Handbook of Civil Engineering Calculations

FIGURE 37 The case of no-stiffeners corresponds to k - 5. Check the original assumptions for doubling the shear strength To dou ...

FIGURE 38 Try a single stiffener plate Using the plate in Fig. 39, which is 3.5 x 0.25 in (8.89 x 0.635 cm), the moment of iner ...

Working with the "Flexural Strength Parameters" table in the Appendix of the AISC LRFD Specification, the compactness of the bea ...

Mnx = Mpx-(Mpx-MJ^-^} \Ar-Ap/ The properties Sx and Zx of the cross section in Fig. 40 must now be calculated. Sx = —, where c = ...

Determine the welded section flexural strength Determining flexural strengths, we obtain tfff_p^_ 361^x0691^ =4107kip.ft(5569k ...

FINDING THE LIGHTEST SECTION TO SUPPORTA SPECIFIED LOAD Find the lightest W8 in A36 steel to support a factored load of 100 kips ...

*«/„ - *W, = No, - °-^9 °X2^X36kSi - 73.4 kip-ft (99.5 kNm) which is also the tabulated value for (J)^Mp f°r a W8 x 28 in the Be ...

ex = e cos 45° = 6 in x 0.707 = 4.2 in (10.7 cm) ey = e sin 45° = 6 in x 0.707 = 4.2 in (10.7 cm) M -=*-• = 1001 S 8 ^ 4 " 2 in ...

Mx = 200 kip-ft (271 kNm); My = O; single-curvature bending (i.e. equal and opposite end moments); and no transverse loads along ...

Pu = required compressive strength (based on the factored loads), kips <$><Pn = design compressive strength, kips (kN) ...

Original frame = nonsway frame + sway frame for Mn, fa M" FIGURE 43. Frame models for Mnt and M 7 ,. where K= 1.0, / is the mome ...

tions due to end moments M 1 and M 2 are in opposite directions, then M 1 IM 2 is negative; otherwise M 1 IM 2 is positive. (2) ...

For a W14 with KL = 15 ft m = 1.0 and U= 1.5. Substituting in Eq. [8.2], we obtain PUJB =^8 °0 + 200 x 2.0 + O = 1200 kips (5338 ...

Determine the design flexural strength To determine <feMn* (me design flexural strength), refer to the Load Factor Design Se ...

Check minimum wall thickness of pipe, Fig. 45: t > D -*- = 6.625 in / 36 ksi f. = 0.083 in (0.21 cm) V 8£ V^8 x 29,000 ksi v ...

= 51.4 ksi (354. IMPa) The modified modulus of elasticity for composite design is Em = 29,000 ksi + 0.4 x 3267 ksi x 2 ^ 8 ' 9 . ...

DETERMINING DESIGN COMPRESSIVE STRENGTH OF COMPOSITE COLUMNS Determine the design compressive strength of a W8 x 40 (A36 steel) ...

Determine the modified yield stress and modulus of elasticity Determine Fmy and Em: Ar Ac Fmy = Fy + clFyr — +c 2 f^—- AS As wh ...

(^ 0 Pn = 768 kips for this case is also tabulated on p. 4-73 of the AISC LRFD Manual) The 768-kip design strength is considerab ...

Afy = 11.8 in 2 x 36 ksi = 425 kips (1890 kN) Vh = 425 kips (189OkN) Find the number of shear studs required The nominal streng ...