(a) (b) A325-N or A325-X or A490-N A490-X Shear plane Threads iNcluded in shear plane Threads eXcluded from shear plane (a) (b) A325 N = 48 ksi A490 N = 60 ksi A325 X = 60 ksi A490 X = 75 ksi

(b) (a) (c) (d) d + d + OVS STD d + d + d + 2.5d SSL LSL STD: Standard hole SSL: Short-slotted hole OVS: Oversized hole LSL: Long-slotted hole Note: Sizes given are for d = , , and

(a) (b) 2.5d d + Slotted holes transfer vertical shear. Horizontal freedom permits rotation. Load Direction LSL (a) (b)

(a) (b) (c) Bolt Washer Nut Shank length Thread length Hexagonal nut head A325 Nominal Diameter, d F H Bolt length Bolt Nut Washer (a) (b) (c)

(a) (b) (d) (c) (e) Bolt in Single Shear Bolt in Double Shear Shear plane (Faying surface) (a) Bolt in Single Shear (b) (d) (c) Bolt in Double Shear (e)

Faying surfaces Washer Washer Nut Threads Head Hole clearance Grip Stick through Bolt length

(b) (a) Projected Area For min. edge distance in line of stress, see AISC Spec Sect. J 3.5 Projected Area (b) (a) Effective thickness in bearing

The top picture shows the bearing failure of the plate The top picture shows the bearing failure of the plate. Note that bearing failure ony occurs in steel plates and nothing happens to bolts. In bearing failure tha bolts travel in the weak plate meking a circular hole as an elliptical one. The bottom picture shows the tension failure of bolts. The tension failure can result in fracture as shown in first second or fourth bolt or as a yielding failure and/or thread failure as shown in the third bolt.

Check all spacing and edge-distance requirements. The tension member is a PL x 6. it is connected to a -inch-thick gusset plate with -inch-diameter bolts. Both components are of A36 steel. Check all spacing and edge-distance requirements. Compute the nominal strength in bearing. PL X 6 PL ; OK 36 ksi ; OK 3 2 1 36 ksi A36 steel ;

b FOR BOLT 1 So = 20.18 k FOR BOLT 2 OR 3 S – h = 2.75 – 1.813”

DESIGN BEARING STRENGTH = 2 X 34.26 + 20.18 = 88.54 NOMINAL BEARING STRENGTH = kips NOTE: NOMINAL (FAILURE) BEARING STRENGTH = DESIGN (SAFE) BEARING STRENGTH 0.75 WE USE = 0.75 RESISTANCE (SAFETY) FACTOR FOR BEARING STRENGTHS

Check all spacing and edge-distance requirements. The tension member shown in Figure 1 is a PL x of A242 steel. It is connected to a -inch-thick gusset plate (also A242 steel) with –inch-diameter bolts. Check all spacing and edge-distance requirements. Compute the nominal strength in bearing. 2 1 (WEAK PL)

} } CODE MIN ‘S’= 2.66d= 2.66 X 0.75= 2.0” OK ACTUAL ‘S’= 3” CODE MIN = 1.25” [TABLE J3.4] ACTUAL = 1.5” OK BEARING STRENGTH OF BOLT 1 2d= 2 x 0.75” = 1.5” So =1.2 x 1.094 x x 70= 34.46 kips

BEARING STRENGTH OF BOLT 2 2d= 1.5” = 2.4 x 0.75 x x 70= 47.25 kips TOTAL BEARING STRENGTH = 2 X 34.46 + 4 X 47.25 = 257.9 kips

Check all spacing and edge-distance requirements. A C8 x 18.7 is to be used as a tension member. The channel is bolted to a -inch gusset plate with -inch-diameter, A307 bolts. The tension member is A572 Grade 50 steel and the gusset plate is A36. Check all spacing and edge-distance requirements. Compute the design strength based on shear and bearing. Compute the allowable strength based on shear and bearing. A 307 BOLTS A572 GR 50 2 1 t = 0.487” G.P. IS WEAK C8 x 18.7 t = ,

} } CODE MIN ‘S’= 2.66 x = 2.33” OK ACTUAL ‘S’= 3” CODE MIN [T J3.4] = 1.5” ACTUAL = 2” OK BEARING STRENGTH OF BOLT 1 2d= 1.75” = 2.4 x x x 58 = 45.68 kips

BEARING STRENGTH OF BOLT 2 2d= 1.75” =2.4 x x x 58= 45.68 kips TOTAL NOMINAL BEARING STRENGTH= 6 X 45.68= 274.1 kips TOTAL NOMINAL SHEAR STRENGTH= SHEAR STRENGTH X SHEAR AREA = 24 X 6 X X = 86.58 k DESIGN STRENGTH= 0.75 X 86.58 = 64.9 kips

Determine the number of -inch-diameter, A325 bolts required, based on shear and bearing, along line a-b in Figure 2. The given loads are service loads. A36 steel is used. Assume that the bearing strength is controlled by the upper limit of 2.4dt . Use LRFD. DIAMETER A 325-N BOLTS PL ALL a D = L = b G.P. IS WEAK 2L4 x 4 x

NUMBER OF BOLTS NEEDED= DESIGN LOAD DESIGN STRENGTH NOMINAL SHEAR STRENGTH OF ONE BOLT = 48 X X X 2 = 42.4 k NOMINAL BEARING STRENGTH OF ONE BOLT = 2.4 X X X 58 = 39.15 k NUMBER OF BOLTS NEEDED= DESIGN LOAD DESIGN STRENGTH Ieb`ildcb`dp3` BOLTS USE 2