10. Shear


Recall from ‘Mechanics of Elastic Materials’:

(1)   \begin{equation*} \sigma_{xp1xp1} = (\sigma_{x1x1})_{max} = \frac{\sigma_{xx} + \sigma_{yy}}{2} + \sqrt{\left(\frac{\sigma_{xx} - \sigma_{yy}}{2}\right)^2 + \tau_{xy}^2}\end{equation*}

Using our notation in the above diagrams, with all quantities taken as positive:

(2)   \begin{equation*} f_1 = \sqrt{\tau^2 + \left(\frac{f_{pc}}{2}\right)^2} - \frac{f_{pc}}{2}\end{equation*}

(3)   \begin{equation*} f_1 = f_{cr} \Rightarrow \tau = \tau_{cr} = f_{cr}\sqrt{1+\frac{f_{pc}}{f_{cr}}}\end{equation*}

As the axial compressive stress increases, the shear stress required to cause cracking will increase and the inclination of the cracks to the longitudinal axis will decrease.

Initial web-shear cracking for prestressed members:

V_{cw} = (3.5\sqrt{f'_c} + 0.3f_{pc})b_wd_p + V_p, \text{ psi      (1)}

where V_p is the vertical component of the PT

(f'_c \leq 10000 psi)

3.5\sqrt{f'_c} is essentially f_{cr},  and adding 0.3f_{pc} captures the idea that a greater PT force increases shear capacity.

Initial flexure-shear cracking for prestressed members:

V_{ci} = (0.6\sqrt{f'_c})b_wd_p + \frac{V}{M}M_{cr}, \text{ psi      (2)}

\frac{M_{cr}}{M} > 1, here, since we’re assuming uncracked behavior

Note:   M_{cr} is NOT constant along the length of the beam, except for beams with tendons of constant eccentricity.

d_p < .8h, else use .8h in the above equations

e.x. – estimating diagonal cracking shears – stirrups considered effectively absent


Web-shear cracking:

V_{cw} = (3.5\sqrt{f'_c} + 0.3f_{pc})b_wd_p + V_p

where f_{pc} = \frac{P}{A}=\frac{12*.153*152}{686} = 407 psi

d_p = y_t + e =8.72 + 15.57 = 24.29 > 0.8h = 28.8 \Rightarrow \text{ take } d_p = 28.8"


V_p = 12strands*.153\frac{in^2}{strand}*152ksi\frac{24.03-14.95}{30'*\frac{12 in}{ft}} = 6.96kips

V_{cw} = (3.5\sqrt{5000} + 0.3*407)*8*28.8 + 6960 = 92.1 kips

We now need to determine the uniform load w that causes a shear of  92.1 kips at a location of 28.25' from midspan.

Note:  The beam can be assumed to rest on a bearing pad of 8”.

50 strand diameters is .5*50 = 25''.  The distance of 50 strand diameters is typically taken from the far end of the beam, while the distance at which BMDs and SFDs typically start, assumes the origin to be at the center of the support i.e. the center of the bearing pad, rather than the far end of the beam.  Thus, 25 - \frac{8}{2} = 21'', or 1\frac{3}{4}, which is the distance from the center of the support, which is the distance of most interest to us.

This is how we obtained the value of 28.25' from midspan for the following calculations.

Now, imagine looking at a FBD of the middle (2*28.25)ft:

(4)   \begin{equation*} \frac{w(28.25*2)}{2} = 92.1 \Rightarrow w = \frac{92.1}{28.25} = 3.26 klf\end{equation*}

Flexure-shear cracking:

V_{ci} = 0.6\sqrt{f'_c}b_wd_p + \frac{V}{M}M_{cr}

V = \frac{w(60')}{2} - wx = 30w - xw

M = \frac{w(60')}{2}x - \frac{wx^2}{2} = 30w*x - \frac{wx^2}{2}

The moment-to-shear ratio at a distance of x = 1 \frac{3}{4}" from the center of the bearing pad:

(5)   \begin{equation*} \frac{M}{V} = \frac{30w*1.75 - w*\frac{1.72^2}{2}}{30w-1.74w} = \frac{30*1.75 - \frac{1.75^2}{2}}{30-1.75} = 1.80ft = 21.7in\end{equation*}

6\sqrt{f'_c} = 424 psi

-\frac{P}{A} - \frac{Pe}{S_b} + \frac{M_{cr}}{S_b} = .424 \Rightarrow M_{cr} = 6616in*kips

V_{ci} = 0.6\sqrt{5000}*8*28.8 + \frac{6616*1000}{21.7} = 315 kips

where 8 in = b_w, 28.8 in = 0.8hand 21.7 in = \frac{M}{V}

The load w, required to cause this shear is:

w = \frac{315}{28.25} = 11.2 klf

We can repeat the above calculations for a number of different locations along the length of the beam, and tabulate.

