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Chapter 5
Steel Shear and Torsion

Paul W. McMullin

5.1 Stability

5.2 Capacity

5.3 Demand vs Capacity

5.8 Where We Go from Here

Shear is fundamental to how bending members resist load. It holds the layers of a member together, causing them to act as one. This chapter will focus on how we size beams for shear force. We will look at the direct shear method and introduce tension field action.

Let’s take a moment and conceptually understand the fundamentals of shear behavior. You are likely familiar with the action scissors make when cutting paper or fabric. The blades are perpendicular to the material, going in opposite directions. This creates a tearing of the material like that shown in Figure 5.1. In beam shear, the action is similar, but the movement of material is parallel to the length of the beam. The top portion moves relative to the bottom, illustrated in Figure 5.2. Shear stress is zero at the top and bottom edge, and a maximum at the middle, shown in Figure 5.3.

Shear strength is fundamentally tied to bending strength and stiffness. If we take a stack of paper and lay it across two supports, it sags (Figure 5.4a), unable to carry even its own load. If we glue each strip of paper together, we get a beam with enough strength and stiffness to carry a reasonable load, as shown in Figure 5.4b.

5.1 Stability

Web buckling can occur when it becomes too slender, creating a reduction in strength. This is handled in two ways, direct method and tension field action. For direct shear, we multiply the shear strength by the web shear coefficient C_v, discussed in the next section. This keeps the stresses low enough to avoid buckling the web.

Figure 5.1 Shearing action similar to scissors

Figure 5.2 Shearing action from bending

Figure 5.3 Shear stress distribution in (a) rectangular cross-section and (b) wide flange

Figure 5.4 Paper beam with layers (a) unglued and (b) glued Lego™ figures courtesy Peter McMullin

In tension field action, the web buckles, but we add web stiffeners to prevent collapse. To understand this, we look at the shear force on a small portion of the web, illustrated in Figure 5.5a. The shear forces are parallel to the applied load, and balance each other. Turning the force block 45 degrees, the forces resolve into compression, causing the web to buckle, and tension, providing strength—opposite of what we want in concrete. Figure 5.6 shows this phenomenon in a test specimen.

Figure 5.5 (a) Shear force in a beam web and (b) equivalent tension and compression forces

Figure 5.6 Tension field action in test specimen

Reprinted with permission, Structure magazine, September 2014

Figure 5.7 Tension field pseudo truss

Tension field action creates a pseudo truss. The buckled web creates diagonal tension members, while the stiffeners provide vertical compression elements, illustrated in Figure 5.7. This approach is highly efficient and often used in deep highway girders, where the webs are slender. Sometimes the sun hits these girders in a way that you can see a slight ripple in the web.

5.2 Capacity

5.2.1 Direct Shear

In the direct shear method, we proportion the web to carry the forces without buckling. Nominal shear strength V_n is given by

V_{n} = 0.6 F_{y} A_{w} C_{v}

(5.1)

where

F_y = tension yield strength, lb/in² (MN/m²)
A_w = web area = dt_w
C_v = web shear coefficient.

The 0.6 accounts for the fact that shear yield occurs at approximately 60% of the tension yield strength. The web area is taken as the thickness of the web times the full member depth. The web shear coefficient depends on the slenderness, as follows:

For rolled I-shaped members:

\begin{matrix} \frac{h}{t_{w}} \leq 2.24 \sqrt{\frac{E}{F_{y}}} \\ C_{v} = 1.0 \end{matrix}

(5.2)

For all other shapes, except round HSS:

\begin{array}{l} When \frac{h}{t_{w}} \leq 1.10 \sqrt{\frac{k_{v} E}{F_{y}}} \\ C_{v} = 1.0 \end{array}

(5.3)

\begin{matrix} \frac{h}{t_{w}} \geq 1.10 \sqrt{k_{v} E / F_{y}} \\ C_{v} = \frac{1.10 \sqrt{k_{v} E / F_{y}}}{h / t_{w}} \end{matrix}

(5.4)

where

h = clear distance between flanges, in (mm), shown in Figure 5.8.

t_w = web thickness, in (mm)

E = modulus of elasticity, 29,000 k/in² (200 GN/m²)

F_y = tension yield stress, k/in² (MN/m²)

k_v = 5 for h/t_w <260

= 10 for web stiffeners spaced at h

= 25 for web stiffeners spaced at h/2.

