Monther Dwaikat

1
68402 Structural Design of Buildings II61420
Design of Steel Structures62323 Architectural
Structures II
Tension Member Design

• Monther Dwaikat
• Assistant Professor
• Department of Building Engineering
• An-Najah National University

2
Table of Contents

• Typical Tension Members
• Introductory Concepts
• Design Strength
• Effective and Net Areas
• Staggered Bolted Connections
• Block Shear
• Design of Tension Members
• Slenderness Requirements

3
Tension Members

• Applications
• In bridge, roof and floor trusses, bracing
  systems, towers, and tie rods
• Consist of angles, channels, tees, plates, W or S
  shapes, or combinations

4
Typical Tension Members

• Tension chord in a truss

5
Typical Tension Members

• Cables

• Ties

6
Tension Members

• Commonly Used Sections
• W/H shapes
• Square and Rectangular or round HSS
• Tees and Double Tees
• Angles and double angles
• Channel sections
• Cables

7
Introductory Concepts

• Stress The stress in the column cross-section
  can be calculated as
• f - assumed to be uniform over the entire
  cross-section.
• P - the magnitude of load
• A - the cross-sectional area normal to the load
• The stress in a tension member is uniform
  throughout the cross-section except
• near the point of application of load
• at the cross-section with holes for bolts or
  other discontinuities, etc.

8
Design Strength
9
Introductory Concepts

• For example, consider an 200 x 10 mm. bar
  connected to a gusset plate loaded in tension
  as shown below in Fig. 2.1

20 mm hole diameter
200 x 10 mm plate
Fig. 2.1 Example of tension member
10
Introductory Concepts

• Area of bar at section a a 200 x 10 2000
  mm2
• Area of bar at section b b (200 2 x 20 ) x
  10 1600 mm2
• Therefore, by definition the reduced area of
  section b b will be subjected to higher
  stresses
• However, the reduced area therefore the higher
  stresses will be localized around section b b.
• The unreduced area of the member is called its
  gross area Ag
• The reduced area of the member is called its net
  area An

11
Steel Stress-strain Behavior

• The stress-strain behavior of steel is shown
  below in Fig. 2.2

E 200 GPa Fy 248 MPa
Fig. 2.2 Stress-strain behavior of steel
12
Steel Stress-strain Behavior

• In Fig. 2.2
• E - the elastic modulus 200 GPa.
• Fy the yield stress Fu - the ultimate stress
• ?y is the yield strain ?u the ultimate strain
• Deformations are caused by the strain ?. Fig. 2.2
  indicates that the structural deflections will be
  small as long as the material is elastic (f lt Fy)
• Deformations due to the strain ? will be large
  after the steel reaches its yield stress Fy.

13
Design Strength

• We usually determine the strength capacity of
  any structural element based on possible
  scenarios of failure!
• Possible failures of a tension member include
• Yield of the element
• Fracture of element
• The stress of axially loaded elements can be
  determined as
• The stress is therefore a function of the cross
  sectional area thus the presence of holes will
  change the stress.
• Bolted connections reduce the area of the cross
  section.

14
Design Strength

• A tension member can fail by reaching one of two
  limit states
• excessive deformation
• fracture
• Excessive deformation can occur due to the
  yielding of the gross section (for example
  section a-a from Fig. 2.1) along the length of
  the member
• Fracture of the net section can occur if the
  stress at the net section (for example section
  b-b in Fig. 2.1) reaches the ultimate stress Fu.
• The objective of design is to prevent these
  failure before reaching the ultimate loads on the
  structure (Obvious).
• This is also the load resistance factor design
  approach for designing steel structures

15
Load Resistance Factor Design

• The load resistance factor design approach is
  recommended by AISC for designing steel
  structures. It can be understood as follows
• Step I. Determine the ultimate loads acting on
  the structure
• The values of D, L, W, etc. are nominal loads
  (not maximum or ultimate)
• During its design life, a structure can be
  subjected to some maximum or ultimate loads
  caused by combinations of D, L, or W loading.
• The ultimate load on the structure can be
  calculated using factored load combinations. The
  most relevant of these load combinations are
  given below

16
Load Resistance Factor Design

• 1.4 D
• 1.2 D 1.6 L 0.5 (Lr or S)
  
• 1.2 D 1.6 (Lr or S) (0.5 L or 0.8 W)
  
