Lecture 5 – Page 1 of 9

Lecture 5 – Tension Members (cont.) Block Shear

Tension members may also fail in connections due to Block Shear. A failure plane along a path of connected bolts (or weld) exists, consisting of a combination of tension acting perpendicular to the line of force, and shear acting parallel to the line of force as shown below:

Lecture 5 – Page 2 of 9

From AISC p. 16.1-112, the available strength “Rn” for the limit state of block shear along a parallel shear plane(s) and perpendicular tension plane(s) is the following:

Rn = 0.6FuAnv + UbsFuAnt < 0.6FyAgv + UbsFuAnt

ASD Available block shear rupture strength =

nR

LRFD Available block shear rupture strength = Rn

where: = 0.75 (LRFD) and = 2.00 (ASD)

where: Anv = net area subjected to shear (in2) Ant = net area subjected to tension (in2) Agv = gross area subjected to shear (in2) Ubs = 1.0 if a single line of bolts parallel to load = 0.5 if multiple lines of bolts parallel to load = see AISC p. 16.1-352

Example 1 (ASD) GIVEN: A steel angle L6x6x½ using A36 steel is subjected to a service tensile load of 94 kips. REQUIRED: Determine the ASD allowable block shear rupture strength, and compare with the applied load as shown.

Lecture 5 – Page 3 of 9

Anv = net area subjected to shear (in2) = (Matl. Thickness)[Lv – (# bolts)(Bolt dia. + 1/8”)] = ½”[10” – (2.5 holes)(¾” + 1/8”)] = 3.91 in2 Ant = net area subjected to tension (in2) = (Matl. Thickness)[Lt – (# bolts)(Bolt dia. + 1/8”)] = ½”[2½” – (0.5 holes)(¾” + 1/8”)] = 1.03 in2 Agv = gross area subjected to shear (in2) = (Matl. Thickness)(Lv) = (½”)(10”) = 5.0 in2 Rn = Available block shear = 0.6FuAnv + UbsFuAnt = 0.6(58 KSI)(3.91 in2) + (1.00)(58 KSI)(1.03 in2) = 195.8 Kips Check if Rn < 0.6FyAgv + UbsFuAnt < 0.6(36 KSI)(5 in2) + (1.00)(58 KSI)(1.03 in2) < 167.7 Kips

ASD Available block shear strength =

nR

= 00.2

7.167

Kips

ASD Available block shear strength = 83.85 Kips Since 83.85 Kips < 94 Kips applied service load, it is NO GOOD

10” = length of shear plane = Lv

3½” 2½” = width of tension = Lt

Angle gage “g” see AISC p. 1-46

Lecture 5 – Page 4 of 9

Tension Rods and Bars

Threaded rods and bars are often used as exposed architectural tension members. Typically they are attached to the structure by means of welding or bolted clevises and bolts.

Roof support tension rods at the Patcenter, Princeton, NJ

The nominal tensile strength, Rn, of the cross-sectional area through a tension rod is expressed as follows:

Dun AFR 75.0

where: Fu = Ultimate stress (KSI) from AISC p. 2-39 AD = gross cross-section area of rod (in2) = from AISC Table 7-18 p. 7-83

ASD Available Tensile Strength =

nR where = 2.00

LRFD Available Tensile Strength = nR where = 0.75

Clevise

Turnbuckle – used to splice & tighten rods

Lecture 5 – Page 5 of 9

Example 2 (LRFD) GIVEN: A threaded tension rod is used to support a service dead load = 9 kips and a service live load = 23 kips. Use A36 steel. REQUIRED:

1) Determine the smallest-diameter threaded rod using LRFD. 2) Determine the smallest clevise to support the LRFD factored load. 3) Determine the smallest turnbuckle to support the LRFD factored load.

Step 1 – Determine LRFD tensile load:

From inspection, use Load Factor 2 from AISC p. 2-8:

Pu = 1.2D + 1.6L = 1.2(9 Kips) + 1.6(23 Kips) = 47.6 Kips

Load

Turnbuckle

Clevise

Rod

Lecture 5 – Page 6 of 9

Step 2 – Determine the smallest diameter threaded rod:

LRFD Available Tensile Strength = nR where = 0.75

= [0.75FuAD] Substitute Pu = 47.6Kips = [0.75FuAD]

Solve for AD:

]75.0[ u

uD F

PA

)]58(75.0[75.0

6.47

KSI

KipsAD

AD = 1.46 in2 From AISC p. 7-83:

Select db = "8

31 Diameter bolt → Gross area = 1.49 in2 > 1.46 in2

Step 3 – Determine the smallest suitable clevise:

From AISC Table 15-3 p. 15-14: Use No. 4 Clevise → LRFD avail. strength = 52.5 Kips > 47.6 Kips

Step 4 – Determine the smallest suitable turnbuckle:

From AISC Table 15-5 p. 15-16: Use 1½” Dia. turnbuckle → LRFD avail. strength = 52.5 Kips > 47.6 Kips

NOTE: Must use 1½” Dia. rod since turnbuckle strength is not as much as rod strength!!!

Lecture 5 – Page 7 of 9

Sag Rods

Sag rods are commonly used for light-gage metal roof systems to LATERALLY support roof purlins and wall girts as shown below:

20’-0

”

See “Example 3” Truss

Truss

Lecture 5 – Page 8 of 9

Example 3 (ASD) GIVEN: A steel building has trusses spaced 20’-0” apart. Sag rods are to be used to laterally support the roof purlins. Assume the purlins and sag rods contribute 2.0 PSF of dead load, in addition to a roof dead load of 13 PSF and a roof snow load of 27 PSF on the horizontal projection of the roof. Assume 5/8” diameter A36 steel for the sag rods. REQUIRED: 1) Determine the actual ASD tensile load on the top inclined sag rod. 2) Determine if the 5/8” diameter sag rod is acceptable for the top inclined sag

rod.

Step 1 – Determine ASD tensile load on top inclined sag rod:

a) Dead load of purlins & sag rods = 2.0 PSF

b) Dead load of roofing = 16.0 PSF

c) Snow load acting on roof surface:

27 PSF

27 PSF PSF6.2510

3

Lecture 5 – Page 9 of 9

d) ASD uniform load acting on top inclined sag rod: w = (2.0 PSF)+(16.0 PSF)+(25.6 PSF) = 43.6 PSF

e) Component of load parallel to sag rod = PSFPSF 8.1310

16.43

f) Total tension load on sag rod:

Total Load = Trib. Area(load parallel to sag rod) = (10’-0”)(37.9 ft.)(13.8 PSF) Total tensile load = 4230 Lbs. = 4.23 KIPS

Step 2 – Determine if the 5/8” diameter sag rod is acceptable for the top inclined sag rod:

ASD Available Tensile Strength =

nR where = 2.00

= 00.2

75.0 Du AF

where: Fu = Ultimate stress (KSI) from AISC p. 2-39

AD = gross cross-section area of rod (in2) = from AISC Table 7-18 p. 7-83

ASD Available Tensile Strength = 00.2

)307.0)(58(75.0 2inKSI

ASD Available Tensile Strength = 6.68 KIPS > 4.23 KIPS OK