1. (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

2. (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

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

4. (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)

5. (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)

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

7. (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

8. 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.

12. 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 ;

13. b
FOR BOLT 1 So
= k
FOR BOLT OR 3
S – h = 2.75 – ”

14. 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

15. 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)

16. } } 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 x x 70= kips

17. BEARING STRENGTH OF BOLT 2
2d= 1.5”
= 2.4 x 0.75 x x 70= kips
TOTAL BEARING STRENGTH = 2 X X 47.25
= kips

18. 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 = ,

19. } } 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 = kips

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

21. 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

22. 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 = k
NUMBER OF BOLTS NEEDED= DESIGN LOAD
DESIGN STRENGTH
Ieb`ildcb`dp3`
BOLTS
USE 2