• 1.2 D 1.6 W 0.5 L 0.5 (Lr or S)
  
• 0.9 D 1.6 W
• Step II. Conduct linear elastic structural
  analysis
• Determine the design forces (Pu, Vu, Mu) for
  each structural member

17
Load Resistance Factor Design

• Step III. Design the members
• The failure (design) strength of the designed
  member must be greater than the corresponding
  design forces calculated in Step II
• Rn - the calculated failure strength of the
  member
• ? - the resistance factor used to account for the
  reliability of the material behavior equations
  for Rn
• Qi - the nominal load
• ?i - the load factor used to account for the
  variability in loading to estimate the ultimate
  loading

18
Design Strength of Tension Members

• Yielding of the gross section will occur when the
  stress f reaches Fy.
• Therefore, nominal yield strength Pn Ag Fy
• Factored yield strength ?t Pn
• ?t 0.9 for tension yielding limit state

19
Design Strength of Tension Members

• Facture of the net section will occur after the
  stress on the net section area reaches the
  ultimate stress Fu
• Therefore, nominal fracture strength Pn Ae Fu
• Where, Ae is the effective net area, which may be
  equal to the net area or smaller.
• The topic of Ae will be addressed later.
• Factored fracture strength ?t Ae Fu
• where ?t 0.75 for tension fracture limit
  state

20
Net Area

• We calculate the net area by deducting the width
  of the bolts some tolerance around the bolt
• Use a tolerance of 1.6 mm above the diameter hole
  which is typically 1.6 mm larger than the bolt
  diameter
• Rule

b
21
Design Strength

• Tensile strength of a section is governed by two
  limit states
• Yield of gross area (excessive deformation)
• Fracture of net area
• Thus the design strength is one of the following

Load Effect
YIELD
FRACTURE

• The difference in the f factor for the two limit
  states represent the
• Seriousness of the fracture limit state
• The reliability index (probability of failure)
  assumed with each limit state

22
Important Notes

• Why is fracture ( not yielding) the relevant
  limit state at the net section?
• Yielding will occur first in the net section.
  However, the deformations induced by yielding
  will be localized around the net section. These
  localized deformations will not cause excessive
  deformations in the complete tension member.
  Hence, yielding at the net section will not be a
  failure limit state.
• Why is the resistance factor (?t) smaller for
  fracture than for yielding?
• The smaller resistance factor for fracture (?t
  0.75 as compared to ?t 0.90 for yielding)
  reflects the more serious nature consequences
  of reaching the fracture limit state.

23
Important Notes

• What is the design strength of the tension
  member?
• The design strength of the tension member will
  be the lesser value of the strength for the two
  limit states (gross section yielding net
  section fracture).
• Where are the Fy Fu values for different steel
  materials?
• The yield ultimate stress values for different
  steel materials are dependent on type of steel.

24
Ex. 2.1 Tensile Strength

• A 125 x 10 mm bar of A572 (Fy 344 MPa) steel is
  used as a tension member. It is connected to a
  gusset plate with six 20 mm. diameter bolts as
  shown below. Assume that the effective net area
  Ae equals the actual net area An compute the
  tensile design strength of the member.

20 mm hole diameter
200 x 10 mm plate
A572 (Fy 344 MPa)
25
Ex. 2.1 Tensile Strength

• Gross section area Ag 125 x 10 1250 mm2
• Net section area (An)
• Bolt diameter db 20 mm.
• Nominal hole diameter dh 20 1.6 21.6 mm
• Hole diameter for calculating net area 21.6
  1.6 23.2 mm
• Net section area An (125 2 x (23.2)) x 10
  786 mm2
• Gross yielding design strength ?t Pn ?t Fy Ag
• Gross yielding design strength 0.9 x 344 x
  1250/1000 387 kN

26
Ex. 2.1 Tensile Strength

• Fracture design strength ?t Pn ?t Fu Ae
• Assume Ae An (only for this problem)
• Fracture design strength 0.75 x 448 x 786/1000
  264 kN
• Design strength of the member in tension
  smaller of 264 kN 387 kN
• Therefore, design strength 264 kN (net section
  fracture controls).