The strength reduction factor for direct shear ϕ is 0.9, except for rolled I-shapes, when

\frac{h}{t_{w}} \leq 2.24 \sqrt{\frac{E}{F_{y}}}, it is 1.0.

Figure 5.8 Web height and thickness definitions

Interestingly, most rolled shapes in the Steel Manual have compact webs with C_v = 1.0.

5.2.2 Torsion

Torsional moments occur when loads are eccentric to a members shear center. For wide flange and hollow structural shapes, this is the centerline of the member, for channels it is away from the web, and for angles at the centerline of the leg parallel to the force, illustrated in Figure 5.9.

When the load is not applied to the shear center, the member will twist, like the channel show in Figure 5.10.

For open shapes like wide flanges and channels, it is best to avoid torsion, as they have low torsion strength. Closed shapes, like HSS, carry torsion efficiently. We will look at the strength equations for rectangular HSS. For open shapes, see AISC Design Guide 9.¹ The nominal torsion capacity T_n for HSS is given by Equation 5.5.

T_{n} = F_{c r} C

(5.5)

where

F_cr = critical torsion stress, lb/in² (MN/m²)

C = torsion parameter, in³ (mm³), see Table A1.4

The critical torsion stress is a function of wall slenderness (Equation 5.6).

\begin{array}{l} When \frac{h}{t} \leq 2.45 \sqrt{\frac{E}{F_{y}}} \\ F_{c r} = 0.6 F_{y} \end{array}

(5.6)

Figure 5.9 Shear center for various shapes, indicated by arrow

\begin{array}{l} When 2.45 \sqrt{E / F_{y}} < h / t \leq 3.07 \sqrt{E / F_{y}} \\ F_{c r} = \frac{0.06 F_{y} (2.45) \sqrt{E / F_{y}}}{h / t} \end{array}

Figure 5.10 Channel loaded off shear center, showing deflection

\begin{array}{l} When 3.07 \sqrt{E / F_{y}} < h / t < 260 \\ F_{c r} = \frac{0.458 π^{2} E}{{(h / t)}^{2}} \end{array}

The strength reduction factor for torsion ϕ is 0.9.

Initial Sizing

To help you select a beam large enough to carry a given shear load V_u, Table 5.1 provides the shear strength ϕV_n for various beam sizes and web areas.

Table 5.1 Shear strength for various beam sizes and web areas

5.3 Demand vs Capacity

5.3.1 Direct Shear

Once we have the shear capacity ϕV_n we compare it to the shear demand V_u. If ϕV_u ≥ V_u, the member has adequate shear capacity.

The shear force distribution in a beam depends on support conditions and loading. In a single-span, simply supported beam, with a uniform distributed load, the shear stress is zero at the middle and maximum at the ends. A cantilever is the opposite, with maximum force at the supported end. A multi-span beam has maximum shear at the supports. These are illustrated in Figure 5.11.

5.3.2 Torsion

Similar to direct shear, once we calculate the torsion capacity ϕ_TT_n, we compare it to the shear demand T_u. If ϕ_TT_n ≥ T_u, the member has adequate torsion capacity.

Figure 5.11 Shear force variation at points along (a) simply supported, (b) cantilever, and (c) multi-span beams

5.4 Deflection

Shear action contributes little to bending deflection—usually only 3–5%—and can be ignored. Refer to Chapter 4 for additional discussion on deflections calculations and acceptance criteria.

Torsion deflection, on the other hand, can have a substantial effect on deflections, particularly when other members cantilever from torsion members.

5.5 Detailing

Detailing considerations for shear are similar to those found in Chapter 8 for connections. Keep in mind shear force is a maximum near supports and at the mid-height of the cross-section. Any field modifications for piping or conduit in these areas will reduce shear strength and should be carefully evaluated by a professional engineer.

5.6 Design Steps

Step 1Draw the structural layout, include span dimensions and tributary width
Step 2Determine Loads
- – Unit Loads
- – Load Combinations yielding a line load
- – Member maximum shear demand
Step 3Identify Material Parameters
Step 4Estimate initial size or use size from bending and deflection check
Step 5Calculate capacity and compare to demand
Step 6Summarize the results

5.7 Design Example

5.7.1 Direct Shear Example

Building on the beam example in Chapter 4, we will now check the shear capacity of the wide flange beam where the span has been doubled, due to a column interference, as shown in Figure 5.12. This gives us a beam span L of 60 ft (18.3 m).

