I do not know the exact numbers for oleo stroke and so on, but this is how you would calculate this. A 747-400 weighs 400 tons on take-off and 296 tons on landing, maximum. See here for the source of those figures.
Next is the landing speed, this is approx. 160 knots = 82 m/s. Now let's assume the pilot has misjudged the height and does not flare, but slams the aircraft with the 3° approach angle into the runway. It is supposed to survive this, so let's just continue. This gives you a vertical speed of 4.3 m/s, and at 296 tons this is an energy of 2,750 kWs = 0.76 kWh, which needs to be dissipated by the landing gear. Now I make the assumption that the gear stroke is 0.5 m (those who know please put it in the comments, and I will correct the calculation). We have 0.5 m to decelerate a mass of 296 tons from 4.3 m/s to zero. If we assume a constant deceleration, the force will also be constant and the sink speed will decrease linearly.
The average sink speed during this process is 2.15 m/s, so it will take 0.23 s and an acceleration of 18.5 m/s$^2$. Force is mass times acceleration, so the force is 5,473 kN or 1.23 million pounds. This is just the inertial force to stop the descent. While taxiing to the take-off position, the aircraft will press with 400 tons = 878,000 lbs on the runway, since the wings do not yet produce any lift. This shows that even a hard landing does not stress the landing gear so much - after all, the acceleration is just shy of 2 g, acting on a much lighter plane.
In reality, the landing gear of a 747 is staggered, so the inner main gears will touch the ground first. Also, I expect that the force will not be constant along the whole gear stroke. This will change the details of this approximation, but the general magnitude should not be different.
This answer goes into more details on how to calculate the damage an airplane will do to a given runway or apron.
Now for the braking stresses. The landing field length of a 747-400 is 2175 m, and let's just assume that the pilot forgot to use thrust reversers, aerodynamic drag was switched off that day, and all the braking power had to be supplied by the 16 main wheels. Let's also assume that the pilot uses 1200 m of that field length for braking (I'm just making this up to get to an upper limit of what force will act on the tarmac). Now we need to decelerate from 82 m/s to zero within 1200 m. Linear deceleration means an average speed of 41 m/s, so the whole process takes 29.27 s. Dividing speed by time yields a deceleration of 2.8 m/s$^2$.
To bring the aircraft to a full stop, we have an energy of 995,152 kWs = 276.4 kWh to dissipate along a distance of 1200 m. Using again Newton's second law, we see that this requires a horizontal force of 829 kN = 186,322 lbs, which translates into 51,8 kN = 11,645 lbs per wheel. This is certainly more than what happens in reality, but to put it into proportion: The static load per main wheel at the maximum landing mass is 174 kN = 39,150 lbs (assuming that 4% of the mass is carried by the nose gear). This extreme (horizontal) braking force is still less than 30% of the (vertical) static load, which is well below of the maximum braking coefficient of an aircraft wheel on a dry runway.
EDIT:
CGCampell correctly remarked that emergency procedures at take-off weight will produce the biggest braking loads. Now I will calculate the highest possible braking loads, and for this I need this plot of a polynominal for the braking coefficient, which is the ratio between vertical and horizontal forces before the tire skids. I do not know the source; I collected it somewhere in the past and never found a reason to doubt its validity.
Shortly before the aircraft comes to a stop, the highest friction coefficient is reached, and then little lift is produced by the wings, so the vertical tire loads are those of the static case. At 96% of 400 tons acting on 16 wheels, this is 24 tons = 235,344 N = 52,907 lbs of downward force per wheel. Since the friction coefficient is 1 at low speed, the same load is transferred horizontally from each wheel to the ground, almost five times more than what I approximated for the landing above. Clearly, going to the limits produces much higher loads.
Formulas used in the calculation:
How to calculate the deceleration of a mass $m$ from a speed $v_0$ to $v_1$
Energy has no absolute value, we just add or subtract energy. The amount of energy change $∆E$ between the aircraft at a speed $v_0$ and at a speed $v_1$ is:
$$∆E = \frac{m}{2}\cdot (v_0^2 - v_1^2)$$
The unit of energy is the Joule, which equals the Watt-second or the Newton-meter. Thus 1 J = 1 Ws = 1 Nm = 1 $\frac{\text{kg}\cdot \text{m}^2}{\text{s}^2}$. Metric units are neat, right?
In physics, energy $E$ equals work $W$, and work is performed when a force $F$ travels along a distance $s$. As does the braking force, acting on the moving aircraft:
$$E = W = F_{brake}\cdot s$$
How do we get the rollout distance $s$?
We use the fact that the energy change due to deceleration equals the brake energy:
$$\frac{m}{2}\cdot (v_0^2 - v_1^2) = F_{brake}\cdot s \Leftrightarrow s = \frac{m}{2\cdot F_{brake}}\cdot (v_0^2 - v_1^2)$$
How can we find the time $t$ it takes to slow the aircraft down?
If the force is constant, the acceleration $a = \frac{m}{F}$ is constant, too. So the speed over time $v(t)$ is:
$$v(t) = v_0 + a\cdot t$$
We decelerate, so $a$ is a negative number. But the same works for a positive acceleration, too. To find the time $∆t = t_1 - t_0$ we make use of the fact that v varies linearly over time, such that the average speed is the arithmetic mean between $v_0$ and $v_1$. And time is just distance divided by speed:
$$∆t = \frac{2\cdot s}{v_0 + v_1} = \frac{m\cdot s}{F_{brake}\cdot(v_0 + v_1)}\cdot (v_0^2 - v_1^2)$$
With such formulas it is always a good idea to compare the units on both sides. Yup, seconds, so the result is really a time.