27
Effective Net Area

• The connection has a significant influence on the
  performance of a tension member. A connection
  almost always weakens the member a measure of
  its influence is called joint efficiency.
• Joint efficiency is a function of
• material ductility
• fastener spacing
• stress concentration at holes
• fabrication procedure
• shear lag.
• All factors contribute to reducing the
  effectiveness but shear lag is the most
  important.

28
Effective Net Area

• Shear lag occurs when the tension force is not
  transferred simultaneously to all elements of the
  cross-section. This will occur when some elements
  of the cross-section are not connected.
• For example, see the figure below, where only one
  leg of an angle is bolted to the gusset plate.

29
Effective Net Area

• A consequence of this partial connection is that
  the connected element becomes overloaded the
  unconnected part is not fully stressed.
• Lengthening the connection region will reduce
  this effect
• Research indicates that shear lag can be
  accounted for by using a reduced or effective net
  area Ae
• Shear lag affects both bolted welded
  connections. Therefore, the effective net area
  concept applied to both types of connections.
• For bolted connection, the effective net area is
  Ae U An
• For welded connection, the effective net area is
  Ae U Ag

30
Effective Net Area

• The way the tension member is connected affects
  its efficiency because of the Shear Lag
  phenomenon
• Shear lag occurs when the force is transmitted to
  the section through part of the section (not the
  whole section)
• To account for this stress concentration in
  stress, an area reduction factor U is used

Bolted Connections
Welded Connections
Over stressed
Under stressed
31
Effective Net Area

• Where, the reduction factor U is given by
• U 1- 0.9 (4.7)
• - the distance from the centroid of the
  connected area to the plane of the connection
• L - the length of the connection.
• If the member has two symmetrically located
  planes of connection, is measured from the
  centroid of the nearest one half of the area.

32
Effective Net Area

• Increasing the connection length reduces the
  shear lag effect
• Some special cases govern bolted and welded
  connections

33
Effective Net Area

• The distance L is defined as the length of the
  connection in the direction of load.
• For bolted connections, L is measured from the
  center of the bolt at one end to the center of
  the bolt at the other end.
• For welded connections, it is measured from one
  end of the connection to other.
• If there are weld segments of different length in
  the direction of load, L is the length of the
  longest segment.

34
U for Bolted Connections
OR

• Two major groups of bolted connections
• Connections with at least three bolts per line
• W,M and S shapes and T cut from them connected in
  flange with
• All other shapes
• Connections with only two bolts per line

35
Ex. 2.2 Design Strength

• Determine the effective net area the
  corresponding design strength for the single
  angle tension member in the figure below. The
  tension member is an L 4 x 4 x 3/8 made from A36
  steel. It is connected to a gusset plate with 15
  mm diameter bolts, as shown in Figure below. The
  spacing between the bolts is 75 mm
  center-to-center.

L 4 x 4 x 3/8
db 15 mm
36
Ex. 2.2 Design Strength

• Gross area of angle Ag 1850 mm2 T 9.5 mm
• Net section area An
• Bolt diameter 15 mm.
• Hole diameter for calculating net area 15 3.2
  18.2 mm.
• Net section area Ag 18.2 x 9.5 1850 172.9
  1677.1 mm2
• is the distance from the centroid of the
  area connected to the plane of connection
• For this case is equal to the distance of
  centroid of the angle from the edge.
• This value is given in the section property
  table.
• 28.7 mm.

37
Ex. 2.2 Design Strength

• L is the length of the connection, which for this
  case will be equal to 2 x 75 150 mm.
• Effective net area Ae 0.809 x 1677.1 in2
  1357 mm2
• Gross yielding design strength ?t Ag Fy 0.9 x
  1850 x 248/1000 412.9 kN
• Net section fracture ?t Ae Fu 0.75 x 1357 x
  400/1000 407.1 kN
• Design strength 407.1 kN (net section
  fracture governs)
• (Lower of the two values)

38
Ex. 2.3 Design Strength

• Determine the design strength of an ASTM A992 W8
  x 24 with four lines if 20 mm diameter bolts in
  standard holes, two per flange, as shown in the
  Figure below. Assume the holes are located at the
  member end the connection length is 225 mm.
  Also calculate at what length this tension member
  would cease to satisfy the slenderness limitation
  in LRFD specification.

