Presented by: Civil Engineering Academy

Presented by: Civil Engineering Academy

Structural Design and Material Properties of Steel Presented by: Civil Engineering Academy

Advantages 1. High strength per unit length resulting in smaller dead loads and smaller member. 2. Uniformity of properties with time, except when corroded, resulting in simple analysis in its design 3. Elastic and ductile, hence, can sustain more additional loads before fracture/failure 4. Durable when properly maintained Disadvantages 1. Susceptibility to corrosion 2. Reduced strength when exposed to high temperature 3. Heat Conductor 4. Susceptibility to buckling 5. Fatigue under stress reversals or repeated loadings

Beams Dead and Live Loads Columns Trusses Buckling/Axial/Combined Stresses Bending Deflection Shear Everything is under the Structural Mechanics Section

Yield stress, Fy, is that unit tensile stress at which the stressstrain curve exhibits a well-defined increase in strain without an increase in stress. Tensile strength, Fu, is the largest unit stress that the material achieves in a tension test. Modulus of Elasticity, E, is the slope of the initial straightline portion of the stress-strain diagram. For all structural steels, it is usually taken as 29,000 ksi for design calculations. Ductility is the ability of the material to undergo large inelastic deformations without fracture. Toughness is the ability of a specimen to absorb energy and is characterized by the area under a stress-strain curve.

S shape W shape HP shape M shape C shape L shape MC shape ST shape WT shape MT shape Pipe (P) shape

HSS (sq.) HSS (rect.) PL shape Built up sections

CERM Ex. 58.1

A g = bt Eq. 60.4 CERM A g = gross area Fiqure 60.2 CERM Tension member with Unstaggered Rows of Holes

Example A long tensile member is constructed by connecting two plates as shown. Each plate is 1/2 in x 9 in, A572 grade 50 steel with an ultimate strength of 65 ksi. The fasteners are 1/2 in diameter bolts. Determine the gross area of the plate. Solution: The gross area of the plate is: A g = (0.5 in)(9 in) = 4.5 in²

If rivets or bolts are used, holes must be punched or drilled in the member. As a result, the member s cross-sectional area diameter d h t at the connection is reduced. The load-carrying capacity A of the member may also be reduced, depending on the size and location of the holes. The net area actually available to resist tension is illustrated in Fig. 60.2(b).

b n = net width d h = hole diameter (d+1/8 for errors in hole punching) b =gross width s ² /4g = correction factor s = distance between holes

Fiqure 60.4 Tension member with Uniform Thickness and Unstaggered Holes A n = b n t Eq. 60.6 CERM A n = net area b n = net width t = thickness Eq. 60.7 CERM

Solution: Determine the net area for the plate shown. Assume the holes are for 7/8 dia. bolts. Compute the net Area for desired paths ½ Path 1-1: An = (0.5 in)(8 in)-(2 holes)(0.5 in)(7/8 in+1/8 in buffer) + 0.5( 2² ) = 3.13 in² 4(4) Path 2-2: An = (0.5 in)(8 in) 2 holes(0.5 in)(7/8 in + 1/8 in buffer)+0.5( 5² ) = 3.78 in² 4(4) Path 3-3: An = (0.5 in)(8 in) 1 hole(0.5in)(7/8 in +1/8 in buffer) = 3.5 in² 8

Use simple mechanics of materials equations to analyze individual connectors, but use the AISC design aids for groups of connectors. AISC Sec. J can be used to design weld groups, or the welds can be calculated. See more in the CERM Ch. 45 and 66 for connectors and welding.

1. Check tension on Gross Area 2. Check tension on Net Area 3. Check shear on rivets or bolts 4. Check bearing on rivet/plate 5. Check combined shear and tearing

A plate with a width of 12 in and a thickness of 0.787 in is to be connected to two plates of the same width and half the thickness by 1 in diameter rivets. The rivet holes are 0.0787 in larger than the rivet diameter. The plate is A36 steel with F y = 35.97 ksi and F u = 72.52 ksi. The rivets are A502, Grade 2, hot driven rivets with allowable shearing stress of 21.76 ksi. Determine the maximum load P that can be applied without exceeding the allowable stresses. Ignore block shear. Assume that deformation at bolt hole is a design consideration. Use LRFD.

Diameter of hole = 1 in + 0.0787 = 1.0787 in By tension on gross area of main plate: P = A g F t, where F t = 0.60 F y P = (12 in)(.787 in)(0.60)(35.97 ksi) P = 203.82 kips By tension on net area of main plate P = A n F t, where F t = 0.50 F u A n = [12 in 2(1.0787)]0.787 = 7.75 in 2 0.85A g ( = 8.027 in 2 ) P = 7.75(0.50)(72.52 kips) P = 282.015

By shearing of rivets (double shear) P/4 = 2(A V )(F V ), where Fv = 21.76 ksi for A502, Grade 2, hot driven rivets A v = π(1) 2 /4 = 0.785 in 2 P = 8(0.785)(21.76 ksi) P = 136.65 kips By bearing of rivet on main plate: P/4 = A p F p, where F p = 1.2 F U A p = diameter of rivet x thickness of plate P = 4(1.0787)(.787)(1.2)(72.52 ksi) P = 489.55 kips therefore the safe P = 136.65 kips

Fiqure 60.5 CERM Block Shear Failures LRFD Allowable F t = = 0.3F u ASD Allowable F t = F u / 2.0 Allowable F v = 0.5F u Allowable F v = 0.6 F u / 2.0 P = F v A v + F t A t

Example The 12 in wide, 0.315 in thick steel plate is attached to another plate by three 1 in diameter A490 bolts whose threads are not excluded from shear plane. The plate is A36 with F y = 35.97 ksi and F u = 59.47 ksi. Determine the safe shear block load P which the connection can resist. Solution: P = A v F v + A t F t F v = 0.30F u and F t = 0.50 F u A v = 2(2 1.063/2)(0.315) = 0.925 in 2 A t = [6.3 in 2(1.063)](.315) = 1.315 in 2 P = 0.925in (0.30)(59.47 ksi) + 1.315 in 2(0.50)(59.47 ksi) P = 94.71 kips

LRFD (Load Resistance and Factor Design) Tensile Strength: ASD (Allowable Stress Design) Design tensile strength = t P n Allowable tensile strength = P n / Ω t t = resistance factor P n = nominal tensile strength Ω t = safety factor P n = nominal tensile strength

Tension yielding in the gross section (A g ) LRFD (Load Resistance and Factor Design) ASD (Allowable Stress Design) P n = F y A g, ; = 0.9 P n = F y A g ; Ω t = 1.67 Tensile rupture in the net section P n = F u A e ; t = 0.75 P n = F u A e ; Ω t = 2.0 Where A g = gross area A e = effective area A n = net area F u = specified minimum tensile strength

Nominal tensile strength: Pin-connected Tensile Rupture On The Net Effective Area LRFD (Load Resistance and Factor Design) ASD (Allowable Stress Design) P n = 2tb eff F u ; = 0.75 Shear Rupture On The Effective Area P n = 0.60 F u A sf ; t = 0.75 P n = 2tb eff F u ; Ω t = 2.0 P n = 0.60 F u A sf ; Ω t = 2.0 Where A sf = 2t(a + d/2) t = Thickness d = pin diameter a = shortest distance from edge of the pinhole to the edge of the member measured parallel to the direction of the force F u = specified minimum tensile strength

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