See the example in Chapter 4 for steps 1 to 2b. In this case, we don’t have the point loads on our beam.

Step 2c Determine Member Shear Demand

We are only concerned with the maximum shear, which occurs at the ends. Using the formulas in Appendix 2, and w_u = 2.38 k/ft (34.2 kN/m), we see half the total beam load is supported at each end.

\begin{matrix} V_{u} = W_{u} \frac{L}{2} \\ = 2.38 k / ft (\frac{60 ft}{2}) \\ = 71.4 k \end{matrix}

\begin{matrix} = 34.2 kN / m (\frac{18.3 m}{2}) \\ = 313 kN \end{matrix}

Step 3 Material Parameters

Because we are a using a wide flange beam, our yield stress is

F_{y} = 50 {k/in}^{2}

F_{y} = 345 {MN / m}^{2}

Figure 5.12 Shear example layout

Step 4 Initial Size

Choosing a W30 × 116 (W760 × 173), which is reasonable for bending, we know

\begin{array}{l} d = 30 in \\ t_{w} = 0.565 in \\ t_{f} = 0.85 in \end{array}

\begin{array}{l} d = 762 mm \\ t_{w} = 14.4 mm \\ t_{f} = 21.6 mm \end{array}

This yields a web area A_w of

\begin{matrix} A_{w} = d t_{w} \\ = 30 in (0.565 in) \\ = 16.95 {in}^{2} \end{matrix}

\begin{matrix} = 762 mm (14.4 mm) \\ = 10.973 {mm}^{2} \end{matrix}

Step 5 Capacity

We now find the nominal shear capacity V_n. First, though we need to know what C_v is. We begin by finding h/t_w. We can conservatively take h as

\begin{matrix} h = d - 4 t_{f} \\ = 30 in - 4 (0.85 in) \\ = 26.6 in \end{matrix}

\begin{array}{l} = 762 mm - 4 (21.6 mm) \\ = 676 mm \end{array}

\begin{matrix} \frac{h}{t_{w}} = \frac{26.6 in}{0 .565 in} \\ = 47.1 \end{matrix}

\begin{matrix} \frac{h}{t_{w}} = \frac{676 mm}{14 .4 mm} \\ = 46.9 \end{matrix}

Note the small difference is due to roundoff error in the conversions. We now compare this to

2.24 \sqrt{\frac{E}{F_{y}}}

\begin{matrix} 2.24 \sqrt{\frac{29, 000 k / {in}^{2}}{50 k / {in}^{2}}} \\ = 53.9 \end{matrix}

\begin{matrix} 2.24 \sqrt{\frac{200, 000 MN / m^{2}}{50 MN / m^{2}}} \\ = 53.9 \end{matrix}

Because

\frac{h}{t_{w}} \leq 2.24 \sqrt{\frac{E}{F_{y}}}, C_{v} = 1.0 and ϕ = 1.0.

We can now calculate nominal shear capacity.

\begin{matrix} V_{n} = 0.6 F_{y} A_{w} C_{v} \\ = 0.6 (50 k / {in}^{2}) 16.95 {in}^{2} (1.0) \\ = 508.5 k \end{matrix}

\begin{array}{l} = 0.6 (345, 000 kN / m^{2}) 0.010973 m^{2} ( \\ = 2, 271 kN \end{array}

\begin{array}{l} Because ϕ = 1.0, \\ ϕ V_{n} = 508.5 k \end{array}

ϕ V_{n} = 2.271 kN

Comparing this to the shear demand V_u, we see our beam is OK. Dividing the demand by capacity, we get a demand–capacity ratio of 0.14, meaning the web is only 14% utilized. Low DCRs for shear are common for wide flange bending members.

Step 6 Summary

Our W30 × 116 (W760 × 173) beam works.

5.7.2 Built-up Web Shear Example

Because our web works so easily, let’s now look at a built-up girder. Following the previous example, we jump to Step 4, and determine the member geometry.