db 20 mm
75 mm 75 mm 75 mm
39
Ex. 2.3 Design Strength

• For ASTM A992 material Fy 344 MPa Fu 448
  MPa
• For the W8 x 24 section
• Ag 4570 mm2 d 201 mm.
• tw 6.2 mm. bf 165 mm.
• tf 10.2 mm. ry 40.9 mm.
• Gross yielding design strength ?t Pn ?t Ag Fy
  0.90 x 4570 x 344/1000 1414.9 kN
• Net section fracture strength ?t Pn ?t Ae Fu
  0.75 x Ae x 448
• Ae U An - for bolted connection
• An Ag (no. of holes) x (diameter of hole) x
  (thickness of flange)
• An 4570 4 x (203.2) x 10.2.
• An 3623 mm2

40
Ex. 2.3 Design Strength

• What is for this situation?
• is the distance from the edge of the flange
  to the centroid of the half (T) section

41
Special Cases for Welded Connections

• If some elements of the cross-section are not
  connected, then Ae will be less than An
• For a rectangular bar or plate Ae will be equal
  to An
• However, if the connection is by longitudinal
  welds at the ends as shown in the figure below,
  then Ae UAg
• Where, U 1.0 for L 2w
• U 0.87 for 1.5 w L lt 2 w
• U 0.75 for w L lt 1.5 w
• L length of the pair of welds w
• w distance between the welds or width of
  plate/bar

42
Ex. 2.3 Design Strength

• The calculated value is not accurate due to the
  deviations in the geometry
• But, U 0.90. Therefore, assume U 0.90

43
Ex. 2.3 Design Strength

• Net section fracture strength ?t Ae Fu 0.75 x
  0.9 x 3623 x 448/1000 1095.6 kN
• The design strength of the member is controlled
  by net section fracture 1095.6 kN
• According to LRFD specification, the maximum
  unsupported length of the member is limited to
  300 ry 300 x 40.9 12270 mm 12.27 m.

44
Special Cases for Welded Connections
45
Special Cases for Welded Connections

• For any member connected by transverse welds
  alone,
• Ae area of the connected element of the
  cross-section

46
U for Welded Connections
OR

• Two major groups of welded connections
• General case
• W,M and S shapes and T cut from them connected in
  flange with
• All other shapes
• Special case for plates welded at their ends
• Any member with transverse welds all around ONLY

47
Ex. 2.4 Tension Design Strength

• Consider the welded single angle L 6x 6 x ½
  tension member made from A36 steel shown below.
  Calculate the tension design strength.

42.4 mm
140 mm
48
Ex. 2.4 Tension Design Strength

• Ag 3720 mm2
• An 3720 mm2 - because it is a welded
  connection
• Ae U An
• 42.4 mm for this welded connection
• L 152 mm for this welded connection
• Gross yielding design strength ?t Fy Ag 0.9 x
  248 x 3720/1000 830 kN
• Net section fracture strength ?t Fu Ae 0.75 x
  400 x 0.72 x 3720/1000 803 kN
• Design strength 803 kN (net section fracture
  governs)

49
Design of Tension Members

• The design of a tension member involves finding
  the lightest steel section (angle, wide-flange,
  or channel section) with design strength (?Pn)
  greater than or equal to the maximum factored
  design tension load (Pu) acting on it.
• ? Pn Pu
• Pu is determined by structural analysis for
  factored load combinations
• ? Pn is the design strength based on the gross
  section yielding, net section fracture block
  shear rupture limit states.

50
Design of Tension Members

• For net section fracture limit state, Pn 0.75 x
  Ae x Fu
• Therefore, 0.75 x Ae x Fu Pu
• Therefore, Ae
• But, Ae U An
• U An - depend on the end connection.
• Thus, designing the tension member goes
  hand-in-hand with designing the end connection,
  which we have not covered so far.

51
Design of Tension Members

• Therefore, for this chapter of the course, the
  end connection details will be given in the
  examples problems.
• The AISC manual tabulates the tension design
  strength of standard steel sections
• Include wide flange shapes, angles, tee
  sections, double angle sections.
• The gross yielding design strength the net
  section fracture strength of each section is
  tabulated.
• This provides a great starting point for
  selecting a section.

52
Design of Tension Members

• There is one serious limitation
• The net section fracture strength is tabulated
  for an assumed value of U 0.75, obviously
  because the precise connection details are not
  known
• For all W, Tee, angle double-angle sections, Ae
  is assumed to be 0.75 Ag
• The engineer can first select the tension member
  based on the tabulated gross yielding net
  section fracture strengths, then check the net
  section fracture strength the block shear
  strength using the actual connection details.

