Question Regarding Hip Joint Reaction Force on Standing Model

Hello,

I have run a very simple case of two leg stance using the standing model for validation of another model. When looking at the Hip Joint Reaction Force, it is saying that the value is 365 N. The model is set to 75kg, so shouldn’t the hip joint reaction force be 1/3 of the body’s weight, roughly 245 N?

Thanks,
Yogesh.

Hi Yogesh,

If you run the model with no muscles the reaction force you see in the joint will be 100% associated with the mass, but when having the muscles in the model any torque around the joint will lead to muscle contractions which will increase the reaction force, this is most likely what you see.

Best regards
Søren

Hi Søren,

Thanks for the response.

We have a single leg stance model that Moonki has given us. I have run a scenario that does 90 degrees of knee flexion.

We are getting 2167 N at 75kg body weight, which follows the theory that it should be around 3 times the body weight. Shouldn’t this value then be much higher due to having torque around the joint?

Thank you,
Yogesh.

Hi Yogesh,

The joint torques will lead to significantly higher loads in the joints due to the muscle contractions.

It also correct that for many human activities the loads may reach 3-4 bodyweight.

I have not fully understood the task your model is doing, but it sounds as it could be in the right ballpark.

Best regards
Søren

Hi Søren,

I am trying to understand your answer. So for two leg stance, do you think people simplify muscle forces when drawing free body diagrams and calculating joint reaction forces? According to the literature the JRF at the hip joint during two leg stance should be 1/3 the body weight. Does that mean that AnyBody has more accurate muscle forces compared to the simplified free body diagram leading to higher joint reaction forces that is greater than 1/3?

Please advise.

Thank you,
Yogesh.

Hi Yogesh,

In a stick figure model with no muscle actuators but only torque generators in the joints, you will see lower forces than in a model with muscle actuators.

If there is not a perfect balance in the model so non-zero toque in hips it will lead to higher forces than in a model with no muscles.

You can try to simulate both situations by disabling the muscles in your model then you will get a model with joint torque actuators and you can compare the results.

Obviously the way you drive the model is important so balancing the upper body above the hip joints will reduce the loads in the hips.

Best regards
Søren