etermine the LRFD design strength of the member below, A 36 Steel, 7/8-in bolts:

Question

etermine the LRFD design strength of the member below, A 36 Steel, 7/8-in bolts: [20-points) 2 1 in 3 in 3 in 4 in 4 in 5. Select the lightest C section that will safely support the service tensile loads PD 65 k and PL 50 K·The member is to be 14 ft long and is assumed to have two lines of holes for 3/4-in bolts in the web. Assume that there are at least three holes in each line 3 in on center. Use A36 steel. [20- points] Select the lightest w12section available to support working tensile loads of PD = 120 k and PV 288 k.The member is to be 20 ft long and is assumed to have two lines of holes for 3/4-in 0 bolts i each flange. There will be at least three holes in each line 3 in on center. [20-points] 6,

Explanation / Answer

ASTM A36

Fy= 36 ksi

Fu= 58 ksi

L 6X3 1/2 X 3/8 PROPERTIES

Ag = 3.42 in2

rz =0.763 in

y(bar) 2.02 in

x (bar) =0.781 in

tensile yielding

Pn = Fy*Ag = 36*3. 42 = 123.120kips

LFRD

t = 0.9

t Pn= 0.9*123.12=110.8 kips

U= 1- X(BAR)/l = 1- .781/8 = 0.9023

An= Ag - (dh + 1/16) t = 3.42 -(7/8 +1/16)3/8 =3.068 in2

Ae = An *U= 3.068*0.9023=2.7686 in2

Pn= Fu * Ae = 58*2.7686 =160.58 kips

t = 0.75

t Pn= =0.95*160.58 =120.43 kips

Available tensile strenght is governed by the tensile yielding limit state

design strength is 110.8 kips

thanks

please do comments if any query.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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