53
Design of Tension Members

• Additionally for each shape, the code tells the
  value of Ae below which net section fracture will
  control
• Thus, for Grade 50 steel sections, net section
  fracture will control if Ae lt 0.923 Ag
• For Grade 36 steel sections, net section fracture
  will control if Ae lt 0.745 Ag
• Slenderness limits
• Tension member slenderness l/r must preferably be
  limited to 300 as per LRFD specifications.

54
Slenderness Requirements

• Although tension elements are not likely to
  buckle, it is recommended to limit their
  slenderness ratio to 300
• The slenderness limitation of tension members is
  not for structural integrity as for compression
  members.
• The reason for the code limitation is to assure
  that the member has enough stiffness to prevent
  lateral movement or vibration.
• This limitation does not apply to tension rods
  and cables.

55
Steps for Design of Tension Members

• Steps for design
• Calculate the load
• Decide whether your connection will be welded or
  bolted
• Assume U factor of 0.75
• Determine the gross area of the element
• Assume An 0.75 Ag

56
Steps for Design of Tension Members

• Steps for design
• Choose the lightest section with area little
  larger than Ag
• Calculate, Ag, An, U and Ae for the chosen
  section
• Check
• Check slenderness ratio

57
Ex. 2.7 Design of Tension Members

• Design a member to carry a factored maximum
  tension load of 350 kN.
• Assume that the member is a wide flange connected
  through the flanges using eight 20 mm diameter
  bolts in two rows of four each as shown in the
  figure below. The center-to-center distance of
  the bolts in the direction of loading is 100 mm.
  The edge distances are 40 50 mm as shown in the
  figure below. Steel material is A992

20 mm
50 mm
100 mm
40 mm
40 mm
Fy 344 MPa Fu 448 MPa
50 mm
100 mm
58
Ex. 2.7 Design of Tension Members

• Select a section from the Tables
• Ag 3501000/(0.9344) 1130 mm2.
• Assume U0.75
• Ag ? 3501000/(0.45448)1736 mm2
• Try W8x10 with Ag 1910 mm2.
• An 1910 4(23.25.2) 1427 mm2.
• x 24.6 mm (students have to compute it)
• U 1 24.6/100 0.754
• Gross yielding strength 591 kN, net section
  fracture strength 362 kN

59
Extra Slides
60
Block Shear

• For some connection configurations, the tension
  member can fail due to tear-out of material at
  the connected end. This is called block shear.
• For example, the single angle tension member
  connected as shown in the Fig. 2.3 below is
  susceptible to the phenomenon of block shear.
• For the case shown above, shear failure will
  occur along the longitudinal section a-b
  tension failure will occur along the transverse
  section b-c.
• AISC Specification on tension members does not
  cover block shear failure explicitly. But, it
  directs the engineer to the Specification on
  connections

61
Block Shear
Fig. 2.3 Block shear failure of single angle
tension member
62
Block Shear

• Block shear strength is determined as the sum of
  the shear strength on a failure path the
  tensile strength on a perpendicular segment.
• Block shear strength net section fracture
  strength on shear path net section fracture
  strength of the tension path
• OR
• Block shear strength gross yielding strength of
  the shear path net section fracture strength of
  the tension path
• Which of the two calculations above governs?

63
Block Shear

• ? Rn ? (0.6 Fu Anv UbsFu Ant) ? (0.6 FyAgv
  UbsFu Ant)
• ? 0.75
• Ubs 1.0 for uniform tensile stress 0.5 for
  nonuniform tensile stress
• Agv - gross area subject to shear
• Agt - gross area subject to tension
• Anv - net area subject to shear
• Ant - net area subject to tension
• Fu - ultiamte strength of steel
• Fy - yield strength of steel

64
Block Shear

• Failure happens by a combination of shear and
  tension.

