes to --- s2 Anet=Agross-Aholes + t t=1/2 [in] * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged) 3

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] A P 1.25 [in] B D U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC 1.875 inches squared. [pause] The last term we need to solve for is, --- s2 Anet=Agross-Aholes + t t=1/2 [in] * 4 * g Aholes= # of holes * t * (dhole+2 * tdamaged) Aholes= 1.875 [in2] 3

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P 1.25 [in] U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC the diagonal term, s squared t, over 4 times g. For the given failure path, --- s2 Anet=Agross-Aholes + t * 4 * g Agross=5.75 [in2] diagonal Aholes= 1.875 [in2] term

Find: φ * Rn [lb] + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P 1.25 [in] U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC there are 2 diagonal segments, A B and B C. Therefore there will be 2 diagonal terms, --- s2 Anet=Agross-Aholes + t * 4 * g Agross=5.75 [in2] diagonal Aholes= 1.875 [in2] term

Find: φ * Rn [lb] + + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC in the equation. [pause] After plugging in the stagger lengths, and --- sAB2 sBC2 Anet=Agross-Aholes + + t t * * 4*gAB 4*gBC Agross=5.75 [in2] diagonal Aholes= 1.875 [in2] terms

Find: φ * Rn [lb] + + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC gauge lengths for these two paths, as well as the thickness of the beam, --- sAB2 sBC2 Anet=Agross-Aholes + + t t * * 4*gAB 4*gBC Agross=5.75 [in2] diagonal Aholes= 1.875 [in2] terms

Find: φ * Rn [lb] + + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC t, both diagonal terms equal, --- sAB2 sBC2 Anet=Agross-Aholes + + t t * * 4*gAB 4*gBC Agross=5.75 [in2] diagonal Aholes= 1.875 [in2] terms

Find: φ * Rn [lb] + + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 fracture path  ABC 0.625 inches squared. [pause] And the net area, for the beam, equals --- sAB2 sBC2 Anet=Agross-Aholes + + t t * * 4*gAB 4*gBC Agross=5.75 [in2] 0.625 [in2] 0.625 [in2] Aholes= 1.875 [in2]

Find: φ * Rn [lb] + + 2.5 [in] 3 [in] 1.50 [in] t=1/2 [in] A P U= 0.85 1.25 [in] C dhole=19/16 [in] tdamaged=1/32 [in] L 6 x 6 x 1/2 Anet= 5.125 [in2] 5.125 inches squared. [pause] Now we can solve for the nominal resistence --- sAB2 sBC2 Anet=Agross-Aholes + + t t * * 4*gAB 4*gBC Agross=5.75 [in2] 0.625 [in2] 0.625 [in2] Aholes= 1.875 [in2]

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] of the beam resising a fracture-type failure, which equals, --- fu * Anet * U Anet= 5.125 [in2]

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 2.527 * 105 [lb] 2.527*105 pounds. [pause] And when multiplied by the resistence factor, --- fu * Anet * U fu= 58,000 [lb/in2] Anet= 5.125 [in2] U= 0.85

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 2.527 * 105 [lb] also called the strength reduction factor, the design strength against fracture, --- fu * Anet * U fu= 58,000 [lb/in2] Anet= 5.125 [in2] U= 0.85

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 1.895 * 105 [lb] equals, 1.895*105 pounds. [pause] Since the failure mode with the smaller design strength governs --- 0.75 2.527 * 105 [lb] * φfracture Rn,fracture

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 1.895 * 105 [lb] the design strength of the beam, the design strength, equals, ---

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 1.895 * 105 [lb] 1.863 * 105 pounds. [pause] φ * Rn=1.863 * 105 [lb]

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 1.895 * 105 [lb] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 When reviewing the possible solutions, --- φ * Rn=1.863 * 105 [lb]

Find: φ * Rn [lb] dhole=19/16 [in] fu= 58,000 [lb/in2] tdamaged=1/32 [in] Agross = 5.75 [in2] φfracture=0.75 U= 0.85 φ * Rn=min (φyield * Rn,yield, φfracture * Rn,fracture) 1.863 * 105 [lb] 1.895 * 105 [lb] A) 1.55*105 B) 1.86*105 C) 1.90*105 D) 2.23*105 the answer is B. φ * Rn=1.863 * 105 [lb] answerB