Step 4 Initial Size

Keeping the depth the same as the rolled section in the previous example, we will use a flange plate thickness t_f of 1.0 in (25.4 mm). Based on this, we find h, illustrated in Figure 5.13.

\begin{array}{l} d = 30 in \\ h = 28 .0 in \end{array}

\begin{array}{l} d = 762 mm \\ h = 711 mm \end{array}

We will guess a web thickness t_w of 0.25 in (6.0 mm). Remember, plate thicknesses need to be standard sizes so they are readily available.

This yields a web area A_w of

\begin{matrix} A_{w} = h t_{w} \\ = 28 in (0.25 in) \\ = 7.0 {in}^{2} \end{matrix}

\begin{array}{l} = 711 mm (6.0 mm) \\ = 4, 266 {mm}^{2} \end{array}

Step 5 Capacity

Finding h/t_w

\begin{matrix} \frac{h}{t_{w}} = \frac{28 in}{0.25 in} \\ = 112 \end{matrix}

Figure 5.13 Built-up girder web shear example

\begin{matrix} \frac{h}{t_{w}} = \frac{711 mm}{6.0 mm} \\ = 119 \end{matrix}

Taking k_v = 5.0, we now compare this to

1.37 \sqrt{\frac{k_{v} E}{F_{y}}}

\begin{matrix} 1.37 \sqrt{\frac{(5) 29, 000 k / {in}^{2}}{50 k / {in}^{2}}} \\ = 73.8 \end{matrix}

\begin{matrix} 1.37 \sqrt{\frac{(5) 200, 000 MN / m^{2}}{345 MN / m^{2}}} \\ = 73.8 \end{matrix}

Because \frac{h}{t_{w}} > 1.37 \sqrt{\frac{k_{v} E}{F_{y}}},

\begin{matrix} C_{v} = \frac{1 .51 k_{v} E}{{(h / t_{w})}^{2} F_{y}} \\ = \frac{1.51 (5) 29, 000 k / {in}^{2}}{{(112)}^{2} 50 k / {in}^{2}} \\ = 0.35 \end{matrix}

\begin{matrix} = \frac{1.51 (5) 200, 000 MN / m^{2}}{{(119)}^{2} 345 MN / m^{2}} \\ = 0.31 \end{matrix}

Now to get the nominal shear capacity:

\begin{matrix} V_{n} = 0.6 F_{y} A_{w} C_{v} \\ = 0.6 (50 k / {in}^{2}) 7.0 k / {in}^{2} (0.35) \\ = 73.5 k \end{matrix}

\begin{matrix} = 0.6 (345, 000 kN / m^{2}) 0.004266 m^{2} (0.31) \\ = 273 kN \end{matrix}

Because this is a built up shape, ϕ 0.9,

\begin{array}{l} ϕ V_{n} = 0.9 (73.5 k) \\ = 66.2 k \end{array}

\begin{array}{l} ϕ V_{n} = 0.9 (273 kN) \\ = 246 kN \end{array}

Unfortunately, our capacity ϕV_n is below the shear demand V_u, and our beam does not work. We can jump to the next plate size 5/16 in (8 mm) and know our web will work just fine. Perhaps you can determine the DCR for this thicker plate.

Step 6 Summary

A 5/16 in (8 mm) web plate works without stiffeners.

5.7.3 Torsion Example

Let’s now look at a canopy frame that induces torsion and shear in its supporting member, shown in Figure 5.14.

Step 1 Draw Structural Layout

The canopy is supported by a HSS section on the exterior building grid and cantilevers out to cover the building entrance, shown in Figure 5.14. Key geometric values of the grid HSS are:

\begin{array}{l} L_{t} = 12 ft \\ L_{c a n t} = 6 ft \\ L = 28 ft \end{array}

\begin{array}{l} L_{t} = 3.66 m \\ L_{c a n t} = 1.83 m \\ L = 8.54 m \end{array}

Step 2 Determine Loads

Step 2a Unit Loads

The unit loads are as follows:

\begin{array}{l} q_{D} = 15 lb / {ft}^{2} \\ q_{S} = 60 lb / {ft}^{2} \end{array}