Area failing by tension
Area failing by shear
Shear Fracture
Shear Yield
Tension Fracture
Tension Fracture
Failure Mode 2
Failure Mode 1

• The two possible failure modes shall be
  investigated

65
Ex. 2.5 Block Shear

• Calculate the block shear strength of the single
  angle tension member shown bellow. The single
  angle L 4 x 4 x 3/8 made from A36 steel is
  connected to the gusset plate with 15 mm diameter
  bolts as shown below. The bolt spacing is 75 mm
  center-to-center the edge distances are 40 mm
  50 mm as shown in the Figure below.

db 15 mm
66
Ex. 2.5 Block Shear

• Assume a block shear path calculate the
  required areas

50
db 15 mm
75
75
40
67
Ex. 2.5 Block Shear

• Agt gross tension area 50 x 9.5 475 mm2
• Ant net tension area 475 - 0.5 x (15 3.2) x
  9.5 388.5 mm2
• Agv gross shear area (75 75 40) x 9.5
  1805 mm2
• Anv net shear area 1805 - 2.5 x (15 3.2) x
  9.5 1372.8 mm2
• Ubs 1.0
• Calculate block shear strength
• ?t Rn 0.75 (0.6 Fu Anv UbsFu Ant)
• ?t Rn 0.75 (0.6 x 400 x 1372.8 1.0 x 400 x
  388.5)/1000 363.7 kN

68
Ex. 2.5 Block Shear

• Check upper limit
• ?t Rn ? (0.6 FyAgv UbsFu Ant)
• ?t Rn 0.75 (0.6 x 248 x 1805 1.0 x 400 x
  388.5)/1000
• ?t Rn 318 kN
• Block shear strength 318 kN

69
Ex. 2.6 Design Tensile Strength

• Determine the design tension strength for a
  single channel C15 x 50 connected to a 15 mm
  thick gusset plate as shown in Figure. Assume
  that the holes are for 20 mm diameter bolts.
  Also, assume structural steel with yield stress
  (Fy) equal to 344 MPa ultimate stress (Fu)
  equal to 448 MPa.

70
Ex. 2.6 Design Tensile Strength

• Limit state of yielding due to tension
• Limit state of fracture due to tension
• Check OK.
• Note The connection eccentricity, x, for a
  C15X50 can be found in section property tables.

71
Ex. 2.6 Design Tensile Strength

• Limit state of block shear rupture
• Agt gross tension area 225 x 18.2 4095 mm2
• Ant net tension area (225 - 3(23.2))18.2
  2828 mm2
• Agv gross shear area 2(19018.2) 6916 mm2
• Anv net shear area 2((190 - 2.523.2) 18.2)
  4805 mm2
• Ubs 1.0
• Calculate block shear strength
• ?t Rn 0.75 (0.6 Fu Anv UbsFu Ant)
• ?t Rn 0.75 (0.6 x 448 x 4805 1.0 x 448 x
  2828)/1000 1919 kN

72
Ex. 2.6 Design Tensile Strength

• Check upper limit
• ?t Rn ? (0.6 FyAgv UbsFu Ant)
• ?t Rn 0.75 (0.6 x 344 x 6916 1.0 x 448 x
  2828)
• ?t Rn 2021 kN
• Block shear strength 1919 kN
• Block shear rupture is the critical limit state
  the design tension strength is 1919 kN.

73
Staggered Bolts
74
Staggered Bolts

• For a bolted tension member, the connecting bolts
  can be staggered for several reasons
• To get more capacity by increasing the effective
  net area
• To achieve a smaller connection length
• To fit the geometry of the tension connection
  itself.
• For a tension member with staggered bolt holes
  (see example figure above), the relationship f
  P/A does not apply the stresses are a
  combination of tensile shearing stresses on the
  inclined portion b-c.
• Net section fracture can occur along any zig-zag
  or straight line. For ex., fracture can occur
  along the inclined path a-b-c-d in the figure
  above. However, all possibilities must be
  examined.

75
Staggered Bolts

• Empirical methods have been developed to
  calculate the net section fracture strength.
• net width gross width -
• d - the diameter of hole to be deducted (db 3.2
  mm)
• s2/4g - added for each gage space in the chain
  being considered
• s - the longitudinal spacing (pitch) of the bolt
  holes in the direction of loading
• g - the transverse spacing (gage) of the bolt
  holes perpendicular to loading direction.
• net area (An) net width x plate thickness
• effective net area (Ae) U An where
• net fracture design strength ?t Ae Fu (?t
  0.75)

76
Staggered Bolted Connections

• Stresses on inclined planes are a mix of tension
  and shear and thus a correction is needed.

• All possible failure paths passes shall be
  examined. The path that yields the smallest area
  governs.