Typically, (1) controls near the support and (2) controls away from the support, though not necessarily at midspan, since M_{cr} increases towards midspan for draped or harped tendons.

The previous procedure for calculating V_c (V_{cw} and V_{ci}) must be used if the effective prestress force is less than 40% of the tensile strength of the tendons.  Typically, we are above 40%, in which case the following single formula for V_c can be used:

(6)   \begin{equation*} V_c = (0.6\sqrt{f'_c} + 700 \frac{V_ud_p}{M_u})b_wd \leq 5\sqrt{f'_c}b_wd\end{equation*}

The following procedure will assume that the effective prestress force is greater than 40% of the tensile strength of the tendons.

Typical Shear Design: Step-By-Step

Section dimensions assumed to have been calculated based on other criteria.  Loads known.

\rightarrow Choose stirrup size – if #3, then A_v = 2(.11) = .22in^2

Choose a distance x along the beam.  Calculate  V_u and M_u at our distance x.

We can start with the location of maximum shear force V_u .

Calculate V_c from (0.6\sqrt{f'_c} + 700\frac{V_ud_p}{M_u})b_wd \leq 5\sqrt{f'_c}b_wd

(3.5\sqrt{f'_c}b_wd was the upper limit for RC design)

2\sqrt{f'_c}b_wd, the ‘simplified’ formula for V_c in RC design, can be taken here as a lower limit.

(7)   \begin{equation*} \frac{V_ud_p}{M_u}\leq 1.0\end{equation*}

For pre-tensioned members,  V_c must not exceed V_{cw} within the transfer length.

We want to conservatively account for the lack of prestressing effectiveness (force f_{pc}, and M_{cr}, which depends on the prestressing magnitude and location) near the supports.  There is no f_{pc} or M_{cr}  term in the above equation for V_c, but those terms are present in V_{ci} and V_{cw}, hence the logic behind the aforementioned criteria involving V_{cw}.

If V_u > \phi V_c, then solve for s from \frac{V_u}{\phi = 0.75} = V_c + V_s = V_c + \frac{A_vf_yd}{s}

V_u < \frac{1}{2} \phi V_c \Rightarrow no stirrups needed

\phi V_c > V_u > \frac{1}{2}\phi V_c \Rightarrow solve for s from A_v=50\frac{bs}{f_y} or A_v = \frac{A_{ps}}{80}\frac{f_{pu}}{f_y}\frac{s}{d}\sqrt{\frac{d}{b_w}}, whichever is smaller

Calculate  4\sqrt{f'_c}bd and 8\sqrt{f'_c}bd.

If V_s < 4\sqrt{f'_c}bd, then check s \leq \frac{3h}{4} and s \leq 24".

If  4\sqrt{f'_c}bd < V_s < 8\sqrt{f'_c}bd, then check s \leq \frac{3h}{8} and s \leq 12".

If V_s > 8\sqrt{f'_c}bd, then we need to go back to the very beginning and choose a new section bd.

Check A_v > 50\frac{bs}{f_y}

Check A_v > \frac{A_{ps}}{80}\frac{f_{pu}}{f_y}\frac{s}{d}\sqrt{\frac{d}{b_w}}

If any are violated, decrease s ; take a new s and repeat, starting with a new V_s.  If more than one check is violated, use the smaller s value.

Repeat the procedure for various sections along the beam.  Sections with a different V_u and M_u.

Finding s from s_{max} criteria, then finding A_v, is an alternate approach;

The web steel volume is the same either way.

The former approached outline above in detail is generally used, since having many different stirrup sizes (different A_v), would be a hassle.  The exact arrangement is always up to the design engineer.  Safety and ease of construction should be taken into consideration.