Figure 5.14 Torsion example canopy configuration

\begin{array}{l} q_{D} = 0.718 kN / m^{2} \\ q_{S} = 2.87 kN / m^{2} \end{array}

Step 2b Load Combination

Because this is located in a region where it snows a lot, the snow load dominant combination will control. Multiplying the factored unit load by the canopy width L_t we find the line load w_u,

\begin{matrix} W_{u} = [1.2 q_{D} + 1.6 q_{S}] L_{t} \\ = [1.2 (15 lb / {ft}^{2}) + 1.6 (60 lb / {ft}^{2})] 12 ft \\ = 1,368 lb / ft \end{matrix}

\begin{array}{l} = [1.2 (0.718 kN / m^{2}) + 1.6 (2.87 kN / m^{2})] 3.66 m \\ = 20 .0 kN/m \end{array}

Step 2c Determine Member Demands

We will look at the maximum shear and torsion, which occurs at the ends.

Shear demand V_u is

\begin{matrix} V_{u} = W_{u} \frac{L}{2} \\ = 1.37 k / ft (\frac{28 ft}{2}) \\ = 19.2 k \end{matrix}

\begin{array}{l} = 20 kN / m (\frac{8.54 m}{2}) \\ = 85.4 kN \end{array}

The torsion demand T_u is similar but we add the torque arm l_cant into the equation

\begin{matrix} T_{u} = W_{u} L_{cant} \frac{L}{2} \\ = 1.37 k / ft (6 ft) (\frac{28 ft}{2}) \\ = 115 k ft \end{matrix}

\begin{matrix} = 20 kN / m (1 .83 m) (\frac{8.54 m}{2}) \\ = 165 kN m \end{matrix}

Step 3 Material Parameters

Because we are a using a rectangular HSS member, our yield stress is

F_{y} = 46 k / {in}^{2}

F_{y} = 317 MN / m^{2}

Step 4 Initial Size

The following equation gives us a quick way to get the required torsion parameter C,

C_{E S T} = \frac{T_{u}}{ϕ 0.45 F_{y}}

The 0.45 is an estimate to account for the shear stress reduction and direct shear in the member. We’ll see how close it gets us:

\begin{matrix} C_{E S T} = \frac{115 k ft (12 in / ft)}{0.9 (0.45) 46 k / {in}^{2}} \\ = 74 {in}^{3} \end{matrix}

\begin{matrix} C_{E S T} = \frac{156 kN m}{0 .9 (0 .45) 317, 000 kN / m^{2}} {(\frac{1, 000 mm}{1 m})}^{3} \\ = 1.22 \times 10^{6} {mm}^{3} \end{matrix}

Looking at Table A1.4, we see a HSS of 12 × 12 × 3/8 (HSS 304.8 × 304.8 × 9.5) has a larger C value:

C = 94.6 {in}^{3}

C = 1.55 \times 10^{6} {mm}^{3}

Input geometric information is as follows:

\begin{array}{l} d = 12.0 in \\ t = 0 .375 in \end{array}

\begin{array}{l} d = 305 mm \\ t = 9 .5 mm \end{array}

\begin{matrix} h = d - 4 t \\ = 12 in - 4 (0.375 in) \\ = 10.5 in \end{matrix}

\begin{matrix} h = d - 4 t \\ = 305 mm - 4 (9.5 mm) \\ = 267 mm \end{matrix}

For direct shear, these yield a web area A_w of

\begin{matrix} A_{w} = 2 h t \\ = 2 (10.5 in) 0.375 in \\ = 7 .88 {in}^{2} \end{matrix}

\begin{array}{l} = 2 (267 mm) 9.5 mm \\ = 5,073 {mm}^{2} \end{array}

Step 5 Capacity

We now find the nominal shear and torsion capacities V_n and T_n. First, though we need to look at local web buckling. We begin by finding h/t:

\begin{matrix} \frac{h}{t} = \frac{10.5 in}{0 .375 in} \\ = 28 .0 \end{matrix}

\begin{matrix} \frac{h}{t} = \frac{267 mm}{9 .5 mm} \\ = 28 .1 \end{matrix}

For direct shear, we compare this to

1.10 \sqrt{\frac{k_{v} E}{F_{y}}}

1.10 \sqrt{\frac{(5) 29, 000 k / {in}^{2}}{46 k / {in}^{2}}} = 61.8

1.10 \sqrt{\frac{(5) 200, 000 MN / m^{2}}{317 MN / m^{2}}} = 61.8

Because \frac{h}{t_{w}} \leq 1.10 \sqrt{\frac{k_{v} E}{F_{y}}}, C_{v} = 1.0.

For torsion, we compare h/t to

2.45 \sqrt{\frac{E}{F_{y}}}

2.45 \sqrt{\frac{29, 000 k / {in}^{2}}{46 k / {in}^{2}}} = 61.5

2.45 \sqrt{\frac{200, 000 MN / m^{2}}{317 MN / m^{2}}} = 61.5

Essentially the same limit as direct shear. Which makes sense, as we are looking at local buckling and shear force in the same direction.

Almost there. We can now calculate nominal shear and torsion capacity. For shear we get,

V_{n} = 0.6 F_{y} A_{w} C_{v}

\begin{matrix} = 0.6 (46 k / {in}^{2}) 7.88 {in}^{2} (1.0) \\ = 217.5 k \end{matrix}

\begin{matrix} = 0.6 (317, 000 kN / m^{2}) 0.005073 m^{2} (1.0) \\ = 965 kN \end{matrix}

Applying the strength reduction factor of ϕ = 0.9,

\begin{matrix} ϕ V_{n} = 0.9 (217.5 k) \\ = 195.8 k \end{matrix}

\begin{matrix} ϕ V_{n} = 0.9 (965 kN) \\ = 868 kN \end{matrix}

Now a similar process for torsion capacity ϕ_TT_n. The shape is compact, so we use the following equation for critical torsion stress F_cr

\begin{matrix} F_{c r} = 0.6 F_{y} \\ = 0.6 (46 k / {in}^{2}) \\ = 27.6 k / {in}^{2} \end{matrix}

\begin{matrix} = 0.6 (317 {MN/m}^{2}) \\ = 190 {MN/m}^{2} \end{matrix}

T_{n} = F_{c r} C

\begin{matrix} = 27.6 k / {in}^{2} (94.6 {in}^{3}) (\frac{1 ft}{12 in}) \\ = 218 k ft \end{matrix}

\begin{matrix} = 190, 000 kN / m^{2} (1.55 \times 10^{6} {mm}^{3}) {(\frac{1 m}{1000 mm})}^{3} \\ = 295 kN m \end{matrix}

Applying the strength reduction factor of ϕ_T = 0.9,

\begin{matrix} ϕ_{T} T_{n} = 0.9 (218 k ft) \\ = 196 k ft \end{matrix}

\begin{matrix} ϕ_{T} T_{n} = 0.9 (295 kN m) \\ = 266 kN m \end{matrix}

Because we are checking two failure modes that act at the same place, we use the following interaction equation:

(\frac{V_{u}}{ϕ V_{n}} + \frac{T_{u}}{ϕ T_{n}}) \leq 1.0

Plugging in our numbers we get

{(\frac{19.2 k}{195 .8 k} + \frac{115 k ft}{196 k ft})}^{2} = 0.47

{(\frac{85.4 kN}{868 kN} + \frac{156 kN m}{196 kN m})}^{2} = 0.47

Because we are less than 1.0, our combined shear and torsion member is acceptable. Circling back around to our estimate of C, we see we may have been unnecessarily conservative. However, we have not checked deflection, and it is possible this will control the design, as any rotation will be greatly amplified at the end of the canopy. But for now, we know the tube is strong enough.

Step 6 Summary

The HSS12 × 12 × 3/8 (HSS304.8 × 304.8 × 9.5) works for strength.

5.8 Where We Go From Here

This chapter provides the basis of the direct shear method and torsion for closed shapes. From here we can expand our understanding of shear in slender web elements and tension field action. This provides reductions in web members of built-up girders. We can further understand torsion in open shapes by consulting texts on advanced mechanics of materials,² which use the soap film analogy to find torsional shear stress.

Notes

1. Seaburg, P.A., Carter, C.J., Torsional Analysis of Structural Steel Members, Steel Design Guide 9 (Chicago: American Institute of Steel Construction, 1997).

2. Cook, R.D., Young, W.C., Advanced Mechanics of Materials (Upper Saddle River: Prentice Hall, 1985).

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